ere See es Soe MATHEMATICS LIBRARY. ster" Return this book on or before the Latest Date stamped below. University of Illinois Library gon 13 196. Avg eee wav i 7 1978 lo 2 CT 29 nee OCT 29 199 L161—H41 ia a a : ay 4 t Ay. 7 ‘i hi id Ae atk ar a] EN) ee ay ; “tl uf 4) Ry a te Ms nS em Ait ‘ Digitized by the Internet Archive in 2022 with funding from - University of Illinois Urbana-Champaign https://archive.org/details/solutionoftrigon00gask Baa) a wy ah Os ‘Vie ie * SOLUTIONS. TRIGONOMETRICAL PROBLEMS PROPOSED AT ST JOHN’S COLLEGE, CAMBRIDGE, FROM 1829 TO 1846, 7 T 2, } > aa? i OP hes is ~ BY THOMAS GASKIN, M.A. LATE FELLOW AND TUTOR OF JESUS COLLEGE, CAMBRIDGE, CAMBRIDGE: PRINTED AT THE UNIVERSITY PRESS: FOR DEIGHTONS. SOLD BY SIMPKIN, MARSHALL & Co. AND GEORGE BELL, LONDON. M.DcCC. XLVI. ae FE OW Qe, Ye \ si le PREFACE. THE Problems which have been selected for solu- tion, were proposed at St John’s College to students at the end of their fourth term of residence, and are well calculated for the application and illustration of the use of Trigonometrical formule. The time allowed for the solution of each paper being limited to about three hours, the author has adopted the method which the nature of each example most naturally suggested, in preference to the employment of analytical artifices by which many of the results might have been obtained more concisely. The solutions of a series of Geome- trical Problems proposed at St John’s College to students of the same standing, consisting chiefly of examples in plane co-ordinate Geometry, have been prepared upon a similar plan, and the author hopes to be enabled to publish them about the commencement of the ensuing Michaelmas Term. In every case he has endeavoured to point out the form in which the student would be expected to pre- sent the solution to the Examiner. CAMBRIDGE, May 24, 1847. ERRATA. Page 15, line 4, in Denominator, for sin A sin B read cos A cos B. Page 26, line 1, for shadow read shadows. f A gf * TRIGONOMETRICAL PROBLEMS. ST JOHN’S COLLEGE. Dec. 1829. (No. L) 1. Derive the tangent of an angle, and express it in terms of its sine; also shew that if two opposite angles of a quadrilateral inscribed in a circle (one of its angles being 90°) be produced to meet, the theorem for the tangent of the sum of two angles may be briefly deduced from the figure. 2. Ina plane triangle, the length of the perpendicular let 6* sin C + c* sin B fall from the Zz A on the side (a) = a Cc 3. Shew that if sinw = sina.sin(w + y), then sin @ sin y UE a ene ie ee 1 — sina cos y Also if tana = cosa tan y, then a. tan” 5 ain 2y tan (y — v7) = — 1 + tan? 5 oe 2y 4, Eliminate 0 between the equations m =cosec 6 — sin@; n = sec 0 — cos 8. 5. Given the distances, from the angles, of the point at which the sides of a plane triangle subtend equal angles; find the sides and its area. 6. Assuming the truth of Demoivre’s formula for integral values of the exponents, shew that it is true for fractional values ; and explain the apparent absurdity of one side of the equation having more values than the other. iE 2 TRIGONOMETRICAL PROBLEMS. Fhe aU tan = = tant 2 and tan =2 tang; then @ is an arithmetic mean between a and [. sree 1 + sin 2a 8. Shew that sin (45 + wv) = Sy and thence that 2 cos (60 os = = We 20. We * Rg a ..eV 2 where the (2) occurs » times, and the upper or lower sign is used according as ” is even or odd. m — 1 9. If tan@= tant 2 and cos’ = Pe then 2 ™ = F eos 68 + (sin OE 10. The sides of a triangle are in arithmetic progression, and the distance of the centres of the inscribed and circum- scribed circles is a mean proportional between the greatest and least; shew that the sides are as the numbers Ne aig pn REG er ere ao \ a ie t tan a — ?) = tan’- , then the value of @ may be approximated to by the equation sin 3a — &c¢. sin 2a + ge CPS Live, Piao eee 12. If from one of the angles of a regular polygon of n sides, lines be drawn to all the angles, shew that their sum 0 . . = a cosec” we 2a being the length of a side. n 13. The bearings of two objects, B and C, (one of which is directly south of the other, and at a given distance from it) are observed from a station 4; the observer then moves to a second station D, and takes the angles subtended by AB, TRIGONOMETRICAL PROBLEMS. 3 AC; hence find the bearings of D from B and C, and its distance from A, in forms adapted to logarithmic compu- tation. 14. Apply the exponential expressions for the sine to shew that sin @. Be — sin2@. ue + sin 36. any — &e. = cot~’ (1 + cot 0 + cot’é). 15. Shew that cosnw = cos*a{1— —— Dette Be Cin Wa Uae he hae) tan‘ vw — &c.t. 1:2.3.4 Is this true when (m) is fractional? If not, investigate the true series in that case. SOLUTIONS TO (No. I.) 1. (a) See Hymers’ Trigonometry, Art. 11. sin A cet sin A ; cos A \/1 — sin? A (y) Let ABCD (fig. 1) be the quadrilateral figure in- scribed in a circle, having the angles B and D right angles; pro- duce BC, AD to meet in F’; and let 2 BAC=A, zDAC=B; then — = cosec AF'B = sec FAB = fA | AB’ (3) tan 4 = iG rae pC TNC RA - AD “DC Sai ished 1 Uae pe 4B OO DY AB’ DG (4B 4 FR se at _ wg aS FB. BC =i | AB’ Mey AB? Re FB BC Sal + Sea): TRIGONOMETRICAL PROBLEMS. and tan (4 + B) = . tan (d + B) - tan A = tan BS1 + tan d. tan(4 + BY}, tan 4 + tan B 1e— tania tan ps 2. The length of the perpendicular =bsinC=-csinBs b+e 6 (b sin C) +e (6 sin C) b+e b(bsinC) +e(csn B) Be sinC +c’ sinB b+c¢ 3. (a) Sinw=sina (sin & cosy + cos # siny); ., sin # (1 — sina cos y) = cos & (sina sin y); sin hence COS & v == fan 2 sina sin y 1 — sina cosy +, tan(y — v) = wha 2 sin® Z tan y \ tany — tana tan y (1 — cosa) l+tanytanew 14+ cosa tan’ y e 2 a . 2 sin 5 sin y cos y (ik pe ae) a eat ee 1 + (cos*s — sin’ 5 tan?y cos’y + (cos? ¢ — sin’ ;) sin’y a e 2 e 2 sin 3 sin 2y ree SiGe ie aa (cos? < +sin’ =] (1 + cos 2y) + (cos ¢ sin?) (1 —cos 2y) \ ands” Chit, aes sin 2Y ~ a obs coll 2 COs’ — + 2 sin” — cos 2y 9 9 5 ~ a. tan® 5 sin Qy a 1 + tan® — cos 2y 2 TRIGONOMETRICAL PROBLEMS, 5 u cos’@ sin’@ 4. m= ——-sind=———-; n=; ... mn =sin@ cos0; sin @ sin cos @ « 4 » coss@ sin30 1 1 aid nt nea sint'@ cost@ ~sin3@cos3@ = (mn)3’ . (mn)i (m3 + 3) = 1. 5. Let O be the point within the triangle ABC (fig. 2) such that ZAOB= 2 AOC= 2 BOC=120°", AO=8, BO=07, CO=0'; - BC =a=/BO' + 0C?—2B0.0C. cos z BOC = i/o? + 02450’ ( cos BOC = - 4). Similarly b= A ETE TE Fees S/S +524 d8: and AABC= AAOB + AAOC +A BOC = 400 sin 120 4,8) / ° 3 1 Wr +40" sin 120 + 46’0” sin 120 = sa (50 +00" + 00’). 6. See Hymers’ Triconometry, Arts. 130, 131. p a tan — + tan — 2 2g B a 1 — tan — tan — 2 2 Bo, ae B tan’ + tan? tan - +6 = SRO = NO ryeiriken et Ate (n7+) = ie 1 —tan‘ 3 1—tan’ = where 7 is 0 or any integer, positive or negative; and one value of p= 228. 6 TRIGONOMETRICAL PROBLEMS, f 1 — cos 20 8. (a) Sing = A/ = 20828, lett 0= 454 a; ; 1 — cos (90 + 2a 1 + sin 2a , sin (45 +a) = . ( ) = RNS. : 2 2 wv (3) Hence 2 sin (45 - =1/2-2sine. Put 45- 3 for wv; "28nd —$ { 45-=) = A/ 2—2sin 45-— =\/2-\/2-2sine; 45 aes or sin (45 - > + Ein Gites / \/2 —2 sina, where the vinculum is repeated twice; again, put 45 — 5 for x, 4b ‘ Aye nA and 2 sin (5 - += - 5) = 2— 2-2 sin (45 - 2) where the vinculum is repeated three times; and by continu- ing the operation, we have . AS 45 Ad 2s1n 445 —— + —... + (—1)""! 2 gn- wv = Ve —/2 —.../ 2-2 sina, the vinculum being repeated » times. Or, =. 1 n gn 2sin Fe 4 1+4 v aS ee ted = | goo eee z : (21)? pe a 2sin [39 > be urg]=V7 ON) o Oe eae or 2sin $30 — (- 4)’ (30 -a)t = V2 Via VY 2-2sin wv, and 2 cos }60 + (— 4)" (30 - ai = V9 VJ 2 —...A/2-2sin a. Ve 24/22 en If w= 0, 2cos }60 + (— 4)" 30} = the vinculum being repeated 7 times. TRIGONOMETRICAL PROBLEMS. 7 A 1—tan?@ cos?@ — sin3@ De COR Uy) tee a ; aes ? 1+tan?@ cos3@ + sint@’ dha ips he ‘wate tee @ — 2 sin? 0 cos3 9 + sin’ ,) Mit stl COS OP al a | p cos? @ + 2 sin? 9 cos? @ + sin3@ — — cos’ @ — sin’ @ cos? 9 + sin’ 6 cos’ § + sin’ 0 (cos? 9 + sin’ 9)? a (cos? 0 + sin? 6)3f ” (by multiplying the numerator and denominator by cos’ @ + sin? 0) 4 h Z el ence m = ——;——_- 5, - (cos? 9 + sin3 0)$ (cos? 8 + sin? 0)3 10. Let a, b, c be the three sides of the triangle; AR, r the radii of the circumscribed and inscribed circles respectively, © the distance between their centres; then 0? = R? -2Rr, We pas abe F Ti SiG SOO — ay VSS -aA(S-D(S-c) Rr a we, ne ace mic te aT ee Gere eRe but since the sides of the triangle are in arithmetical progres- sion, @ + c= 26; 3b ac . S=—, and 2Rr= —. 2 oO A. Again, o=ac; .. R?-2Rr=ac, or Rta; a’ b*c? Apres actrees TN EZ oy A, an 16.5 (S — a)(S — b)(S —c) 64a€ _ S(S — a)(S - b)(S — 0) = abbte’, 3b b (2a—b or 32 (= - 4) -3-( ) = abe, ys 2 Z TRIGONOMETRICAL PROBLEMS. and 4(3b —2a)(2a — 6) = a(2b— a), or 4(8ab — 4a” — 3b’) =2ab-a’; -, 15a? — 30ab + 126° = 0, or 5a°-— 10ab+ 46° = 0, 3 hence 5a? — 10ab + 5b? =b®, and \/5(a—b) =+b; ay ee Cc ich A/ on here 8 a co) x + pt a tan® — + tan-— tan — : 4 i an = Aira a Gk GRRL GLEE c 1 — tan’ — Tost a tan = 2tan (5 - @), and a, relWE a \ tan — sIn — cos (= Rae 2 2 g : ( ) au a an |{--— cos— sin {- — @) coszsin (5-9) Bea D) 3, and sin(a— o)=3sing; sin co) or e(@-P)V=I — em (4-9) N=1 3(e? V1 _ e~ PV-1), 3(e2PV-i _ iy. TRIGONOMETRICAL PROBLEMS, ay (3 ii e—4N=1) e2PV=1 — e@V=1 4 CF Oe SN KT or e2?V=1 = beat iat 1 p-anv-—1 1+ ae and 2p/-1 = log, Qa + 4 ¢*V=1) — log, ( ce 4d e-4V-1) Oe pe Let — d(etv=1 =e e-4N=1) 5 a ee See (e2¢ 4! OG Tah ena eN a) 1 ae 53 (e84V> _ e- 84NV=1) — &e., oo | = and dividing by 2\/—1, we have | eas 1 torte 1 tte g= -sina--.—sin2a+—>.— sin3a — &e. 3 Ss 12. Let ABCDEF (fig. 3) be a regular polygon of n sides, 7 the radius of the circumscribing circle; then the angles subtended at the centre by 4B, AC, AD, &c. drawn Qn 6 4A COO from A are —, —, —, &c. respectively ; n n n O77, 29 Th « ABz=2rsin-—-, AC=2rsin—, AD= 2rsin—, &e. n n and > the sum of all these lines {5 pe BPE peaked Fir ea = 2rsin — + sin — + Sin — +... + SsIn —————__?.. n n n n bar nee OT . (rn -1)7 Now let PR a ae aap Ae Tie)? ‘ n n Wes pean SoH . “ar _ nw .. 2cos — S' = sin — + sin — + sin — +......+ sIn — n n n n = ms i 1b 9 _ (n-1)7 = § —sin— + sin — + § — sin ———— n n 10 TRIGONGMETRICAL PROBLEMS. Bas Ot? | sin — Tv oor hence S$ (1 — cos z| =sin—, and §S'= ; n n ey 2 sin* — Qn 297 sin — AB 2a 90 ~D=2rs =- a te eee 2 Sea 4 n 2 sin* — 2 sin* — 2 sin* — 13 Let 2ABC (fis. 4) =a, _2ZACB = Bitthen if-( he due south of B, 7 —a and # will measure the bearings of B and C respectively from the south; also let 2 ADB =y, ZADC = 0; “BCe&a, 2DCB=0;... wACD = (6 - 9), and sin (9 + + — 0) a sina STIttGet ae Giant 3 G — tet a eS sin (ry — 0) sin (a + (3) sin (3 — 0 + é) sin 6 _ on {(0-d)+y} _asina sin {G-(0-9)} sin (yy — 0) ~ sin (a + B) — sin 6 ; sina sin (ry — 6) sind sin (a + ay sina sin (ry — sind sin (a + 3 DC =a and DC = AC. or sin {(@-0) + y} = .sin (8 + 6 — 8), and sin {(8+-+)-(8+6-6){ = DH, acetic 0), hence | eect ae sin (3 + 7) cot (B+ 8-0) ~ cos (B + 7) = SET sin a sin (ry — 0) a sind. sin (a + B)sin(B ++) _ GOED .. m may be determined by logarithmic computation; and sin (8 + ry) cot (8B + 6 — 0) = cos(B +) + sin (B 4+ ry) cot et ee PY SE I ia ox ave ce es sin @ ; eae ae nipuein (Batok Let TRIGONOMETRICAL PROBLEMS. 11 which determines (3 + 6 — 0 in a form adapted to logarithmic computation; hence @ is known, and 2 DBC = 7 — (0+ 4 — 0) is known. Also AC .sin ACD asina sin (3 — 0) . AD= = ‘ .sin ADC sin (a + (3) sin awivile ae 14. Let 2cos9 = te a»/—1sin@ = w= and if the sum of the series = ‘< we have nae 1 5 a I 2\/-1S8 =sin9 (2 | —4sin’é (a -—) + 4sin’@ [w-—) &c. # v \ & vu =sin@.a—- 4 sin’@ atl +. + sin’ @ . a &e. athe ee ace th Ae aw sin 1+asin0 = log, (1 + # sin 6) — log, (1-4 "") = log, wv sin 0 wv 2 ( mie 0 ; + {@+-—)] sin Wok 1+ wavsin@ Pees ctl a) _e PS a oe ee — : sin 0 ” e2V-18 4 iy. 1+ (a =| sin 0 we @ } cot S§ 2+ 2cos0sin@ 1 + cos 0 sin @ hence , and cot § = esi 7 2./—isind. sin 0 sin’ @ =1+cot?@+cot0; .«. S =cot~'(1 + cot 6 + cot’é). 15. (a) See Hymers’ Triconometry, Art. 136. (3) When x is a fraction whose denominator is m we have (cos @ +\/ — 1sin 9)" = cosn(2rm + 0)+ Vf — 1sinn(2rmr +) where 7 is any integer from 0 to m— 1; 12 TRIGONOMETRICAL PROBLEMS. and (cos @ + \/ — 1 sin 0)” = (cos 0)" (1 + \/ = 1 tan 6)" = (cos 0)" ((S + 8’\/ — 1), where nm.n—~i! m.(m—1).(m7—-2).(u-3 Ss a RAY prs ed cen ee Se) Pile yg 152 1.2.3.4 m.(m—-1).n-2 We tan 6 ee ae ee Le2.t . cosn(2ra + O)++/ —I1sinn (217+ 9) =(cos0)"(S+4/ — 1 iS): and equating the possible and impossible parts, cos n(2r7 + 0) = (cos 0)". S; sinn(2ra + 0) = (cos 0)". 8’, cos 20 = cosn(2r7 +0)cosn(2r7) + sinn (277 + O)sinn (277) (cos 0)" $cos (2rnum) S' + sin (2rna) S’t....(1) Sin 20 = sin 2 (2ra + 0) cos2rnm —cosn(2ra + 8) sin2rn7 (cos 8)” {cos (2rm7) S” — sin (2rn7) St....(2) Equations (1), (2) are the correct expressions for cos 20, sinn@ when 7 is fractional. ST JOHN’S COLLEGE. Dec. 1830. (No. II.) 1. Prove that in different circles, the angles at their centres are directly proportional to the arcs which subtend them, and inversely proportional to the radii; and explain the : arc ) meaning of the formula z = (=) rad, 2. Prove the following formule and adapt them to radius (7). sin (4 — B) sin (4 + B) 1 tan* A — tan’ B= , (1) cos’ 4 .cos’ B TRIGONOMETRICAL PROBLEMS. 13 1 sin’ A 1 ( Tee cos 4 er th (2) 2 1+ tan 5) : 2 (3) sin 34 + cos 3A 2sin2 A + ) : Hi. O88 TT —__—_——_ 1, and cosw=+(1—-—cosa) = +2 sin’> T a If a> =e 1-—cosa>1, and cos v= £(1+4cosa)= +2 cos*—. ro) ~ | . ° Tv The two latter values of w are impossible when a<—, 2 T e s ° and the two former when Or i » since in either case cos # would be greater than 1. 3. acos@+bcos(0+a) = (a+ bcosa) cos @ — bsinasin @ ’ a+bcosa : a +6cos = bsin a (Fo cos 6 - sin 8) ; let MS Behe easy ty bsina : bsina . acos 8 + bcos (0 + a) =b sina (cot B cos @ — sin 8) ae =a cos (0 + B) = b sina cosec B. cos (6 + B) = / a + 2ab cosa + b®. cos (0 + B). 28 TRIGONOMETKICAL PROBLEMS. 4. Let O (fig. 10) be the centre of the circle whose radius = r, ABC any triangle described about the circle, and - touching it in the points a, b,c; join OA, OB, OC, Oa, Ob, Oc, then A ABC = AOAB + AOAC + A BOC =1(AB.Oc + AC.Ob + BC . 0a) = = (AB +AC + BC); *. since r is constant, A ABC «(AB + AC + BC). b 5. Let A be the given angle, and — = 5 c B-C b-c A A then tan = cot— = 4cot—; b+¢ 2 2 B- A pe tan Cem ia loge consemces alae 2 = [ tan 57°. 54’ — 3 log? = 10.2025255 — .90309 = 9.2994355 = L tan (11°. 16’ +0). Now Ltan11°.17/ — Ltan11°. 16’ = 6588, L tan (11°. 16’ + 6) — L tan 11°. 16’ = 1139; Vom d 180 ; = 0 ee os 588 B— C ; , ra 11°.16’.10” and B—C = 22°. 32’. 20”. B+C=180-A= 115°, 48’; +, 2B = 138°. 20.20", and B = 69°. 10'.10”", C = 46°. 37’. 50”. 6. (a) See Hymers’ Triconometry, Appendix, Arts. 17, 18. 5e0 (8) Log .756 = log u 100 = log 75.6-2=—14.8785218=1.8785218. é; 75 oO ies log .0756 = log ee 2.8785218 $ TRIGONOMETRICAL PROBLEMS. 29 *. log (.0756)3 = 4(— 3 + 1.8785218) = — 1 + .6261739 = 1.6261739. 7. at = (eloeet) = e#l8e4 = e*, or w = log,a. x. 8. Let AB (fig. 11) be the arc; draw AM perpendicular to OB; then AC = AM; and sector AOC = me vile = rate ce at ae = A AOB. Ps 2 -, sector COB = sector AOB — sector AOC sector AOB — A AOB = segment ACB. 9. If h be the height of the hill; then A = ascent in 3 miles, and 1 foot = ascent in 5 feet; *, 1 foot : 5 feet :: Amiles : 2miles, or h = at e 3 mile. 3 Let / be the length of the road; then / = ascent in / miles, Greltoot, asl siteet. <= re: ol: - J=12h = 4miles. 10. a=csnB=bsinC, B=csin A, y=bsin A; a’ besinB.sinC 6b ec be ‘ By be.sind.sind aa a : Cc fe Oe, similarly ea = = 5 and a ae a a a” wie. aye be ac’ ab ay a(i meee’) Fi , - Bead SEE NES (« rs) 11. acos’?@ +bsin’@ = (etal) a (2D Ps 30 TRIGONOMETRICAL PROBLEMS. a+b a—b 1 where a’ + 3? = ao ET sores and w+ — = 2c0s20; 1 Tels Fee B_Va-Vb a=h(/at+vV/d), B=t(Va OL ioe cep hence log, (a cos’@ + bsin’@) = log, (a + 32) (« a =) wv v x simea so) -4( 8) (23) 248) led) = atogea +2{F cos20 - 3 (2 ) cos404+ 24 3 (2) cos 66~ &e.| a = 2log,a + log, (1 +P) + log, (1 je) = 2log,a + aff (2 cos? 9 — 1) — Ae, (2 cos? 20 — 1) +4 (F) (2 cont 8-1)— Bel = 2log.a- 2{2-3 (2) 4 (2) - &e.| 2 3 ie fF cos’ @ — 4 (2) cos? 20 + 4 (F) cos’ 36 — xe. a a a = 2 flog. — log, (1 +E) tea {(2) cos’*§ —4 (2B) costo6 Ke \a ooeRN / 2 = 2log, (. = B) + +4 (2) cos’ 9 —4 (2) cos’ 20 + ke = 4log, (rosy) + af (F) cos’*@ — 4 (2) cost 24 +4 (BY costs 0 - xe}, col TRIGONOMETRICAL PROBLEMS. 31 12. Let AD (fig. 12) be the direction of one of the walls, AB a vertical line from A along thé wall; C the shadow of the point B; then AC is the shadow of AB, and is therefore a meridian line: draw CD perpendicular to 4D; then CD is the breadth of the shadow = 6; also z BCA = sun’s altitude =a, and z24CAD =the inclination of the first wall to the meridian line 4C = 8; AB the height of the wall = a, “, AC =acota, and CD=b=acotasin#. Similarly, since the inclination of the second wall to AC is 90-8; 6, =a,cotacosp; et bi\* as a (cot aji: or cota = VJ (2) a (=)", 1 Ver Oy tie ab and by division A = a cot 3, or cot 3 = a ‘ ST JOHN’S COLLEGE. Jan. 1833. (No. IV.) 1. Prove the following formule : (1) tan A + cot 4 = 2cosec 24. (2) tan A + sec A ere or 2) tan 2 cot 4 + cosec A ( Lae * Q0 (3) 2cos A = eAN=1 4 e-AN-1; and explain how the last formula can have any meaning. 2. Find the general forms which express all the values of A in the following trigonometrical functions: (1) sin da (2) cosA=+1. (3) cosd+sind=1/2. 32 TRIGONOMETRICAL PROBLEMS. wv bu : 8. Vers~! —.— vers~! — = vers-'(1 — 6): required the a a wv value of —. a 4. Given A = sec 0 + cosec 0 (tan 0)* §(cosec 0) + 1}, B = tan 6 — (tan 0)° § (cosec 6)? + 1, prove by eliminating 0 that 43 — Bi = 23. 5. Prove that sin@ + sing =sin(@+ ) can only be a true equation when (6+ @) is either 360° or some multiple of 360°. 6. As the roots of a quadratic can always be found by the common algebraical method, state the use of the method which requires the trigonometrical tables. 7. If the cosine of an angle be capable of being ex- pressed by an algebraical series of powers of the angle, shew clearly ad priori, that no negative or odd powers of the angle can enter the series; and in the case of the sine, that no negative or even powers can enter the series. 8. 2 a“ Mm TT .\ ° Now if m be a multiple of 4. Sin (= + 0} = sin @; 0 r+0 m—1)7r+0 *, tan —.tan —— sits = 1 whatever be (6), m m m 7 T 5 7 Qa 4m — 3) put @=—, and tan —.tan —.tan — ......tan eres =} 4 4m 4m Am 4m when m is a multiple of 4. . (mar [iim =2n, sin CS + 6) = (— 1)"sin@, 7 igre m — 3) and tan ——..tan— ...... ems ) a=°( = ALY", 4m 4m 4m Tt If m=2n+1, and Ea > . (mr . 37° DTG Lo sn +6) = sin (ma + — } = cosna.sin— =(-1)’sin—; 2 A, 4 4. sin 6 Pepe le (EET): oy his sin saa + 0) 2 7 oy) 4m — 3 or tan —. ue see Biad lence tire JON Am 4m 4m 4m when m is of the form 2n +1. Hence if m be of the form 4” or 4m +1, P=1; if m be of the form 4” + 2 or 472+ 3, P=-—-1, which is the form in which the problem ought to have been proposed. 12. Let C (fig. 20) be the common centre of the two circles; 4 the given point in the outer circle; P,, P,, &e. the points of division of the inner circle ; 48 TRIGONOMETRICAL PROBLEMS. 2 1G - 2P,CP, = 2 P,CP,, &e.= at wig Let ACR Ss - LACP,=B +a, 2ACP; = B+ 2a, &e.; and AP, = R? + 7° —2Rr cos p, APZ=R?+r°-2Rrcos(B +a), &e.; or AP? + APS Ss... HAP Hn (Rr) —2Rr {cos 3 +c0s(B +a) +cos(B+2a)+...+¢08[B+(n—1)a]f. Let S' =cos B +.cos(3 +a) +cos(B+2a)+...+c0s {3 +(m—I1)a}; . 2S'cosa = cos (3 + a) + cos(B + 2a) +... + cos (B+ na) + cos(B — a) + cos 3 + cos (B +a) +... + cos{B+ (n—-2)a} = S' —cos 3 +cos (8 +na) +S + cos (B-a) —cos {8 +(n—1)a$ = 29, since ma=27; “.S=0; and AP) + AP,’ + APS + we. + AP,’ =n (RP? oR r)s ST JOHN’S COLLEGE. Dnc. 1834. (No. VIL) 1. Derinr the sine, cosine, and tangent of an angle, and from the definition prove directly that sin(4 +B) =sin 4 cosB + cos 4 sin B, A + B being greater than 90°, and cos 4 + sin A SE a) cos A — sin 4° 2. Given cos” A + cos (n — 2) A'= cos A, find A: and cos U — € given cosv = , shew that 1 —ecosu vd) 1+e U Si ee tan —. 1—e 2 TRIGONOMETRICAL PROBLEMS, ~~ ——i«OQ , 3. Prove that tan! hice = sin-'e; and if ey Sal ; eee ry =(m-—1) tan {sin een where » is a very small fraction, +n prove that r = (m — 1) tan x (1 — 7 sec’ z) nearly. 4. If sin(b+x)+sin a }(sin6)*—(sina)*} + sin {2a-(b-«)} = }(cos 6)* — (cos a)*} sin (4 + 2x), shew that sin (a + 6) cot (a — 3 | w= sec7} 5. A and B are two observed angles of a plane triangle which have small unknown errors; find what is the form of the triangle which will give sin (4+) with the least error. 6. Explain clearly the object of employing subsidiary angles in calculation; and apply the method in calculating the value of J a +6+ / a — 6, the values of a and b being known. 7. State how the approximate ratio of the circumference of a circle to its radius is best found: and assuming the value ‘of the ratio, find the number of degrees, minutes, and seconds subtended by an are which is equal to its radius. 8. By what means can the radius of a circle be deter- mined of which a proposed portion of a railroad is a segment, without observing any angles ? 9. A, B, and C are three stations, and from B a base BD is measured along BA; also the angles CBD, BDC, BCA are observed, as well as the depression of A and the elevation of B with respect to C; from these data find the heights of B and C above the horizontal plane passing through A, and the distances of A, B, and C, from each other. + 50 TRIGONOMETRICAL PROBLEMS. 10. Assuming that (cos 0 + f/ —1sin 0)” = cos m@ + \/—1 sin mé when m is an integer, prove that (cos 9+ 4/ —1 sin 0)” = cos—(2rm + 0) #\/ —1 sin os (2r7+0), n in which ¢ is 0, or any positive integer less than 7. 11. Prove fully that the magnitude of an angle can be most accurately determined by the tables from its sine or cosine, according as the cosine or the sine is the greater of the two. 12. If the perimeter of a triangle be four times the base (a), shew that the polar equation to the locus of the centre of the inscribed circle is p $1 + (sin p)’} =acos@; one extre- mity of the base being the pole, and @ an angle measured from the base. SOLUTIONS TO (No. VI.) 1. (a) See Hymers’ Triconomery, Art. 11. (8) Let the angles BAC, CAD (fig. 21) be denoted by A and B, where 2 BAD= A + B is greater than 90°; in AD take any point P; draw PN, PQ perpendicular to AB, AC; _ and QR, QM perpendicular to PN, AB respectively ; then 2 PQR = 90° — 2 AQR = 90 — 2 BAC = 90 — A, and sin (4 + B) =sin 2 PAB PN PR+iQM PR Bs. Qu AQ = SO Ts ——_ ~ AP AP PQ’ AP * AQ’ AP = sin (90 — 4) .sin B + sin d.cos B = cos A sin B + sin A cos B. TRIGONOMETRICAL PROBLEMS. 51 hy, tan 4 cos 4 cos 4 + sin A ie nite iain d= cos. sin Ae ; ~ eos A (y) tan(45 + 4)= 2. (a) CosdA=cosn 4+ cos(n—2) d=2cos(n—1) Acos 4; T “. cos4=0, or A= Alte Ce T and 2cos(m—-1)A=1;3 «. cos (nm — 1) d = 4 = cos; (6m+1)7 -1)A= fig = ——_____—__, (n ) 2m 1 2 or A 3(n~1) where m is any integer. cos %@ —e cos 8 = —————__;; (2) 1—ecosu 1-—cosv 1—ecosu—(cosu—e) (1+ ¢e)(1 —cosw) C—O = ‘"1+cosv 1—ecosu+cosu—e (1—e)(1+cosw)’ vo il+e u (3) ieee U and fae ie tan?’-, or tan—= a/ tan —. —eé 2 2 1-—e 2 e e 3. (a) Let 0 = tan~' ———; ... tan? = ———., /1-e /1-e hte 1-—eé 1 Bg Oe Weheen and cosec? @ =——— + 1 =; e e e 1] ° ° - coseecO@=+-, and sind=+e, or @=sin7' +e. Oe sin 3 (Pe abl Ld ly BEN ei erence tee iy a ee 24 A) af sit sin? s (1 +2) 1) sin % 4) sin &- ae =(m— I ———————— _- FSS == early / (1 +n) — sin?z /1+2n — sin?s 4—2 52 TRIGONOMETRICAL PROBLEMS. sin & tan z = (m -— 1) ———————- = _ (m - 11) ———_——_———— a/ cos? ¥ + 2n ( LW ao nen = (m — 1) tans (1 + 2m sec’ z)~3 = (m — 1) tan x (1 — 7 sec’ z) nearly. 4, Since cos’ b — cos’ a = sin® a — sin’b, by transposition, we have sin (> + v) + sin {(2a — 6) + w} = (sin? a — sin’ db) Ssin (a +22) + sina}, or 2 sin (a + x) cos (a — db) = sin (a + 5) sin (a — b) 2sin (a + x) cos a; * sin(a+a)=0, or a+v=mn and «=mr-a; : sl b or cot (a — 6) =sin(a+b)cosaw; .. secw - and # = sec™' pip pats®) : cot (a — b) 5. Ssin(d + B) = cos (4 + B)(64 + 5B) and will be least for particular values of oA and dB when cos (4 + B) is least, or when 4 + B is nearly 90°; - £C =180—(4 + B) = 90, and the error will be least when ZC is nearly = 90. 6. By means of a subsidiary angle, an expression may frequently be reduced to a form adapted to logarithmic com- putation. In that case the actual value of the result may be more readily obtained than by a direct arithmetical compu- tation. Thus, in the proposed example, it will be necessary by the common process to extract the square roots of a+ 6, b and a —6; but if we make cos@ = — (from which @ can be a readily determined), TRIGONOMETRICAL PROBLEMS. 53 sires camersny —_ 0 6 Va +b+ 0 —b =A/ 20. (cos = +sin =) _ 0 = 24\/a. cos (45 - =) ; OB log (\/a +b +/a-— b) = log2 + dlog a + log cos (49 -*), which will therefore be known by simple addition; hence a+ be v/ a—b will be the number corresponding to the logarithm, and may be found from the table of logarithms. 7. See Hymers’ Trigonometry, Arts, 147, and 9. 8. Let a wheel of given size be rolled along the railroad, and make 7 revolutions; then, if a be the circumference of the wheel, the length of the portion of railroad over which it has passed = ma; measure the length of the chain joining the two extremities, and let it =¢; this will be the chord of the arc whose length is na; and if 7 be the radius, and @ the angle subtended by the arc at the centre, <0 2sin- ‘= 2 c Ga at (5) E Cc na 1 By e od = = — a ‘enue’ oO petees = eS ~—-. as e Br 1 9 VG ko) 2 ar 2r 6\ar] ? nay? na)? ” a ae and PER aCe ny nearly. ~ 24(na—c) 2 * 6(na —c) F DB .sin BDC 9. In the triangle DBC (fig. 5), EC in(CRD LADO) 3 BC sin CBD BC sin BCA = — , and AB =——_-————_——. sin(CBD + BCA) sin (CBD + BCA) -. the distances of 4, B and C from each other are de- termined. 54 TRIGONOMETRICAL PROBLEMS. Again, let the depression of 4 below C =a, and the eleva- tion of B above C = B; .*. the altitude of C above 4 = AC sina, and the altitude of B above C= BC sin 8; hence the height of B above the horizontal plane passing through 4 = AC sina + BC sin B, which is therefore determined. 10. See Hymers’ Triconometry, Art. 131. 11. When a small change 64 is made in an 2A, the corresponding changes in the sine and cosine are respectively Osind=cosA.04, dOcosdA=-—sind.odd; _ d(sin A) | (d cos A) idm Ht AS mang Now since the tables are only carried to a certain number of decimals, the registered values of sin 4, cos 4, &c. differ from the correct values by small quantities dsin A, dcos A respectively ; .°. 6.4 will be less in the first equation or second, according as cos A is greater or less than sin 4; since if the tables be carried to » places of decimals, the extreme values 1 of Ssin A, dcos A, &c. are - 0” 12. Let BC (fig. 22) be the base, O the centre of the inscribed circle, 7 its radius; draw OM perpendicular to BC, B and join OB ; then BO =p and 24 OBC = pers also pcosp= BM =S—-b=2a—-6, and psing@ =r; but ac , ac , ; ace ee DH ens a 2ap sind = — sin2g, and ih 8a—b 0 = ae a Sea NET ye cosp = TPP cose; “. p(2—- cos’ pd) = acosd, or p(1 +sin’ Pd) = acos ©. TRIGONOMETRICAL PROBLEMS, 55 ST JOHN’S COLLEGE. Decemser, 1835. (No. VII.) 1. Frnp the values of x in the equation 21/ wv — a = sin A, 2. Prove the following formulse, cosec 4 = A + cot’ A; cos 2.4 + cos 2B = 2cos(A4 + B) cos (A — B), (= A J} + sin A ee ee 4 1 — sin A and shew that if 4+ B+ C=28S; cos 2,8 + cos 2(S' — A) + cos 2(S' — b) + cos 2(S8 — c) = 4cos 4 cos Bcos C. 3. Find the value of tan 36 to four places of decimals 4. Ifa, 6, c be the sides of a triangle of which the angle C = 90°; shew that ve —b fe ae = and cos (2A = B) = = (3c* ~ 4°). b+aV/-1 J Prove also that A =———— loo, ———-__—.. any eat Re pi Nay tan 5. The area of a regular pentagon : the area of the isosceles triangle used in its construction, (Eucuip, B. tv. Prop. xi.) as 4/5 : 1. 6. The sides a, 6, ¢ of a triangle are as the numbers 4, 5,6; find the angle B. : Given log,,2 = .3010299, Lcos 27°. 53 = 9.9464040, logy, 5 = .6989700, L cos 27°, 54’ = 9.9463371. 56 TRIGONOMETRICAL PROBLEMS. 7- Explain the use of subsidiary angles. Shew that the equation cos (a.+ #) = cosa sinw + sinb may be reduced to the two cos (b + v) = cos @ ; tan @ = sin b sin (45 + a) OS a cos 45. Cos @ ; C 8. If cos(a — #) =cosacosb, where # is very small; prove that | yD aE © = —2cotasin®—-([1 — cot’?a@ sin’ -| very nearly. 24 oi How is the number of seconds contained in w to be found from this formula ? 9. If in a regular polygon of m sides, straight lines be drawn from the extremities of a side (a) to those of any other side, so as to cross each other; shew that the locus of their : : : ; ; a Q9r intersection is a circle whose radius = F cosec —. n 10. A person wishing to determine the length of an in- accessible wall, places himself due South of one end, and then due West of the other, at such distances that the angles which the wall subtends at the two positions each = a°; prove that if (a) be the distance between the two stations, the length of the wall = atana. 11. If in the three edges which meet at one angle of a cube, three points A, B, C be taken at distances a, b, ¢ from the angle respectively ; the area of the triangle ABC formed by joining the three points with each other =4 Sab? + ace + Be. 12. If a and a’ be homologous sides of two similar tri- angles described, one about and the other in a circle, and Beads A, B, C be the angles of either, then = = 4sin = sin S sin Pi ’ a 13. If (0) be a small angle containing ”; prove the following formula and explain its use: log, 2 = Lsin 1, it a (10 — Lcos @) — Lsin 12 TRIGONOMETRICAL PROBLEMS. 57. SOLUTIONS TO (No. VII.) 1. 4a?—4a't= sin? A, .. 404 — 4a? 41 =1- sin? A, or , 1+ cos'4 A | 2u7—1= + cos 4, and | ies ware eee eae or sin? —; Pt es or Ae see 2. (a) See Hymers’ TRIGONOMETRY, Art. 18; and Prob. 2, Dec. 1831. (3) 2cos A cos B= cos(4 + B) + cos(4 — B); .. 4cos A cos Bcos C = 2 cos (4 + B) cos C+ 2 cos (A — B)cos C =cos (4+ B+C)+cos(4 + B-C)+cos(4 +C-— B) + cos (B+ C -— A) = cos 2,8’ + cos 28 — C) + cos 2(8' — B) + cos 28 — A). 5+ 1 4 3. Cos.s6 = V2 *, 7 UE rere ae aa fe: sec’ 36 = 6 — 24/5, and tan’ 36 = 5 — 24/5 = 5 — 1/20 =5 — 447213595 = .52786405 ; ca tat36 =. 7 2605 22. FAs 4. (a) In any triangle tan- A ae Panay G A-B a-b ‘- when C = 90, Ce 1, and tan (8) Cos(@d4 —B) = cos {2 4 ~ ie ~ 4)| = COS (s4-7) a Gok G = sin3.d = 3sin A — 4sin? d = 3- —-4— = — (3e" — 4a’). c AE 58 TRIGONOMETRICAL PROBLEMS. meatal SAN la 7 (y) \/—1.tan A =< ae Saag ANE ae 1 6+aV-1 *. 4 =——— log, ———_—_—.. 2./—1 hey al 5. Taking Euclid’s figure, if 7 be the radius of the circle, ‘since 2 CAD = 36°, and 2 ADC = 72°, AC =2rsinzZ ADC = 2r cos 18; then ACAD =A a sin Z CAD = 27°. cos?18. sin 36; fee ae 5Y and Pentagon = P=5. a sin 72 = re cos 18, since it may be divided into 5 equal triangles having their vertices in the centre, and vertical angles = 72°; 5” : 5 A: Ee = cos 18 : 27" cos*18. sin 36 :: 5 : 4cos18.sin 36 1 ~, :: 5: 2(sin 54 + sin 18) 512(Vetl, ven) iS Ey Oe oad Pais: RO ee F 6. In determining the angles we may suppose the sides =4, 5 and 6; 15 5 S.S—b 2°2 B a ps 1 iia ype and cos Oe Rs PT: 3? B 5 : 10 PCOS ape 10 + logy 5 - . logy, 2=10+log,,5 - a logy, 2 = 10°6989700 — °7525747 = 9'9463953, TRIGONOMETRICAL PROBLEMS, 59 / B L cos 27°.53 — L cos—— = 87, L cos 27°. 53’ — L cos 27°. 54 = 669, B 8 ° ” , ” , Ad 7 207. 58 4 =. 60" = 279. 53.8 , and B= 55°. 46’. 16”. 4 669 7. (a) See Prob. vi. Dec. 1834. (8) cosacosa# — (sina + cosa) sina = sinb; sin (45 + a) sin @ + cos a@ ————. = tan @ = : : sin 45 cos a let = cos @ wag ; a COs a ( ., = a(cos # —tan@ sinw) = cos (# sin cos a (cos ¢ ps ee + Pp), sin 6 cos p. “. cos(@ +) = aes 8. Cosacose# + sina sin vw = cosacosb; 9 4 .. COS @ (1 -=) +sina@.v=cosacosbh, cos a(1 — cos b cot a and w= -— cos irene) te x sin @ 2 aad i = — 2cota.sin’ = + 3 cota. a: Modi , — 2cota sin’ 5: put this value as a first approximation w= for w in the second member of the equation ; sold . then w# = — 2cota sin z co! AP ’ This gives the circular measure of #; and since the number of seconds in an unit of circular measure = 206265 ; 60 TRIGONOMETRICAL PROBLEMS, the number of seconds in # _ 6b Sesh f) — 2(206265) cot a sin’ E (1 — cot’ a sin® =| ll ees eu — 412530. cot a sin’ 5 (: — cot’ a sin’ >) , 9. Let AB (fig. 3) be a side of the polygon ABCDEF ; CD any other side; draw AC, BD crossing each other in O; then 2AOB = 4 ADB + 2 DAC =22 ADB (since the angles in the segments upon the equal bases 4B, CD are equal) 29 ¥ = — and is constant. n Similarly, if the points A, B be joined with the extremities of any other side so as to cross each other, the angle included Pore 2 : ; 2% between the joining lines will = —, n Hence O always lies in a segment of a circle upon the base ae 2% ! AB, and containing an angle = —, and the radius of the n circle 10. Let AB (fig. 23) be the wall, C, D the stations such that Z4ACB = 4 ADB=a; produce DA, CB to meet in £, then 2 CED = 90° (:.. D is south of A, and C west of B) ; and since 4 BDA = 2 BCA, a circle may be described about ABCD, AB ' sin ACB = diameter of the circle CD CD CD ee ee eS) ee SS eS or AB=CD.tan ACB =a tana. TRIGONOMETRICAL PROBLEMS. 61 ll. Let AB=a, AC=6, BC =c, then Ae / a + b, heen 3 +e, a \/ BP +c, 47y/ and the area of AABC =“ sin Z ACB p’ pie OE ek Sea ee eats We Coes — a G ft) = 1 n/ 4026? — (a” i? = cy 20h 1, (4 (a? + B) (a +) - QePa=LV/eriee + Be. 12. Let rv, R be the radii of the inscribed in and circumscribed about the same triangle, then bee ae r= 48 sin Py sin ‘3 sin C (See Dec. 1830, Prob. 111), a 2a A B & a Pshel me ee ee sin > sin — sin—. 2sin A’ sin A 2 2 and R= Hence if a be the side of a triangle, and A, B, C its angles, the radius of the inscribed circle (1) 2a A B C = .sin — sin — sin—. oy 2 9 sin A Now, if a’ be a side, and A, B, C the angles of a triangle ‘ / . ° e e a inscribed in a circle whose radius is r; then 7 = —_——, 2sin A a Ro ORY Ages ‘ ’ = — . sin — sin— sin= 2sn4 snA 2 Q Q° a’ Fi snot Sie al OS or — = 4sIn— sin— Ssin—; a ye Q 2 where a’ and a are homologous sides. 62 TRIGONOMETRICAL PROBLEMS. ST JOHN’S COLLEGE. Dec. 1836. (No. VIII.) 1. Resotve sin @ into its factors; and explain its use in the calculation of the logarithms of the sines of arcs. 2. Prove that cosnB — cos(n +2) B=2sin(n + 1) Bsin B. -B a-b C Also 1 tri le t “ t so In an Ylangie tan = COs Ts y 8 2 ath 2 3. If in any triangle ABC, a straight line be drawn from one angle C to the point D where the inscribed circle touches the opposite side, and if in the two triangles so formed circles be inscribed, shew that they touch CD in the same point. ; 4. If on the three sides of the triangle 4BC, triangles similar and equal to the original triangle be drawn in such a manner that the angles which centre at A shall be all equal, and the same at B and C; and if in the three triangles so formed, circles be inscribed, and their centres joined so as to form a triangle 4’B’C’. Prove that if R, R’ be the radii of the circles circumscribed about ABC and A’B'C, area of ABC we R area of ABC RR 5. Having given one side, and the opposite angle = 120°, and the line joining the given angle with the point of bisec- tion of the opposite side, solve the triangle, and find its area. 6. In any triangle, having given the angles A and C, and the side c, and A and B very nearly equal, solve the triangle by approximation. 7. Two opposite sides of a quadrilateral figure are parallel to each other. Having given the area, the diagonals, and the difference of the squares of the opposite sides, find the sides and angles of the figure. TRIGONOMETRICAL PROBLEMS. 63 8. Suppose a semicircle described on each of the three sides of a triangle, and the points of bisection of the semi- circular arcs be joined, find the sides and area of the triangle so formed. 9. Find sin 18°, and calculate its value to 5 places of decimals. Solve the equations : cosec? ne Med 2 eis cosec”@ : 2 2 15 (1 — cos20 cos2) = 17 sin 20 sin 29; sin (0 - p) + 4/ sin (0 +) sin (6 - 9) = (an o-a\con a find tan@ and tan ©. 10. ] ~ . A+ sin 2 cos 2 or — A C AMOS cos — cos — sin o, . At ( A A — =) AC sIn” = |co Os cos ; 2 2 pmeAr-e GC and cos” asin? = = —4 (cos A + cos C) ; . cos(4 + C) = — d(cos 4 + cos C) ; hence cos 4 + cosC = — 2cos(d4 + C) = 2 cos B. : re snB m 5. Since Z A is bisected, 6: ¢:: m:n, or — = — sin. Cy 7 and msin B=msin C = m sin (A + B), - ntanB=msnA+mcos A tanB; m sin A hence tan B = m—mcos A If m>n, let n=mcos@, sin 4 sin A Bee cil cotiecontun oh Ah ti SO Cares 2 sin ? sin ? a z If m cos 2 2 ae : i a Sa Similarly, sin — cos— = —__~-____(cos — cos § + sin - cosy); ; 2 2 2cos 3 cos r+ 2 Q for equation (1) is satisfied by substituting — a, and — + for a and vy; and by the same substitution the expressions for , sin 5 and cos Fe in equations (2) are reduced to the expressions 0 0 for COs 5 and sin = in equations (3); and the result for mea 6 a Q sin = cos © may be derived from the result for sin 3 cos =, by ~ ~ : a putting — > for ~ol1s , and —+v for y; 88 TRIGONOMETRICAL PROBLEMS. ee Otrane 6’ ", sin 2 = sin — cos— — cos — sin — 2 73 eo yale {ice = ——~_—_ (sin — cos yy) = sin — sec P. cos [3 cos y ( B ) 2 p 9, Let S=1+2cos’@ + 3 cost 6+ &c., . & cos’ 8 = cos’ 9 + 2 cos! @ + &c. and S'(1 — cos’ 0) = 1 + cos’ @ + cos*@ + &c. 1 I i ~ 1= ‘cos’ 8! Ory. Bierce 0): = vat Again, let S°= 1-2 tan’6@ + 3 tan*@ — &c. “, Stan? @ = tan? @ — 2 tan*@ + &c. and S$’ (1 + tan? @) = 1 — tan? @ + tan‘ @ — &c. 1 1 foe ee 8S oe cos 1 + tan’ 0 “(1 + tan® 6)? : peCOs.0 : and S'S" = a cot’ é. 10. Let DE (fig. 33)=0, 4 BDE=a, 4 CDE=8, LDEB=iry, £:DECm 0: re sino ~ sin(B +0)’ and AC x DC. sIn ADC be DC. sin ADC i Ssin DAC™ Canin (PDE GEDEO) r. a sin} sin (a — (3) | ~ sin(B + 0) sin(a + 0) ’ also RE asina sin(a +)” TRIGONOMETRICAL PROBLEMS. BE. sin AEB BE.sin AEB eee ne = CARD ADE) _ asina sin(o — y)_ ~ sin(a + y) sin(6 + a)’ and in A ABC, AB, AC and 2 BAC=7 —-(a+0) are known ; .. BC may be determined. ire: (a) Sine =— Rc acre cece aster ee ()e also sing =sin(Q2mr+a), or }(2m+4+1)r— 2}, g2mmt+x)V—1 _ p-@mrt+2) VA) "6 sin @ = — Bate } ‘ el2@m+l)r-ayV=l1 _ g-{@m+i)r—ayvri Or sin v7”? = —— On one 1 Since cos 0441/1 sin@=e9V-1; ezmm V1 _. ie e@mtlrvri ie . (y). and by substituting these expressions in equations (3) and (+), they are severally reduced to equation (a) which is there- fore an equally general form with ((3) and (y). (8) Let sinw-Jsin2e+tsin3e —- &c.=S, aS\/—1 on (er V=1 Me vest i 4 (e22Va1 x e~20N=1) + &e. = log, (1 + e*¥=1) — log, (1 + e7*V=!) .......0. (1) 1 + erv-1 fede x CN log, neta log, et =} (since log, 1 = + 2ma\/—1 where m is 0 or any integer) ; 90 TRIGONOMETRICAL PROBLEMS, Again, let §"= sina + tsin2ga + 4sin 3a + &e. °e DS / 1 2 (eV) ~S tas ef 4 (e2*v=1 a e72u a1) + &c. = log,(1 — e~*¥"1) — log,(1 — e*V")....:.... (8). l= e7~tNv-1 = log, eee = log, (- e7 tV-1) = (2m +1)9\/ -1-aV/ -1, since log,—-1=(2m+ lar -1; TT SS” = mmr + By giving the proper value to m, the true values of S' and SS” will be found for any particular value of «. . ‘ a When ~# increases from 0 to aw, S increases from 0 to 3” la ® ° s EE and S' diminishes from — to 0. 4 / Let w=7+27;3 ig . . , Ta. then S' = — (sina’+4sin2v + &e.) = - ( ; 2 *, as # increases from 7 to 27, # increases from 0 to 7, and C5 ° . ° ° S' changes from — i to 0 and is always negative: in this case v xv DY Lee? ae oe emus where m= —1. When w>2a, lett v=2r+2, . S= sing — 4 sin 2a' + 4 sin 3a’ — &c. ah ah . e 7 — =_—-— 7, or mis still = -1; Q o Similarly, whi When «wv > 37x, Ries — 27. soy aegee te w Hence the greatest value of Sis + So One ee TRIGONOMETRICAL PROBLEMS. 91 When « = 7, there is an apparent absurdity in the result, VIZ. ‘ . Soe 7 sina —dsin2ar+4sin3a — &e. = = 3 but when # approaches to a and is less than w, the value of 7 S' increases and approaches to re When «# is greater than zr, S' is negative, and the nearer it approaches to 7, the nearer is 7 1 S to — a or there is a sudden change in the value of S : T qT . e . from + 3 to — 7 , and the true value of JS is discontinuous. When @ is an odd multiple of a, expression (1) becomes log,(1 — 1) — log,(1 - 1) = - © + &, and therefore fails; the same is true of expression (3) when # is an even multiple of a. In all other cases S may be correctly obtained from equa- tion (2). A general expression for S may be obtained in the follow- ing manner. Let S=asine — La sin 2a + 4 a® sin 3a — &e. - ONa/ ol ub. a(erv=1 ay e-tN=1) = La (e22 Va he, e~2tN=1) + &e. 1 + ae¥ He ee = log Se ee —1 = —_______; Mein 1+ ae" 4 e2SV=1 4 1 2+ 2acosx an Say ee ee Sea ePNV— 1 oa./=1 sine 1 + a@ COS YW or cot 8S =2—— a@ sin & hence a(sina cot S —cosxw)=1, or asin(w—S)=sin§; *, when @=1, sin(a—S)=sinS. ......... (4). The last equation is that from which the true value of S may be determined, and is different from the equation v v- Sa S, or Sieg: 92 TRIGONOMETRICAL PROBLEMS. Since sin S = sin(2@m7 +48), @ v- S=2mr+S, .. Sih ana as before, where m is an integer to be determined from the particular value of a. When @ is an odd multiple of a, the value of S§ derived from equation (4) is indeterminate. For a further discussion of the value of S, the reader is referred to a note at the end of this book. 12. (a) Let 7 be the radius of the given circle, ” the number of sides of the polygon; then the area of the inscribed polygon and the area of the circumscribed polygon = =nrtan—-=8; n ak 8 Pee SID — sin”? = V7 Pe at. is vr Be A Sh SID COS Se : T n n T cos — cos — n n sin” — Similarly, 6 — a, = 2nr* ——— 5 TC. cos — Qn sin? — cos. —— cost — BrwA L 2 b-a 2 sin’? — cos — cos — n T i ae >, vie , A T cost —— — sint —— + sint — sin* — Qn Qn Qn 2n = 4 _§ $$ J = 4] 1 + ———_ ]f,, Eh aT, be ste at 7 cos’ — sin? —— cos — | Qn Qn nN ‘ TRIGONOMETRICAL PROBLEMS. 93 : : . : Tv T Coit which is >4 since sin — and cos — are both positive; and 2n n ‘ et : 7 as ” increases, sin zy becomes indefinitely small, and cos — n n B-A ae ae approaches to 1; .°. yr approaches to 4 as its limit, —a (8) If A, B, a, b represent the perimeters, we have An rue T ; T T A=2nrsin—, B=2nrtan—, a=4nr sin—, b=4nrtan —; n n Qn 2n a tT e 2 T sin —.sin pe Tv T n n o B=-A=2nr.tan — (1 — cos S st Ae YG a n n T cos — ” sin —. sin? ca —~/e . Vv nv Similarly, b —a@ = 8nr. ‘ a cost =— Qn T 9 9 Ww cos — cos” — . cos’ —— — A w ; n 4n and = 2cos— }4h4.cos?*—— |. = b—a oT Qn n 2 cos — cos — n n Ths 1 + cos — 1+ cos — “L) 2 Qn = 3 cos — which is greater than 4, since T T Tv WT T 1+cos— >2cos—_, and 1 +cos — >1 + cos —>2cos—-; n n 2n n n 94 TRIGONOMETRICAL PROBLEMS. es x : Tv 2% and when ym is increased indefinitely cos — and cos — ap- n n Jip 5 approaches to 4 as its limit. — a proach to unity; .-. 13. (a) See Hymers’ Tueory or Eauartions, Art. 15. : ° -- . 2art+ 4 il a (GB) sin b =2"~"siIn Pe sin a P. sin ALN ait sin Dare : m m m m put “, + for ~; see? meee _ (m *, sin ee + p) = 2"-leos wy COs me co 2 m m m sa (Ort eo: and by division Ga lL) ort sin ® a = _ (ma ; sind tee (+9) P=tan Hi tan KON .-. tan m m Now if m be of the form 42 +1, P= LET = + tang; + cos @ if m be of the form 42 +2, P= ue =-1=- (tan); — sin @ if m be of the form 4”, P ee = 1 = (tan); ~ sing 1 + (-— Wd ae hence in every case P= + (tan p) ? sin s—— + —+-. i: Big Teh Say Rey Ties 6 11. The lines which join the centres of four squares described on the sides of any parallelogram form a square, of which the area exceeds the area of the parallelogram by 4th the sum of the squares described on its sides. 12. A circle revolves in its plane about a point O, and from a point P in that plane the greatest and least angles 7 98 TRIGONOMETRICAL PROBLEMS under which the circle is seen are 2a and 2/3, shew that the distance of O from the centre = PO sin @, where sin tan’ & ai £) = — 3 , 4 2 sin 3 Investigate a series for tan v in terms of a, and shew how it may be employed to compute the tangent of an angle containing a given number of seconds 135. If tanw =a, shew that approximately #=4°4934118 14. Prove that the complete solution of #°"—2cos6 #"4+1= : 2\74+6 comer et Wai: ee Cos 4 / — I Bit ; resolve the first mem- n ber into its quadratic factors, and point out the geometrical signification of the result SOLUTIONS TO (No. XI.) 1. Let AB (fig. 34) be the tower whose height = 200 feet, BC the staff = 1 foot, D the point in the horizontal plane, 2BDA=0, £CDA=¢; tl tan @ = ne fe Oy oul, OF deed 51 OR ihe ae AD on ‘De 100 2°01 —2 ‘Ol ee té — = t ? Se ————- ?——_-_ = e an (@ — 0) = tan BDC Ric He BDC nearly ; *, BDC when reduced to seconds 2062°65 of , ” = = 411 nearly =6.51. 502 2, (a) Sin(m +1) 4+sin(n-—1)4 = sin(zd + A) +sin(nA — A) = 2sinn A cos A, * sin(z +1) 4 = 2 sin 2A cos A — sin (7 — 1) A. parce Fe TRIGONOMETRICAL PROBLEMS. 99 (3) See Hymers’ Trigonometry, Art. 63. 2 sin? 27 1—cos54 1—sin 36 tan 27° = ——__—._ = —__ = ——_ (y) ( 2 sin 27 cos 27 sin 54 cos 36 Eee CACTI ee = 4/5 —~-1-49)/(10 - 24/35) 5-1)’ » Sy ey OU Es cos + in— 9. -5 sin — 3 a\* a . et) (1 + tan $ = a COS — 2 1+sina 2(1+sina) a in 1+ cosa cos* — 2 ‘vn sin (a uid sin (a + B) _ sin (a + (3) sin (a mr (>) (2) tan(a+) = cos (a + cos (a + 3) cos (a + (3) sin (a — B) 4 (sin 2a — sin 23) ze sina cos a — sin 3 cos 3 (3) Cosa+siny — sin =sin(B+ +) + siny — sin = 2 COS si + — sn oe ae B+ inP=Y) ote Reltss: B+ in eo 4a, g sets F : ut ean! aly a Similarly, cos 3 + sinsy — sina=4 Cos sin — cos — y> B # 2 Pn, by interchanging a and #; Pies 100 TRIGONOMETRICAL PROBLEMS, \ : : cos cos cosa + sinry — sino Q " cos 34+ siny — sina a+ B Y cos Y cos 4. (a) Ifa be the least value of sin~'a, the values of 0 are 2m7 +a, and (2m+1)a—a, where m is any integer ; eu : a ! T—-a . the values of sin 5 are sin |ma+—), and sin ( m+ ‘ where m is any integer ; ee te aes a or sin— = + sin—, + cos-~, 2 2 2 : ed : . O which will include all the possible values of Dee and are four in number. 0 Also the values of tan | are tan (ma + “) : — TiO a a and tan (mn + ==") SOL otal and cot —. 0 ; Hence us has only two different values. _ lp rae FE nS eet ES (8) PN aR ATS AAD! ee ISS yD > TT hence RUG AEET Tata or Qn+i)r-=, where ” is any positive or negative integer ; T 5% . O@=nrt+—, or M7 + —. 12 12 TRIGONOMETRICAL PROBLEMS. 101 5% Tv Tv Tv eee a i aS go and 9 = (nw +A4B)=ne+ 2 tea (nthedye. pancas. 2 5 ec+6 snC+snB Zeer (4) best Coe ai Bie Cree hs tan 2 Cah mc be, CHB es cb 4A =. cot = cot = tan. —; c—b Pn, c—b e, A+B+C C-B A c+6b A hence tan (—— _ —=| = tan & +. B) = ——— tan—. Q 2 c—b 2 A 10 A (3) If 8e= 76, tan (2 +5) ape ee A A and Z tan (2 +5) =n) ~ 2log2 + Ltan> = 1 — ‘6020600 + 8°7624080 = 9:1603480 = L tan8°.13'.50” + 397, L tan 89.14’ — Z tan 8°, 13’. 50” = 1486; uw re Mae oe SOTO Up te RO SSO Q 1486 = 89,13’, 52”. 6. and B = 8°. 13’. 52”.6 — 3°.18'. 42” = 4°. 55. 10".6 hence ZC = 168°. 27.25". 4. ; oe Ses ‘(m +c) (m —c) 6 Cog Set ae 2\/ab =+/(m +0) (m—c) seo © = msin p> and 4ab = m* sin’; but (@ + b)? = m’; 102 TRIGONOMETRICAL PROBLEMS. . (a — b)’ = m* cos’ d, or a — b = + mcos@, anda+b=m; hence a=m™ cove , and b=m sint E The positive value of sin p will be sufficient for deter- mining the subsidiary angle p- 7. (a) See Hymers’ Triconometry, Art. 139. 1 4.3 (cos@)' = 33 (cos 16+400s20+4. = 4+ (cos40 + 4cos26 +3). (8) 8 4A{1+ aa) = (4 cosec®2 a — 2) (4 cosec* 20 — 2 ( sin? 49 ( )( )> since cota + tana = 2 cosec 2a, 2 — sin® 20 or (sin? 40 + 8) = (2 cosec? 2a — 1 ( : ( aa * ) sin? 20 sin? 46; .. (4sin®2@ cos’2@ + 8) = 4(2cosec?2a — 1) (1 + cos*26) cos* 20, or (1 —cos’260) cos’20+2 = (2 cosec®2a — 1)(1 + cos’ 20) cos® 20; hence by reduction enced! 2a.cos' 20 + cot?2a.cos’ 20 = 1, or cost 20 + cos’ 2a cos? 26 = sin’ 2a; 2 2 — cos*2a+(cos*2a-—-2 *, cos*-20 = aisle Be a Le) 2s and cos’?2@ = — 1, or sin? 2a; . : Tv hence cos 206 = = sin2a, and 20 = 8n7 + G aE 2a : T ; ; or 0 = (4n £1) “ + a, where m is any integer T “ . == (27, + Sy, +: a, where 7 is any integer. TRIGONOMETRICAL PROBLEMS. 108 8. wcosa+ sina = 1, ecosB+a'sinB =1, . wsin (a — 8) = sina — sin, a 1 + tan —- tan — 2 g and a sin (a — 8) = cos 3 — cosa, . at sin B tan— + Rie 2 2 ore. = = . a — a cos p 1+ tan-— tan p 2 2 2 6 6 1 area tian tan 2 S"tan= nyo . 2 p 2 Similarly y= see oe Y= RSC ERS Ss 1 + fans tan -— ] be eary ee eae 2 2 2 2 a B y O) eer} a B ety ( ~ tans tan f) (1+tan3 tan5) a“ (1 -tan3 tan5) (1+ tan$ tan 6 gy’ +y' a B vie vs é ( a. BY (tan§ + tan >) (: +tan5 tan 5) + (‘an3 +tan 5) jee lan tan 3) 2-2 Fs : a = _—"—*;; where S, is the sum of the tangents tan—, Ae si + S's 2 9 tan, tans S; the sum of the products of every three, and S, the product of the four tangents. This expression is symmetrical with respect to a, B, yy, 6, and will therefore re- main the same when they are interchanged in any manner. 9. 4sin Asin BsinC = 2sinC {cos(4 — B) -cos(A + B)} = sin(C + B— A) +sin(4 + C —- B) +sin(44+ B-C)-sin(4+B+C); 104 TRIGONGMETRICAL PROBLEMS. “. 4sin?@ = sin(a+ B-y— 6) +sin(a+ y- B- 9) +sin(3+y-a—-06)-sin(a+B+y¥ —- 380) = sin (24y + 8) + sin (26 + 0) + sin (2a + 8) — sin 30; or 3sin@ = sin (2y + 0) + sin (26 + @) + sin(2a + 0); hence | | (sin 2a +sin 2 +sin 2) cotO = 3—(cos2a + cos 28 + cos2y) ; -, 4sin a sin 3 sin y cot @ = 4(1 + cosa cos 3 cosy), 1 + cosa cos (3 cos or cot@= B x sina sin § sin y Now cos(a+ $+) =-1 cos a cos (3 cos y (1 — tana tan 6 — tana tansy — tan 6B tan y), 1 + cos a cos [3 cos ry I (cos a cos 3 cos ry) (tan a tan 3 tan vy) (cot a + cot B + cot y), sin a sin 3 sin ry (cot a + cot B + cot y)> 1 + cosa cos 9 cos : ; : = cota + cot > + cot sin a sin 3 sin vy B be and cot @ = cota + cot 8 + cot y. Also, cosec’ @ = 1 + cot? @ =1 + cot’a + cot? B + cot’ + of 2(cot a cot 3 + cota cot y + cot B cot ry), Il 1 + cot? a + cot’ B + cot? y + 2 (since cot a cot B + cot a cot y + cot 3 cot y = 1), hence cosec” @ = cosec® a + cosec” (3 + cosec’ +. 10. Let Aa, Bb, Ce (fig. 35) be the three lines bisecting the sides of the triangle ABC, and intersecting each other in the same point O; then if R,, R,, &c. be the radii of the circles circumscribed about AOQc, BOc, BOa, COa, COb, AOb, we have TRIGONOMETRICAL PROBLEMS. 105 aulsi-dity ccaulet epee aad poe: c ~ 2sin AOc 4sin AOc’ ° Asin BOe’ Fae ae idee se aa Aah 4sin BOa’ * 4sin COa’ R Ric a. R yea oe 4sin COB’ Sed WAL) be . RRR; = RRR, eee eesteoetoes sevens (a). Also, Oa =44a, Ob=3Bb, Oc=4Ce, each of the triangles 4OB, AOC, BOC =4 A ABC; and each of the six triangles having the common vertex O A ABC oe Let d be twice the area of one of these triangles; then A —= B0+0c+ Be, 7 9 = AO+ Oc + Ac, 10a Ae =) = AO- BO; r = 1 1 Similarly, A (= - = = BO -— CO, 3 1 1 4 (=~) =co-40; \"5 "6 ‘*, by addition, 11. Let A (fig. 36) be one of the acute angles of the parallelogram ABCD; a, b, c, d the middle points of its sides AB, BC, CD, DA respectively; draw am, bn, cp, dq per- 106 TRIGONOMETRICAL PROBLEMS. pendicular to the sides, and equal to Aa, Bb, Cc, Ad respectively. Join dm, Aq, Bm, Bn, qm, nm; then if AB=a, AD=b, 2 BAD=a, we have Also 4 ABC=7-Aa, vg 7 2MB + + — a = 27; or POT ees Fees a b heey = — Bn=—, .. oS Bm 2? n 2 (mn) The triangles Amq, Bmn are equal, and + absina. ZAmq= 2 Bmn, Tv Lqmn= 4AmB=—; consequently all the sides of the figure gmmp are equal, and all the angles right-angles; i.e. gmnp is a square. Again, area of qmnp = (qm)* C+ d +absina =absina + i(2a° + 26°) = area of parallelogram ABCD +i the sum of the squares of the sides, and q, m, n, p are the centres of the four squares described upon the sides, 12. Let the centre C (fig. 37) be between P and O, then the circle will subtend the greatest angle 2a at P; let the radius of the circle = 7, and CO = a, ~ LC sna=%7, or (PO—«#)sina =r. TRIGONOMETRICAL PROBLEMS. 107 Next, let the centre C’ lie in PO produced, then the circle will subtend the least angle 2/3, and PC" sin Ber, or (PO+ 2)snp-=7, and (PO — 2) sina = (PO + 2) sin fs, sina — sin eo = PO. ——~ = POsing, 2 sina + sin ae Or = seune ine ote (tf). sin 1-—sing ‘ 13. (a) See Hymers’ Triconometry, Art. 138. Cee, ce 17a’ tana=a+—+— + te Ce 3 3.5 Doe One Ore | where a represents the circular measure of the angle whose tangent is required ; ” and if the angle contains 2 seconds, a = ——— 8 ; 206265 ” lees n 1 ( n ) 2 ( n ) SO RRALL 70 Sy te ee ee Pe 206265 3 \206265 15 \206265 Fe a a SP Cee ty erat i C. 315 206265 A (8) If 9=tan@, then tan“'@=@, T or O = (2n + Ns — cot-'@ : iS aich es yr w- (5- at a -&] Sn +1) tan (5) -@n+1) 5 (a rAetba De &e.} , 1 T 1 where 7” is any. integer. 108 TRIGONOMETRICAL PROBLEMS. Now, since tan@ is always greater than @ as long as @ is T T less than —, @ must be greater than aa and the least value of (0) will be found by putting m =1 in equation (1) ; 1 37 1 r Sere asc iergs oS) . : 1 30 As a first approximation, @ + 6 = ia and @ =* nearly ; substitute this value of @ in the second member of the equa- tion, and we have « Da Sore It 20 O4-=—+- (=) = 4°71238 + °00365 = 4-716, 0 2 Cpe) “. @ — 47160 = —1, or 0 — 47160 + (2°358)* = 4°560164 ; hence @ — 2°358 = 2°1354; or 0 = 4°4934... The successive values of @ will be found by putting n=2, 3, &c. in equation (1); .°. as a first approximation, 1 37 oT Tv 9+—-= — ee yet ea iy 9 Ome ( ioe ft but @ + = tan 0 + cot 8 = 2 cosec 20, T “. cosec 20 = (2n + 1) 3 nearly ; from which the approximate value of @ may be determined for any value of 7. 14. See Hymers’ Tuerory or Equations. (Art. 15.) ST JOHN'S COLLEGE. Dec. 1840, (No. XII.) 1. Finp the degrees, minutes, and seconds, in the angle whose circular measure is .1; also the values of cosma and tan-'(— 1)” (m an integer) ; and prove the formule, TRIGONOMETRICAL PROBLEMS. 109 cos B — cos A (a) = tand}(4+ B).tand(4 - B). cos B + cos A cosec 2A 1 + tan’? 4 1+cosec2d (1+ tan 4)” (6) (y) sin 34 = 4sin 4. sin (60° + 4) .sin (60° — A). 2 If 4+ B8+C= 180, then cos’?4 + cos? B + cos’C + 2 cos A cos B cosC = 1. Also, find # in the equation = -1 ante! —1 sec-* a — sec”!6b = sec Foe 4 a Shew that # cannot be real in the equation tan’v=tan (v—«), unless sin a be less than 4. 187 3. Prove that (4) = .00000042366, having given log 3 = .4771213 and log 4.2366 = .6270227. Also find A in the equation Lsin A = 9.9358921, having given L sin 59°. 37’.40” = 9.9358894 and diff. for 10” = 124; also find L sin 59°. 37’. 42,18. 4. Prove that L sin(@ +h) — Lsin @ = phcot 6 — Euh’ cosec’ 0, h being very small, and uw the modulus of the common system ; and hence shew the necessity, for the first 5 degrees of the quadrant, of having tables of log sines computed to every second. 5. Prove a priori that sinm A when expressed by sin A will have one or two values according as m is odd or even; and cosnA expressed by cos A will have only one value, a 110 TRIGONOMETRICAL PROBLEMS. positive integer. Shew also that tan 4 expressed in terms of sine 4.4 will have four values, and find them. 6. A triangle may be divided into four others, three of which are isosceles. 7. If the sides of a triangle are a cos 4, bcos B, ccosC, where a, 0, A, &c. are the sides and angles of a triangle, then its angles will be the supplements of 2.4, 2B, 2 8. Find the area of a trapezium which has two sides parallel to one another, in a form adapted to logarithmic cal- culation, 9. If A, B’, C’ be the angles which the sides of a triangle subtend at the centre of the inscribed circle, then will 4sin A’. sin B’. sin C’ = sin A + sin B + sin C. 10. The circle described through any three of the centres of the four circles that can be drawn touching the sides of a triangle has a constant radius, and its value is twice the radius of the circumscribing circle of the triangle. 11. Prove by means of the exponential expressions for | sine and cosine, that 2 tan & tan 2a =———_—., and cos2m#=cos’w — sin’ a, 1 — tan’ @ and also that ; h h’ tan~'(# + h) — tan~'w = sing.sing.— — sin’s.sin 2% = + &e. 1 where # = cots. 12. Generalize Demoivre’s Theorem so as to furnish all m the values of (cos@ + 4/—1 sin 0)"; and shew that it gives the same values whether the expression be taken to mean S/ cos 0 a BARC sin Gis, or n/ (cos 0 ae (= sin gyri TRIGONOMETRICAL PROBLEMS, pt 13. Shew that 49°.17’ is the angle corresponding to the least positive value of @ in the equation cot 0 = 0. 14. The perpendicular distances of the angle A of a triangle from two lines at right angles to one another drawn through C, are e — cosa, \/1 — e’ sina, and the perpendicular distances of angle B from the same lines are e — cos 3, and /1-e& sin 3; prove that if a+b+e=2(1-cosd), and a+b-c=2(1 —cos@), then P-O=a-—, and cos$(P+ 0) =ecosd(a+t f). SOLUTIONS TO (No. XII.) 1. (1) The arc whose circular measure is unity contains 206265 seconds, .*. an are whose circular measure is *1 contains 20626°5 seconds or 5°. 43’. 46”. 5, (2) Cosma + /—1sinma = (cos 7 + Reel sin 7)” =(-—1)"; but sinma=0; .. cosma = (—1)”. (3) Tan7'(-1)"a=n7 + (- 1)" tan“1a, since tan (— 1)”6 = (— 1)”. tan @}, . tan-}(-— 1)" = naw + (- 1)" tan“! 1 = m7 4 (- Dae A+B, A-B 2 sin sin cos B — cos A oo) 2 (a) cos B + cos A A+B vAl i Bs 2 cos cos 2 oF i = Fatt tan 112 TRIGONOMETRICAL PROBLEMS. () cosec 2A 1 1 1+cosec24 1+sin2d (cos 4 +sin A)’ ( sec A \) 1 + tan’? A » Al} tand/, 6 Gis tan), ; F LEAN : (y) Sing = 27"? sin? sin sapere sin = Let 6 = 34, m= 8, *, sin 3A = 2? sin A sin (60 + A) sin (120 + A) = 4sin A sin (60 + A) sin (60 — A). 2. (a) 2cosAcosB =cos(A + B) + cos(A — B); *. 4cos A cos B cos C = 2 $cos(A + B) + cos(A — B)t cosC = cos(4 + B+ C)+cos(4+ B-C) +cos(4 + C-— B) +cos(B + C — A) — 1+ cos(m — 2C) + cos (a7 — 2B) + cos (a — 24) —1 +1 —2cos C +1 —2cos’ B +.1 — 2 cos’ A; -, cos’ A + cos’ B + cos’ C_ + 2cos A cos B cos C = 1. (i5)) Wet sec. "a ==, nec ad 115, eenn 0, oe ey p; b a ays f} — p =a- iE ] b a cosa =—-, cos B = 5, cos =-, cos@ = -; 1 a “. cosa cosd = cosBcos#, anda+p=3+4+0; or sina sing = sin B sin 0; cos ; sin ‘ hence cos d = £ cos@, sing = — sin @, COS a sin a cos’ 3 sin’ 3 and cos? @ + sin?@ = 1 = ——— cos? @ + ——= (1 — cos? @): ; Pp P cos? a oat ais cos 0) ; TRIGONOMETRICAL PROBLEMS, 113 2 Cee) ee 2 cos’ B sin?a — sin’ Bcos’a a or, ——____._——— .cos’ 6 = sin’a — sin’ B; cos” a r} Q«[ b hence cos? @ = cos*°a, and cos@ = — = + cosa = + e av = &k ab. tanv + tan’ a (y) tan’ # = tan(v —a), .. tan(2v—a@) = Eas) dpbiage ened 1 — tan’ « tan (2@ — @) Yr eee tan 2x tan 2v — tan (2 — a) tan2v+tan(27+a) 7° =aaele Qo? sin {2a —- (2v-a)} pe ie eS ee dk esi de® sin (4a = sin §2v+(2@-—a)t 7° af Bais 8) or and is less than +. 187 ye 187 (-4771213) 3 ; ] (;) = —— log3 = — 187 ——_—__+ (a) ne 3 14 8 4 14 892216831 2 14 44°6108415 ‘ . 2 Segre nae 6°3729773 = 7°6270227, 187 1\'* = 42366 (=) =f am 00000042366. (3) Lsin A — L sin 59°. 37’. 40" = 27, ” and eS See , 18; «. A = 59°. 37.42.18. 124 (y) Difference for 2” = 248; for 1” = 1:2; for 08” = 1 nearly ; , difference for 2”.18 = 27 nearly, and L sin 59°. 37’. 42”.18 = 9°9358921. 114 TRIGONOMETRICAL PROBLEMS. 4. See Hymers’ Triconomerry, Art. 88. 5. (a) Sin A =sin(@mz7 + A) =sin(2m+1)r7—- 4; *, sin2.A when expressed in terms of sin 4 will contain the different values of sin nm (2ma+A), and sinn }(2m+1) 7-4}; hence all the different values will be included in the forms sin 2A, and sin(m27 —”A). If be odd, both of these values are equal to sinz 4; but if 2 be even they are equivalent to sin 2.4, and — sin7 A. (3) Cos A = cos(2mm + A), and all the different values of cosmA will be included in the expression cos (2m7 + A) = cosnA. Hence there is only one value of cos” A when is an integer. (y) Sin4A=sin(2n7+44), or =sin {(2n+1) 7-44}; .. tan 4 when expressed in terms of sin 4.4 will contain all the 2n7r +4A (Q2n+1)7r7—44 values of tan aha and tan | s (Ors On NT 7 T : tan (=" + 4) and tan ea + e - 4) \. the former of which 3 as will contain the two values tan 4, tan iS + 4) , and the latter T 38% the two values tan ( de 4) , tan (= _ A) - Hence the four values of tan A when expressed in terms of sin 44 will be 3 tan A, tan ls + A) ean & - 4), and tan (4 ~ 4) : 2 4 4 6. Let ABC (fig. 38) be the given triangle, B and C two of its acute angles; bisect 4B in D, and with centre D and radius AD describe a circle, cutting BC in E; then since 2zAEB in a semicircle is a right angle, 2 AEB is greater than 2 ACB, and &# will therefore fall between Band C. Join AE, ED, and since 4 AKC is greater than 2 ACE, AC is greater than AE; cut off AF = AK, and join HF, then the triangle ABC TRIGONOMETRICAL PROBLEMS. 115 is divided into four triangles BDE, ADE, AEF, EFC of which the first three are isosceles, since DE = DB, AD = DE, and AF = AE. If AC be bisected in G, the A ABC will be divided into the four isosceles triangles BDE, ADE, AEG, EGC. 7. Let the angles of the triangle be 4’, B’, C’, sn B’ bcosB sinBcosB sin2B in 4’ acosA sinAcosA sin2/4° then sinC’ sin2C sin 4d’ sin2/4’ Similarly, sin(4'+ BB’) sin2C mewm cinet ot sin 264 n B' sin 2 C' or cos B’ + re ees ‘Ai sin2.4’ B’ sin2B ,. * sin2C cos. B + cos a sin 2.4 sin 24’ or cos B’ sin2 A + cos A’ sin2B = sin2C: and sin B’ sin2 A — sin A’ sin2B =0; *, adding the squares, sin’ 24 + sin’2 B+ 2cos(4' + B’) sin2.A sin2B = sin?2C; or 2sin 24 sin2B cos (A’ + B’) sin’ 2(4 + B) — sin? 24 — sin? 2B = sin2B jsin (44 + 2B) —sin2B} 2sin2Bsin2A cos (2.4 + 2B); th ll . cos (4’ + B’) = cos2(A4 + B), or cos CY = — cos2(A +B) = —cos2C, and C’ = 7-20. Similarly, 4’=7-2A, Bo =7-2B. 8—e 116 TRIGONOMETRICAL PROBLEMS. 8. Let a, 6, c, d (fig. 39) be the four sides taken in order, a, c the two parallel sides, and 6 less than d; then if p be the perpendicular distance between the parallel sides, Jd - p+ /b —p*=a-c, or Bo pte (a0) -2(0- VE=P +E pt seo) at piano ee Let p=dsin9, and (d —b)(d 4B) = (4 — c) tan® op; np oie known, and 2(a—c)dcos@=(a-—c)’ sec’, or cos @= ie hence @ is known, and the area of the quadrilateral (a+ c)d sin®@ Q p =(a@+c)- = (a+ is known in a form adapted to logarithmic computation. 9. Let O (fig. 2) be the centre of the circle inscribed in the triangle ABC, then ZAOB=C'=7- (= eA, 4435) D152 Te eC =—+ —; Q 2 T A ao B + —, and Ba 4 —; ie ik Bu 3 similarly, A’ = . 4sin 4’ sin B’ sin C’=2 sin A’ {cos (B’— C’) — cos (B’+ C’)} = sin (4’+ B’— C’) + sin (4’+ C’— B’) + sin (B+ C'= A) — sin (4’+ B’+C’) pon ( aoe Sea A+G—£8 mo 2 2 ) +sin (Z 4 2 ) Barc kc panama! * pe Ogee | (89m A+ B+0O + sin (= + “ = ) + sin ( | 2 ‘ga 2 = sin(w — C) + sin (7 — B) + sin (a — A) + sin 2a = sin 4+ sinB + sin C. TRIGONOMETRICAL PROBLEMS. 117 10. Let O (fig. 40) be the centre of the circle inscribed in the triangle ABC; O’, O”, O'” the centres of the escribed circles touching BC, AC, AB respectively; 7’, r” the radii of the two circles whose centres are O', O” ; r ® “6 OG eres O'C = ’ y= ——— ’ (Re eae cos — cos — 24 2 ~ where A represents the area of the triangle ABC; . 0’'o" a A A CG Ac (s paar sec > = — a So. Ral 2 2 (S' - a)(S' — 6) cos © 7 Also 400’, BOO” are straight lines, since the circles whose centres are O’, O” touch AB, AC, and AB, BC respectively ; A B FO OO AOU O Bee (= S =| x mt Ol GO, = ~ - hence the radius of the circle circumscribing O'OO" 0'O" 0'o" Ac esi O04 € 2 cos = 2(S - a)(S - 5) cost C » A.abe abe _ abe a ~ 2,5'(\S—a) (S—b)(S—c) 24 besindA sind where FR is the radius of the circle circumscribing the triangle ABC: In like manner it may be proved that the radius of the circle passing through any three of the points O, O', 0”, O'’=2 R. IO hy ar vs —1sin2e er *V=1_ e-2#N=j Piao) eT tage abst 2 5, 2 cos 2a er? V=L 4 p=? =I (e V-i_. e-*# V=1) (e” Vol e-*V=l) 9g at sin v@.2 cosa 7 ip (age (Oe eae — 22tane 2 tan ae = me ee eer ge Oe RAL, 2 2 — sec’ v 1 — tan’ « 118 TRIGONOMETRICAL PROBLEMS, | (GN eran) (e°V-1_ ere lye (2 cosa)?-+(2\/ —1sina)? ae 9 Wren eS =2cos*w —2sin’w; .. cos 24 =cos’ w — sin’ a. vut+th—-2«@ (y) tan~* (x § h) — tan7!a = tan-! — h h — tan >. ———— ee tan ; Se ieoa SFr cosec’® = + A cots Laas (Asing)sins | _, & sin 3 ; = tan~* ———_—_ , 1+hsinz coss 1+ coss ls eS : k sin & where K=h sins) =@;3 .. tan d@ = —————;; ( p; . 1+ & coss AG... =a ae = ee bap Ne Le kev —— 1 + ke*V-1 ~ eo vV=l — aati “aaa or 2H Ga 1= log, 1 + ke-*N=1 2 = k (e*¥-1 — e~*V-1) aM (e?* Nl — e-**N-1) + &e.; 2 k? “ P=ksnz— = sin 23 + = sin 3% — &«., 2 or tan7'(w+h)—-tan7*vw=sinz. sin # .— —sin’s.sin2%. 5 he 12. See Hymers’ TRIGONOMETRY, Art. 131. 13. @=cot0; .. Osin@ = cos8é; oi O° 6 eg : —-—+ —-W&c.) =1- —+— - — os we (0 ie G:) mh mp ae 30° 56 6° hence 1 wer epee + eu = Be are SCr Facey 2 Q4 720 TRIGONOMETRICAL PROBLEMS. 119 As a first approximation let =O; .. @ 18 — \/204 rar ng = ‘74 nearly. Substitute this value for @” in the second member of equa- tion (1), 68 24 — 360? +50' = Gud = id (TAB =e 093 5 30 0mes0 or 6 — 72 0? + 12°96 = 12°96 — 4°7814 = 8°1786, and @° = 3°6 — 2°8598 = °7402; . § = 86034, and when reduced to degrees = 49°. £7 14. 6? = (e — cosa)’ + (1 — e’) sin’ a = (1 - € Cos a)’; . b=1-—e cosa; similarly, a= 1—-—e cosB; and c’= }(e — cosa) — (e — cos B)}? + (1 — e*) (sin a — sin 8)’ = (cos a — cos (3)? + (1 — e’) (sina — sin SB)’ = 2-2 cos(a — ) - & (sina — sin #)’ = 4 sin’ a=/ (1 — e” cos” a 2 2 Also a +6 = 2 — (cos @ + cos@) = 2 — e (cosa + cos [3); . cos @ + cos § = € (cosa + cos B), and 2¢ = 2(cos@-—cosq@) or ¢ =cos@ — cos; ; -¢O. a) af oe ee 4 sint 2° sint PAP 4 sins S—P (1 — ot cost 2), 4 cos* ? 0 cos? ? Lak BAe Cos boy iB ahete ety 120 TRIGONOMETRICAL PROBLEMS. -0O ra) 0 —@ : +0 or 1 — rein eae’ — Cos” p+ + cos” P-/ cos” pre 2 2 Q = sin? — e? sin? — B go he ; 2 + - + 3 and cos* ? cos” Be u = e* cos” p cos” iw : f : 2 — BNC OS sir + Cos " cos +9 - + and Ae Meise MI ay Sap ME TA bcs g “4 2 ~ - + Eoeee Op eee cos Bogs ey 2 2 —@0 6 fe 2 Pat = cos — — e€cos B, ST JOHN’S COLLEGE. Dec. 184]. (No. XIII.) ProvE that cos(4 — B) = cos 4 cos B + sin A sin B, 1 and having given that sin 30°=4, find the sines and cosines of “15°, 225°, 300°. Given log 2 = °30103, log3 = 47712, find the loga- 2. Explain clearly the meaning rithms of 72, 45, 1800, ‘00024. of a negative characteristic. TRIGONOMETRICAL PROBLEMS. 121 3. Prove the following formule, nae 2r/ sec? A A sin 2A = ——__._—_—_; sec” A ; 0° sinO<0>0 Bae 0 being between 0 and 3 Pie m Pe Nias oti _12(m—n)(1 + mn) Dalry (14+mn)’-(m—-n)’ 4. If a, b, ¢ be the sides, and dA, B, C the opposite angles of a triangle, then a’ sin2B + b*sin2d = 2absinC; asind + 6sinB-+ecsnC _2CD abcosC+accosB+becosd ab 7) where CD is the perpendicular from the angle C upon the opposite side. 5. Find @ and p in each of the following system of equations, (1) psin‘ @— qsin'@ =p pcos'@ -qeos'p =q J - (2) 0-—g@=sin“'-), 10 eo 2 my) HA Ay e sin? @ cos db + cos @ sin p oy sin’ @ cos’ @ — cos’ @ sin* b 3 6. If lines be drawn from the angles of a triangle 4BC to the centre of the inscribed circle, cutting the circumference in the points D, E, &; shew that the angles D, EH, F of the triangle formed by joining these points are respectively equal mr+A r7r+t+B w+ to 9 5 4 A. "ihe A. 122 TRIGONOMETRICAL PROBLEMS. 7. If the sides of a triangle be divided in D, E, F so that AD” BE CL 1 DB EGR ee prove that the area of the triangle formed by joining the points of section : area of the original triangle :: ”’ —- 3n 43: n’. 8. + o —— =(a' +b? + o?)——__;; a and 2abcosC + 2accosB + 2be cos A =(?+b-C)+(@W+0-W)+(V4+C-e@)=0704+VP +e; asin A + bsin B+ csinC _ 2sin A 2bsind 2CD a RS ees —_—- = “abeosC+accosB+becosA4 a ee ab ab 5. (1) p(cos'@ — sint 6) - q(cost@ — sin‘) =q — p, “. pcos20—qceos2p=q-p, or pcos? =q cos’ d, 9 a and pcost@ — ; cos! @ =4q, or cos@ = ee and cos p = An y ier yl DU Ds 126 TRIGONOMETRICAL PROBLEMS. sin’ 0 cos” Pp _ ao : sin 8 cos @ Een or —- =+2:; hence cos’ 9 sin? Dp ae : cos @ sin p (2) sin 8 cos @ — cos @ sin d = 2 cos @ sin — cos @ sin @ = cos @ sin @ = 545; and sin@cosp=4; .. sin(0+ ) = 735, anon? = sin )_8 in? BT ad ee OL and 20 = sin~'4% +sin7'>),, 2 = sin7' 54%, — sin7' 545. Also, using the negative sign, -8cos@sing@= 75, .. cosé sing = — 35, and sin@cos@= 7, or sin(0@+ d) =<, sin(@- >) =54; et edad & ips ba Sa he ser Steed . 20 =sin7 gig + sin7 io? 2m = sin 35 sin~ vo: 6. Let O (fig. 41) be the centre of the circle inscribed in the triangle ABC; join OA, OB, OC, meeting the circle in SBIR Ut then #2MOR = ait ee. and £ODE=4 (4 - ] when v = 2° 11. Draw Ac, Ab (fig. 45) perpendicular to AC, AB respectively ; Bc, Ba perpendicular to BC, BA; and Ca, Cb perpendicular to CA, CB; these perpendiculars will form the hexagon Ac BaCb, and since the angles BAb, BCD are right angles, the points 4 and C are in the circumference of the circle described on the diameter Bb; hence A, B, C, 6 lie in the circumference of a circle which will be that described TRIGONOMETRICAL PROBLEMS. 129 about the A ABC: .. the point 6 lies in the circumference of the circle described about ABC. Similarly, a, ¢ are points in the same circle; .*. a circle may be described about the hexagon dc Ba Cb. Join Ce; then z CAc = = and z AcC = 2 ABC= ZB; *. Ac=bcotB; similarly, Be =acot 4, and zAcB=r-C; -, MAAcB=4absinC cot A cot B; or AAcB = cot A cot B AABC, AAbC = cot A cot C AABC, ACaB= cot B cotC AABC, .. hexagon 4c BaCb = (1 + cot A cot B + cot A cot C + cot B cot C) A ABC. But 44+ B+Cern7, *, tan 4 + tan B + tan C = tan JA tan B tan C, or cot dA cot B + cot A cot C + cot Bcot C=1; hence hexagon de BaCb=2 A ABC. 12. See Hymers’ Triconomertry, Art. 134. 13. Let O (fig. 46) be the centre of-a circle, 1 its radius; CD an arc of 120°; bisect the are in C’, and join OC, OC’, OD; in OC take OE oe and join DE meeting OC’ in FE’; let zODE =80@; then 2 DOE'=60, 4 DOE = 120, — OE hameel e kor ae ee oes : | and OD - sin(60 + @) OD _ sin(120+@) _ sin (60 — 0) | 130 TRIGONOMETRICAL PROBLEMS. OD OD _ sin(60 + 0) —sin(60-@) _ 2cos60sinO _ OEE IOR sin 9 sin 9 and pee nN, Pea neers or if ayn OE'= i, OH n n+1 Now 4,4, (fig. 47) = 120, 4,4, = 60; A, A, = 120, AA; = 60, &c. Y r *, since O4,= -, OB, = 53 — eels r similarly, since OB, = Fe OB, = 14. Let AB (fig. 48) be an object whose height is n feet miles; BT'a tangent to the earth’s surface, r its radius ~ 5280 = 4000 miles nearly, then BT? = BA(2r + BA) =2rBA nearly, 8000 x ” 3n 3n ; = —___— = — nearly, «. BT = w/e miles, 5280 2 2 15. a'"4+1= (a - 2a cosa + 1) (a — 2a cos3a +1)... Ja” — 2 cos(2nm — 1l)a +1}. Now if m be not greater than 2n, wnt Ax + B P a a wert. a —2ecos(2ptil)at1. Q’ 2n where Qia* -2acos(2p+1)at+1} =a" +1, a” sty eter Re) ce v’—2Qecos(2p+1)at+1 + PSa* —2acos(2p + 1)a+ 1}. TRIGONOMETRICAL PROBLEMS, 131 Let cos(2p + 1)a+/%—1sin(2p+1)a be put for & in this equation, . cos(m — 1) (2p +1)a + \/— 1sin(m — 1)(2p + 1)a 2n—1 = (Aw +B) alee \, 2u —2cos(2p +1)a ve" + 1 since. ———_____________ a” — 2x cos(2p4+1)x+1 d, (v°" + 1) d,| xv — 2x cos(2p + 1)a +1] B B nae” (4 + =| —n (4 + =| ? wv BY is a vanishing fraction, and therefore = @—cos(@p+i)a ./oisn@p+ija’ or if (2p +1)a= 8, — sin(m — 1) sin B + \/—1 sin B cos(m — 1) =—nSA + B(cos B — / — 1 sin B)}, and equating the possible and impossible parts, sin 3 sin(m — 1)3 = n(4 + B cos {3) ; sin 3 cos(m—1)B=nBsinB; .«. nB=cos(m — 1)3; nA = sin(m — 1) sin 8 — cos(m — 1) cos 3 = — cosm/3; or 2B = cos(m — (2p+l)a; nA=—cosm(2p +l)a; and by giving to p the values 0, 1, 2...... ”2 —1, we have nu" cos(m—1)a—xcosma cos3(m—1)a—x# cos38ma v4 v’—2x cosatl v’—24 cos 3atl1 et) a Cee Care Se uv —2vcos(2n—1)atl 9—2 132 TRIGONOMETRICAL PROBLEMS. 2n—m—1 hd VX cos (2nm—m—l)a-—wxvcos(2n—-mMm)a 1 ele BS | wv —Zxrcosat 1 cos3(2n —m—1)a-—x#£cos3(2n—m)a wv — 2x cos8at1 i cos (2n — 1) (2n—-m—-1)a—axcos(2n—-1)(2n—m)a a“? — 2x cos(2n -1)at+1 _ v@cosma—cos(m+1)a xcos8ma—cos3(m+ 1)a a? — 2x” cosa+t+1 a? —2xv cos3at+1 ci wv cos (2% —1)ma — cos(2n—1) (m+ l)a + . xv —~2ecos(2n-1)at1 ; since 2na = 7; nae"-'4+ naP™-™-! cos(m — 1) a — cos(m + I) a oe” + 4 uv’ —2x cosa +1 cos 3(m — 1) a — cos38(m+1)a Es av” — 2x¥ cos3a+1 cos (2% — 1) {(m — 1) a — cos (2” — 1) (m+ 1)a sf wv — 2x cos(2n—1)a+1 sina sinma sin 3asin3 ma a —2xrcosa +1 we — 2H c0s38a+ 1 sin (2% — 1) asin (2% — eee) a” —2xecos(2n—-1)a+1 . n am! a yen m-i And S the required sum ae (“| , 2 7 hes | Next, let m>2n = 2rn +m’, then, sin{m(2p+1)a} =sin{@p+1)rat+m(p+t1)a} = (—1)"sinm’ (2p 41)a; TRIGONOMETRICAL PROBLEMS. 183 t sina.sin m’ a sin 3a sin 3m’'a PS Cn ee wv —2xH cosa +1 av —2xcos3a+t+1 sin (2m —1)asin(2” — 1) ma uv” — 2x cos(2n—-—1)a+t1 = aS) n (amie ee) (1). — aa(o li cm 2 ee eet ( ) Qu hikes a If m be either 27”, or any multiple of 27, each term of the series vanishes, and S§ = 0. ST JOHN’S COLLEGE. Dec. 1842. (No. XIV.) 1. Expxiain the different measures of angles and the advantages of each. Find the number of grades in the angle whose circular measure is unity. Prove that sin (4 + B) = sin A cos B + cos A sin B for all values of 4 and B. 2. Prove the following formule : sec A 1) Cosec 4 = ——————.. () \/ sec? 4 — 1 A 2 (2) Versd =2 (sin 5) (3) 2 Vers (A — B) = (sin 4 — sin B)’ + (cos A — cos B)’. 0 0 (4) Tan (= — =) + cot (= - =) = 25008. (5) Cos? o + cos’ (« — 0) + cos’ (o — d) + cos’ (c — W) 0+o+ = 2 +2 cos 0 cos d cos fy, where ¢ == 2*¥, Find the tangents of 36° and 374°. 134 TRIGONOMETRICAL PROBLEMS. 3. Define a logarithm. Explain the use of tables of logarithms, and find log, e. 4. The sides 6, ¢ of a triangle, and the included angle 4 are known. Solve the triangle. If A receive an increment OA, the position of 6 remaining as before, shew that B, C will receive decrements sin B cos C sin C cos B ; Th Teg Os OA 5 i hae tay a OA respectively. = 0 0 ' T Lee Bie 0 being less than Bt Ss] 5. Prove that F Dae ‘ sin@ 2165 Laie dee : q a avin riven ——— = ——, shew at contains 3 degrees BS fa) 2166” ea nearly. eae _sing . If tan ses (p+), shew that tang = 2 ee i = sin oy) } m i n m nN cos@ sin@ DP; cos@ sin@ = cos 26. 7. In the triangle ABC, AC =2BC. If CD, CE re- spectively bisect the angle C and the exterior angle formed by producing AC, prove that the triangles CBD, ACD, ABC, CDE have their areas as 1: 2: 3: 4, 8. If0+o+W=7, prove that aor) 10 a Wei nee 8 Q"—] WY tan — + —]} + tan ae eee f eee te ( a & =) ( gn S £)

+p? - 2 (m? +n)? / (p+1)°—4(m?-+n’). AD AC Tee DB (fig. 49) = CB ia - AD=2DB, and AB=AD+ DB=3DB; AAG ER CR aww also », AK=2BE, or BE=AB=3DB; and DE=DB+BE=4DB8. Now the triangles CBD, ACD, ABC, CDE have the same altitude, and their areas are to one another as their bases, BD, AD, AB, DE, or as BD, 2BD, 3BD, 4BD, or as 1: 2: 3 : 4& respectively. 140 TRIGONOMETRICAL PROBLEMS. 8. Ifa+B+yeT7; then tan(a+ 3+) =90, or tana + tan 3 + tany = tana tan B tany. and (17+ 5) =e (=7+%) =F (Ft E)=n, 3 2% Bets as Ras Waly 8) 4 since 0+ O6+W=7;3 hence the sum of their tangents = the product of the three tangents. 9. (a) By putting » = 1, 2, 3, &c., we obtain for the series ’ 2 3 S=1— — &¢. Wa OMe es ew F | ] and sinl=1— i — &¢. 1 RRS Be Pie Bed 1 1 cos 1 = 1 ——— + ———_ — &c. Lye 1 Aa, GAP ; 4 3 at sin 1 + cos 1 =2 (1 - + ———_—_—__—_ — &¢.. = 29; bea RAR js Ra Sate Tees or § = 4(sin1 + cos 1) 1 ( ae 7 ee “)= 5 yl: ge ieee 1 OL) de COS COS LoL = sin + Af 2\ 4 4 ae (3) See Hymers’ Trigonometry. Appendix, Art. 27. 10. Let AB (fig. 50) be the vertical staff, C the centre of the ring in the vertical line ABC, D the extremity of the shadow; then if DE be drawn touching the ring in LF, DE will be the direction of the sun, and CE is perpendicular to DE. Let CE =r, then d4B= AD =8r; and AC =9r, or CD? = AC’ + AD = 819° + 647° = 1457°, and ED? = CD® - CE? = 1447r*,. «. ED = 127. TRIGONOMETRICAL PROBLEMS. 141 Cha, v Hence tan d4DC = 8, and tanCDE = ED = 5, 9 1 Be aE TR i 2 tan dD = 2 12 4, or the sun’s altitude = tan7! 4. 11. (a) Cos 42° = cos(60 — 18) vis sin 18 = VB a aENs ) = 4cos18 + Y— vs, 8 _V10+ 2/5 +\/5(62 29/5) z n/ 10 +4/20 44/18 — 34/20 8 8 V/ 1447213 + 1/3 (6 — 447213) 3°8042 + 271409 ey TE Re ae 8 59451 —_— == "74314, 3°14159 x.42 3°14159 x 7 (3) Circular measure of 42° = ——— 180 30 21°99113 celal = °73303. 30 12. (a) Let AC (fig. 51) = 71, => - PCB, or 4 PCB =~ - 8; : 4 ACP =~ +0, \ and sector ACP = = (= = 0 | = segment AP + A ACP 2 “+ a ACP; y" Y Yr fs —@ = A ACP = 3 in ACP i cos 9, or @ = cos @. 142 TRIGONOMETRICAL PROBLEMS. @? G! Er Maat Pa sts ee as a first approximation, h2 0 aie rhs i aoe or 0? 4 20 = 2, 8B =4/ 8 — 1, and @- @' = ~ (° ; 0+0 oP heb . (0-6) [1+] = ee ee "4 — &c. nearly ; TAVARES Dn wees 9 “ yer por esata eae = (V3 - -1) i oC ; > (§) 1+0' 1.2.3.4 T-4\/3 25-10\/3 log 8) tars 71-8 = (3-1) + Soe mer ey si 15°3013 8 = ‘7389, and @ reduced to degrees = 42°. 20’. 42”. 13. Let abe (fig. 52) be the triangle formed by joining a, 6, c the middle points of the sides of the triangle ABC; a then ab = , ac¢=5; be= ap and the triangle abc is similar ~~ to the triangle ABC, having each of its sides half the eor- responding side of the triangle ABC. Now if 8, 8, be the semi-perimeters of the triangles 4 BC, abe, S S,= 5 and RS = area of A ABC, r,S, = area of A abe; R te RS = 47,8), and vr; = es TRIGONOMETRICAL PROBLEMS, 143 . e T; Similarly... = — = —.7 ist pines: 2 Nee side be sa 7 cain, r = —-—— oe (a ee a3 as ind we oun ebacd asd 2 Similarly, p» =", ps = 14. Let a’, b’, & be the sides of the new triangle; |S” its semi-perimeter, then a’ = 8 — a, b' = S—b, c = S—c, and wie , nh ae ’ bea Se) IG i a bc p an SUS a(S = OCS =e); CU Cie Stag) (So 0) (Soa) ia, TELUS 4 25 ae 3 or 7? = 2op- 15. a — Bb? —17 — p° = 2p (7 cos@ cos d — r sin@ sin *). oe ye ki cpa es: y’ cos’ @ sin r’ sin* @ sin + p sin’ _ 1 and eee ey p sin" ® - Pe erage F Sa! 7% a> b? a* a” - rary, oe sin —=1- —;~ p’ sin’; and rcos@ = a NA ee ae 5 P. a’ a°b b $a BR ¢: 7 LT eae (a -* e p* sin’ p) cies ——- = 2p cosp.a J 1 - 28D _ op sin g (p sin): 2 or (5 uy sin’ @ — 1) p* — = 2apcosh UA Heenan 144 TRIGONOMETRICAL PROBLEMS. 2 a P hence (5 sin * b = cos" ») © ——2 (5 sin’ @ — cos” p)e $l 4 a? cos® p , Sin’ ig Fi bh? ep Bb? (1 rai ae) 2 «. (Fsin* g + cos* ¢) & ——2 & sin’ p + cos! p) & Sa | (a? — 6"), cos? 2 i> be a p Sg pros ze (Fe sin® ep + cos? ) & b= £2/a? — b ey wae © b oe Rs Be hence a 7 p” cos 1=+ sue b p cos ®, ee @—P _ pe or +f au preos gp #2¥ a : pcosd +1, 16. Let P,, Ps......P, (fig. 53) be the points of division Orethe circle: 702 ote O Leeson es OP, =7,. Produce P,O, P.O, &c. to meet the circle in Q,, Q., &c., and let OQ, prs OQ, = Pe2> &c., ie ge7 7) PG oe PA EE (2. TRIGONOMETRICAL PROBLEMS, 145 Now if a be the angle which OP, makes with OC, 20 n 29 Aor cs (7 — pi) = 2b Jcos a + c03 (a+) + COs (a+) fae n n 27 + Cos (14-1 7)h a0, n or 27, = pi; 22: ee 1 oO ler 8) ibe | le rete e 9 © a® — b r, a— Ov? ll and adding (1) and (2), 25 — 17. See Hymers’ Triconometry, Art. 130. Let » be a fraction = By then (sin 6 + \/ — 1 cos 6)" = feos (= = 6) + / —1 sin (= - 0) =cos 7” (2X7 + 5 — 0) +V-1 sinn (2\m + = — 8) =cos {(44 +1)" — 09} +/—1sin } (4 + 1) —n6} STN fa 1S where A may be any integer from 0 to q-1. Now (sin @ + 4/— 1 cos 6)" = («1 — cos? 0 + / — 1 cos 6)" =(1- cos’)? +n(i - cos’@) ” Beran fai 1 n—3 EEE DAG ee (1 — cos’) ? cos?O/ -1 ie 2h2 ; iss n—4 Oe Tees eg 8 A)! cost O + ees he OES. eA n—1 ae (1 - cos’@) * cos’?@ — 10 TRIGONOMETRICAL PROBLEMS. 146 n | n—2 ri SS= (1 coseyy— a (1 — cos’@) * cos’@ om -—2 oe n-4 dl 2 u Ie ) (1 — cos’@) * cos'@ — &e.; - nt (n—1) (n-2) n= S"=n cos 0 {(1 —cos’8) * — Sea |: (1 —cos’@) * cos’O™ n= —2 —3 — 4 a (n 1) © : Be ) (n ) (1 2 cos’) 2 cos'@ — &e.} ; ° e -v n—2 n hence expanding (1 — cos’6)?, (1 — cos?@)”, &e. n nm.n—2 S=1 Ttis ROS A caine, Fe Coe a ee n(n —2)(n —4)...(m — 27 + a) Saray ee ees ys ( ) OV as.2T7. nm.n—-t1 n—2 = — cos’@ 41 = cos’@ + &c. 14;2 (m — 2) (m — 4)...(m — 2r + 2) a ake 527-2 ; mee ae cast) 70, he} n(n — 1) (n — 2) (nm — 8) ON eat | 1.2.3.4 cos'@ 31 — : cos’@ + &c. _, (n-4) (n—- 6)...(m — 27 +2) a4 i ave Qr—4 seul) 2.4...27 —4 cost" 0; Set ee laie en oo ~ ~ n(n pate eto 1) (m DH cos! 6 - be. 13 BAN (nl ble 2.4 TRIGONOMETRICAL PROBLEMS. 147 n(n — 2) (m= 4)0( -2r +2). hehe eee ay a eon, Aer +(- 1) {(@r — 1)(@r — 8)...1 (27 —1) @r—3)...5.3 n-1 | 2.4...297 Piet Ue oe, Oe ee pet ey ee bye 1) (90 8) | NCE Soo ie A cea an ke nay gd 2.4...27—4 2.4 No OA. 2 Oe 2 2 Qr —1)(27r — 8).2.5 (mM -1) (2-38 (Qn — 1) @r— 8).1.5 (1) (n—3) | a 2.4...27 — 4 2.4 2Qr—1 n—} = the coefficient of # in the product of (i+~) ? and (1 +a)? when they are expanded by the binomial theorem = the co- N+2r—2 2 efficient of # in the expansion of (1+) m+2r—-2n+2r—A4 nm+2r-2 2g (eater 2 2: \ 2 | eS o ae g n(n+ 2) (n+ 4)...(m +27 - 2) a 2V4.. 29 ; hence the coefficient of cos’”@ =(-1) _ n(m—2)...m—(2r—2) n(n+2)(n+r)...(n+27-2) 13. 54. (2t—)) 2. A 2T _ n(n? — 2°) (n? — 4°)... }n?— (27 — 2)°} =(- 1) - 5 1 Ra Meats Ree dy 7 he S=1—- 1 Lola tgp Se Oe IES Sao ee Rees eer nae ee) cos’"0+ &e. Eee anes 10—2 148 TRIGONOMETRICAL PROBLEMS. a n—3 and §"= 7 cos@§ (1 - cos’) ® — ees (1 -cos?@) ® cos’@ “ De 8) (i - cos?) ® cos! — &e.} nm—1 nm—l1 n or ———-=1 —- cost 9.4 P= VO 9) cos + Be, m cos 8 2 2.4 NCE gy Siig eta eae cos?" 9 + &c. A i CEE? cos’ {1 — a: cos’0+ &e. 2 ie 2 ai (- inves (n re 3) (n no 5)...(” —2r+4 1) cos’ 76 + &c.? 2.4 5.;27T-— 2 je Ve RUE IA eI BY cos'@ {1 —— 1h) ees Qe 29+ 4.5 (Bayt ee se eee) cos’’~*0 + &c.? OA. er — & m-—-1 (3 nm—2 ey, by ee (5+ ) costo 3 2 24 et ys a ES — 4, eda De ome CON costo + Ke 8.5 Qa 2 2 ie | (2 — 1) (nm — 8)...(m —2r +1) BG) 3.55327 41 “ ip (2r + 1) (27 —1) 2.4...27r —2 (2r+1) @r-1)...3 5 n-2 OLS e 2 (2r +1) (2r-1).. + 7 (= (n=%) | i ee RP EC ORL 2.4,..27 — 4 2.4 TRIGONOMETRICAL PROBLEMS. 149 (Qr-+1)(2r —1)...3 <(@7r-+1) (27 —1).,.5 n—2 and rs Re Zed. oor — 2 ve 2r+1)(2r—1)...7 (mn —2) (nN — 4 @r+1)@r-1)...7 @-)M@-4) 2.4,...27 — 4 2.4 2r+1 n—2 = the coefficient of vw" in the product of (1 +n)? and (1 mea ae when expanded by the binomial theorem = the coefficient of w” n+2r—] is the expansion of (1+) ° m+2r—-1%n+2r—-3 oe ) ——_—— , ———_———_ .... | ————— - r+] Q Q 2 ey 1.2.0.9 (7 +1) (m + 8)...... (n +2r —1) * dea or , hence the coefficient of cos?”@ , (m—1)(n—3)...m—(2r—1) {rere ae $.5...(27 + 1) Pee lied 9 =(-1) (nm? — 1°) (n? — 3°)... §n? — (27 — 1)*} 1.2.3.,.(27 + 1) , =(-1)' n? ie 1? (n? _— i) (n?— 8*) + ‘=n COS fi =- cos’@ cos*@—&c.... auc ON Fras 1.2.3.4.5 @ 9) f - 8)... fr - ag eV (= 1} 1.2). Shor 1) ye RY also cos { (4h + 1) — ~ 0} = §; sin $(4A+ I) - nOi = S'; . cosv@ = cos| (4X + 1) - }(4 + 1) =" — 70} | n n ; = cos (4X + 1) =. S+sin G0 +> 8" (1) 150 TRIGONOMETRICAL PROBLEMS. : 6 NWT Vas sin2@ = sin[ (4A + ar — {(4r 4+ 1) ioe nO} | = sin (40 +1) <=. S'—cos(4A 7 1). (2). All the different values of cos ”@ and sin ”@ will be deter- mined from equations (1) and (2) by giving to X the values Oe US q-—1, or any q consecutive integers. A may be either positive, or negative. na ne, Now cos (27 + 1) Saris (4X + Pes ; if » be even: and 7 nN cos (n—1) (2r+1) > = COs 5 (4A-+1) — -S -rn)} =sin(4h+1)— -, cos (4. + 8+ sin (40 + =, S' nT ora : ps Chia) Ta 8 ck cos (% — 1) (27 +1) 5.8; If x be odd = (QA —1); cos (27 +1) — = cos (4 - 1) =; n and cos (7 —1) (@r+1) > = cos {(4N— 1) = ~= - (2A — 1) 7} = — sin (4A — =; wey, n and cos (4A — N.S sin (4A — 1) n = cos (27 + 1) aS + cos(m — 1) (27 + O) a. TRIGONOMETRICAL PROBLEMS. 151 Hence equation (1) is made to agree with the result given in the problem, both when 7 is even, and odd. In the latter case — )d has been put for X. This result may be obtained more simply by the aid of the differential calculus, in the following manner : Let cosn@ = a +a, cos@ + a, cos’ @ +,..... + a, cos’ @ + a,,, cos’ *'@ + a,,,cos' t? 8 + &e. ... (1); .. by differentiation, nsinn@ = a,sin@ + 2a, cos@ sin @ +...... + ra, cos'~'@sin @ + (7 + 1) 4,4, cos’ 8 sin re (Phi 2) Gree COS. Os GO -Rc.yts eee? (2), and differentiating again, n* cos n@ = a, cos @ + (2 a, cos? 0 — 2.1.42) +...00. + $7" a, cos’ @ — 7 (r — 1) a, cos’? 6} + {(r + 1)? a,,,cos"*' @ — (7 + 1) ra,,, cos"-'@t a { (7 + 2)° a,42 cos’*? 6 — (7 + 2) (% +1) a,42 cos’ 0%, &e. = —2.1.a, + (1’a, — 3.2a;) cosO +...... + Sra, — (r + 1) (7 + 2) G42} cos” 0 + &e., but n? cosn@ = n?a,. + n’a, cos 0 +......+n' a, cos’ @ + &e. ; and equating coefficients of cos" @, we have n” posh, y Pay (+ IVF 2) Gage = Ms OF Ory = — Ee as hence ad, = fare fo a,=— 152 TRIGONOMETRICAL PROBLEMS. ahig soni SRG n? (n® — 2°) (n? — 4°)... §n® — (ar — 2)°} : L Yeparces oT y nn” — 1° mn” — 3? Also a,> — 3 A,>5 As want PP ° n? — (2r —1)° Qor+] ar or an er een HES y By Qr (2r + 1) Brean les We aid ae n? — (2r —1)° OT Als a= (- 1)' iter Wh ered ery ote Cre Dy Oar ales is the 2r+1 To find ’Gy.7a;.. La equations (1), (2) put cos@ = 0, or O= (4A 4 ye (A being any integer, positive or negative) ; T A T ‘3 cos (4A + 1) 5 = ap; nsinn (40 +1) > = m1; nT ; nT ; or cos” 8 = cos (4A +1) mike S' + sin (4A + Ber . S” as before. ST JOHN’S COLLEGE. Descemssr, 1843. (No. XV.) 1. Prove geometrically the expression for sin 0 in terms of cos2@ and sinzg@. Explain geometrically the two values in the former case, and the four values in the latter. Give any new proof of the formula sin(4 + B) =sin A cos B + cos A sin B, and shew that log, m = log, 6 log, c log, d ... log, m. 2. If on the sides of a triangle are inscribed three tri- angles whose vertical angles are equal to those of the triangle respectively opposite ; shew that the triangle formed by join- TRIGONOMETRICAL PROBLEMS. 153 ing the centres of the circles circumscribed about them, is in every respect equal to the original one. The centres of the escribed circles of a triangle are joined, forming another triangle, this is treated similarly, and so on ; shew that the radii of the circumscribing circles form a geo- metric progression whose common ratio is 2. 3. Solve the equations, ° 4(v — a Q acest) = 2m —2chd7' 1+ 2° yr wea (sw —1)(a@ +1)t=4sa", s =cota+cot B + cot y, cos 70+7cos@=0; tan : : . T ¢=sin2a + sin2 + sin2y, Cte BY = oa 4. Ina plane triangle whose parts are a, b, c,- A, B, C having given (a, A, b+), or (c, A, a + b) in each case, solve it by the aid of a table of logarithms. Also if the angles and sides receive small variations, prove that cOB + b cos ASC = 0, a, b being constant. cosCob+cos Bdc=0, a, A being constant. 6b tan A = 0C.8, a, B being constant. 5. In the following identical equations deduce the latter side from the former, sin O(1 + sin 8) (1 — tan “) Ce et COS. Oia ee ee ee é 1 + cot — 2 sin @ + sing + siny — 1 = (cos — sin "| (cos ps sin ®) (cos — sin ) V2. 2 2. 2 2 2 < 154 TRIGONOMETRICAL PROBLEMS. (tan @ + tan d + tan yy) (cot 0 + cot + cot Wy) = 1 + cosec @ cosec @ cosec Wp, O+p+yas- sin 9+sin p+siny =4sin ais Piagee sin ans +sin(@+p+y). 2 6. On the side of a mountain, at a distance from any extent of horizontal ground, and with a level and chain only, find the inclination of that part of the mountain to the horizon. 7. In each of the two following sets of equations, put the former into the form of the latter, h cos(@ + a) + cosa = 0, (1 + hcos@) Stan(O + a)} + (h? +h cos 0) tana =0, ... (1). AA Ot SD) GN Oral) Lire heat Caton Was rac 7? — QQ? y? — ce’ cos (3 + a) cos(3 — a) anti, We ed i ae cos’ a cos” (3 COS” ry Sieh wees tats Se ; : F—b'e® s—ace s—a’?h where cos*a + cos’ 3 + cos’ ry = 1; 9 € ¢ € s : a’ cos’ a + 6° cos’ 3 + ¢ cos’ y = (=) pres C2). Be ss Also if a+ B+y=7, and f(a, B, y) ~ 3 cosee”—* sec Pig a 2 2 =+(1 + cosa cosy) (1 +cos 3. cosa) cosce —— then f(aBy) +f (By) + f(yaB) = 1. TRIGONOMETRICAL PROBLEMS, 155 1 xv xv BP ISVP SHINS # Be shew etna. eo = es! Re, 1—- 1+ 2—- 8+ 4-—- 54+ 6- and if m be even, a oP: 8ar\2 (m —1)7)* — = {| cosec —)]} + { cosec —] +&c. + ¢cosec —————+} ; 2 2n 2n 2n and sum the series, cos 20 cosec? 20 + 2 cos 40 cosec? 49 + 4cos 84 cosec? 89 + ... + 2"-! cos 2"0 cosec’ (2"0). 9. The angle AOB which two objects A and B in the same plane with O subtend at the central point cannot be observed, and so another station P distant from it by a small quantity / is selected, from whence the angle of deviation APO(@) is observed, as well as APB(C); OB(5) and the angle A4BO(A) are also measured ; shew that the reduction to the centre of the station in minutes is nearly h sin C sin(4 — @) : 3438). b sin A ( ) he a Oo oF .,(2n-—1)7 Pelt os sins eb = sciInhn —— 9. - gnc he An An An shew that = Go) = n(n — 1). Prove this when n = 3. 11. If we have the quantities 0 24 +0 4a + 0 2(n —1)7r +0 oe ae CR ee COS por a 6 ge) COS 22 n nr nr nr 156 TRIGONOMETRICAL PROBLEMS. and if we multiply the m" power of each into the p' power of each of the others; the sum of all such products =p, +p, cos0 + p,cos20 +... + p, cos 70, . : : : = r being the greatest integer contained in - , and Spee p, ... p, being independent of @. 12. From the point of intersection of two rectangular tangents to a circle, secants are drawn dividing the right angle into 27 equal parts, and perpendiculars on each pair of tan- gents at the points of section. Shew that the product of all these perpendiculars yen) 2 y being the radius of the circle. 13. which are equivalent to +sin@ and + cos 0. _ All the values of cos2@ are included in the formula cos 2rq + 20, therefore all the values of sin@ when expressed in terms of cos20@ will be included in the form sin (r7 +6) where 7 is any integer, which is equivalent to + sin@. Hence in this case there are only two values of sin 0. TRIGONOMETRICAL PROBLEMS. 159 (3) Let ABC be a triangle, b a then c=bcos 4 +acosB, or —- cosd4+-cosB=1; C c sin B aes “3 4 cos sin C sin A A+-— cosspaB=1; sin C hence sin B cos A + sin A cos B = sinC = sin(A + B). (y) Let m=0'; .. log,m = zlog,b = log b log,m ; similarly, log,m = log,c log,m, &c. ; “. log m = log,b log,c log.d... log,;m; 2. (a) Let ABC (fig. 25) be the given triangle, a, b, c the vertices of the three triangles described on the bases BC, AC, AB respectively; A’, B’, C’ the centres of the circles described about the triangles aBC, bAC,cAB; R the radius, and O the centre of the circle described about the triangle ABC; join BA’, BC’, AC’; then BA’= radius of the circle described about a BC CB ake CB s sn BaC 2 sin BAC — similarly, BC =R; Rh, ol and £CBA' = 4 (9 - CA'B) = 4 (x - 2CaB) =4 (0-24); similarly, 2 ABC’ = 4 (4 -2C); . LA'BC'= £CBA'+ £ ABC + £4 ABC=1-A-C+B =2B, and 4’C’=2BA' sind ABC =2R sin B=b: in like manner it may be proved that d’B’ = c, and B’C’=a; ‘. the triangle 4’B’C’ is similar and equal to the triangle a bC: (BG) Let A,, B,, C, (fig. 55) be the centres of the escribed circles touching the sides BC, AC, AB respectively; then 160 TRIGONOMETRICAL PROBLEMS. A,B,, A,C,, B,C, bisect the exterior angles of the triangle ABC; and AA, bisects 2 BAC; £ A,AQ, =~. 2 Also 4AC,B=7 — (C,4B + C, BA) a—A ee Ge -7~( PU re Wan ey eon 5). B ewe Hey 2 Vb ein d Op BY ‘oa os — 4 c cos — 2€ COS — B and A,C,=AC,sec AC, 4, = PUT MT as = 4K cos— ; sin = cos = hence if R, be the radius of the circle circumscribing 4, B,C), 4R Hae A,C, 2 ———— =4R, or R,=2R. sesh ER ES TITER B cos a Similarly, if R, be the radius of the circle described about the triangle formed by joining the centres of the escribed circles of the triangle 4,B,C,, R.=2R,, &c. 3. (a) 2 COS 70 = (2 cos 6)' a7 (2 cos @)° howe: Ven ct eee A ote Orie Oy i 2g . ia ) See eee) 2 (cos 70 + 7 cos @) = (2 cos 8)’ — 7 (2 cos @)° + 14 (2 cos 6)? = (2 cos 9)’ } (2 cos 0)! — 7 (2 cos 0) + 144 =0; TRIGONOMETRICAL PROBLEMS. 161 hence (2 cos 0)? = 0; or 0 = (Qn + 1) = also (2 cos 0)' — 7 (2 cos 0)? + 14=0; Cite (3 oo (2 cos 7), = La A “. cos@ = + Se heat 2 (3) Let v= tan Ps chd-! eae: =; and 2m —2chd7! var =2r7r—2(r — 2) = 40 ; _, 4(tan @ — tan’ A) 1 tan p = 40, henee tan 4 _ 4 t — 4 : : (tan p tan® QP) _ Senna” an b tan® 1 + tan’ 1 — 6 tan’ + tan' ’ x tan @ — tan’ d = 0, or tang =0 or £1; and 1 + tan’d = 1 — 6 tan’ + tan’; . tan‘@ =7 tan’@; «. tang =0 or ba/T: hence the values of v are 0, £1 and +/7. ; AS se — ve + se - 1 =— a’; (y) ; 1 4 1 of lee v+e--=0 Seat y 11 162 TRIGONOMETRICAL PROBLEMS. Now s = cota + cot B + cot y = cota cot 9 cot y, since tan(a + $+) = infinity ; and ¢ = sin2a + sin2 + sin2y ae cos 3 Scos(a — yy) + cos (a + y)} = 4 cosa cos B cos ¥y ; 1 A. et eas tana tan B tansy + seca sec PB secy; s but sin(a + 6 + y) = (tan a+tan 3+tan y—tan a tan @ tan +) cos a cos 8 cos y=1; 1 4 ae Pitegin Gana On BEN E89 f- s also tana tan 3 + tana tansy + tan PB tany = 1; 1 and — = tana tan tany; s hence the equation becomes wo — (tana + tan 3+ tan) a +(tana tan3+tana tany+tanGtany)a—tanatanGtany = 0; which has for its three roots tana, tan 3, and tan. 4. (a) a = 6b’ +c’? —-2becosdA = (b +c)? — 4be Sides 9 ~ © A = = m' — 4be cos’— , if b+c=m; (0 — Cc) = /m — 4be = AV mt = (nt ~ a8) 004 2 2 2 2 / ’ + = > * fee, — 2 a” sec — m* tan = asec— V/ 1 — —sin*—: 2 Q 2 a 2 ~~ ie et J let —sin— = sing; a Q TRIGONOMETRICAL PROBLEMS. 163 A A b —_ t= » =" = —— zi : CH SEC A cos @ = m tan ; cot @ ; and b+c=m; Sa: 2 A aia m sin (= +9] asin (=+¢) asin (5+) et eS OF = iy of. sin A cos — sin p sin — cos — Q 2 7 asin -_-— @ 2 SUC 60 0se ee sin A m b+e sinB+sin(B + A) Also, — = = -- — a a sin A ; A A , A 2 sin (B ie *) cos— sin (2 + =| a 2 ee ‘ sin A cr f ; sin — ae A m A . sin B+=) =" sine i 1 : ( “) = "sind, (1) from which B may be determined. A m A Similarly, sin (c as = ees SIN — occ kes 2), y> . 2 a 2 ( ) Equations (1), (2) though apparently identical, will either of them give two different values, which are the values of B Bite. Lor A’ A sin (c +5) = sin (n- B-4+5) = sin peas 2 2 he \ If B and C be first determined from equation (1), asin B asin C 5 ecte 7 ee ee sin A sin A ” which are therefore known. 164 TRIGONOMETRICAL PROBLEMS. Denn a Sin ee m U3) cee ae ees RGREY bil Fes hs c sin (A + B) A + c to , epee AE: ee cos 5 £7 = © cos (<2 - -4), or sin€ = “sin (S44): m 24 Q m 2 Cie C m-—ccosdA +m, =.c (cos 4 4+.cot — sin A); hencecot— =| es 2 2 e sin A Me 2m sin and if c cos A = mcos cot — = ———_—_ ?> Q FT i from which C is determined, and B = 7 — (A + C) is known. To find a and 6b, we have a’? = b? +c? —2bc cos A, or a —b? = m(m — 2b) = &* — 2becos A; (m +c) (m—c) Pane * m? —c® = 2b(m—ccos A), or b = 4m sin and a =m -—b is therefore known. (y) asin B =bsin A; -.acos BOB =bcos A0A = —b cos A(SB + SC), since 4d =27-—(B+C), and dA=—- (0B 4+0C); hence (a cos B + bcos A) 3B + bcos ASC = 0, sees eee (1). Again, asinC =csin 4; .«. acosCoC = sin Adc; or cOB + bcos AdC = Onn. ec noece also, OB +S5C =0; and asin B = bsin 4; “. — acos BSC =sin Adb; hence cos Bdc + cos Cbb = 0...(2). eS... 165 TRIGONOMETRICAL PROBLEMS. Lastly, bsin d =asin B; .. bcos AOA + sin 46) = 0, and 4+C=27-B8B, or 644+6C =0; o. tan 406 = — bdA = DOC. ......00ccceee. (3). 5. (a) Cos@ = cos’@ + cos@ (1 — cos 8) we ,0 eee) = cos’@ + 2 sin*— { cos? — — sin’ a 2 2 2 ne ( Od etx Ai Bria, 2 sin” — | cos— + sin — cos — — sin — 2 o 2 2 2 2 ; = cos’ @ + ( Ori: 2) cos — + sin — a bi. ee i ke, ; 0 2 sin — cos— (1 + sin 8) (1 — tan =} = cos’ @ + 5 g 2 6 1 + cot — Z ae 8 iyi sin 8 (1 + sin 0) 1 — tan > = cos’ § + ne Anoka (8) 4sinasinB siny = 2 sin f cos (a—-y) — cos(a+ y)§ = sin (3 + y—a) +sin(a ++ —f) + sin(a +B — vy) —-sin(a+ B+). Let B+y-a=8, aty-B=, at+B—-y=y, ,atBP+ry=9+O+y, 0+ Ces aie emt NG Waise wee 9 a SY Pm om and sin@ + sing@+ sin — sin(@ + @ + Wy) 0+ _O+ ‘= Asin e+ sin th Rdg tPA 2 2: 2 166 TRIGONOMETRICAL PROBLEMS. If 0+ p+ =e, : ‘ t es Wet atk ‘s x © : r_¥) eRe Ne RD Ys 1 =4sin (= =) sin (= ¢) sin (7-5 = sing) + (cos ® — sin £) — Ye Shien cos —sins] paleo Hs ale" 5 sine} =f 2 (coss sin =) (cost — sin ) (cos — sin Y) : 4 2 2 2 2 2 (y) Sin (6 + @ + W) = cos 6 cos ¢ cos yy (tan 9+tan p+tan V/—tan @ tan @ tan )=1; , tan? + tang + tany, = tan 6 tan gp tan Wy + sec @ sec @ sec Wp ; and cot 0 + cot @ + cot Wy = cot @ cot d cot W, since tan(@+@+wW)= &; . (tan 6 + tan @ + tan \y,) (cot 6 + cot d + cot W) = 1 + cosec 0 cosec @ cosec W. The remaining result has been determined in equation (1). 6. Let AD (fig. 56) be a vertical staff whose height = h ; AB a horizontal line along the side of the mountain, deter- mined by the level; C a point on the mountain on the same level with D; measure 4C = b, CB =a, AB =c, draw CE perpendicular to 4B, and CF perpendicular to the horizontal plane passing through AB, and join HF’; then c.CE =2areaof A ABC = 2./8(S — a) (S — b) (S— 0), and if (@) be the inclination of the mountain to the horizon 6h OM HERS he sin@ = — CEP CE 347 S820) (5 =) (Son TRIGONOMETRICAL PROBLEMS. 167 7. (a) (1+ cos0) —A sin O tana =0; a 1+hcos@ CN) (966 can Cees tan@ + tana A sin @ ais a eth ee ee > ete ee = e 1 — tan@ tana (1 +hcos@)’ Lb tang Asin @ h+cos@ or tan (0+ a) = — - : ay)? sin and 1+hcos@=A sin§@ tana; “. (1 +A cos @) tan (0 + a) + (2 + cos 0) A tana = 0. a (cos* y — sin’ (3) __b” (cos’ a — sin’ y) (8) # + x vn =a" frien be c” (cos* 3 — sin’ a) Ao) a SS ES i 5 eal bid or COs age COS (3400 C.COR cy | —— =0; re a vr? — BP ye of . (a? cos’ a + bcos’ 3 + c* cos? y) 1 — §(0 + ¢°) a? cos’ a + (a° +c’) b° cos’ B + (a” + Bb”) c cos’ ry} 7? + a’ be? = 0;...(1) or s*r’—$(b° +0") a cos’a+(a* +c’) B cos’ B+ (a?+b")c cos” yt 9" 2 r hy ee) 2] 2 2 2 2 2 ° + a’ bc? (a cos’ a +b’ cos’ 3 +c? cos Vee hence ‘ s 9 € e ¢ oO 9 ‘ 6 © 3) 1 1 —}(b° + c*) a’ cosa + (a + €*) b’ cos’ B + (a + 6°) & cos’ y}— s ‘ ESaN + a 6° c* (a? cos’ a + b? cos* 8 +6" cos’ ry) = = 0, ...0000.0000-(2) s! : ilee Din eb G which coincides with equation (1) by putting —— for 1; | ; s a’ cos’ a b° cos” [3 c cos’ y ee ee aon = Pai ea). C2) s § s 168 TRIGONOMETRICAL PROBLEMS. cos’ a cos” [3 cos ry AUTRE AS cI os be ANS BS. (y) f(aBy) _ B- a sin cos— Z ne 1 (1 + cosacos 3) (1 + cosa cosry) (1+ cos B cosy) (sin =F cos) (sin? “cost ) (sin? Y cos— “| Leos _ (1+cosacos) (1+cosacosy)(1+cosBcosty) cos ry — cos 8 ~ (cos 3 — cosa) (cosa — cos ry) (cos ry — cos 3) 1408/3 cosry =F cost = 2sin2—F aint *F - cos - cosa, : . a (since 2 sin and similarly for the rest) cos ry — cos [3 os RUE Soa en 1 + cos [3 cos ry where P is a symmetrical function of a, 9B, y, and remains unaltered by interchanging a, 2, y; . f(aBy) +f(Bya) + f (yap) Pea COS a — COS ’y at deal = + - — 1 + cos B cosy 1 + COS a COS 1 + cos B cosa = P}tan(y — p) + tan(O — W) + tan(¢ - 4)}, (putting cosa = tan@, cos=tang, cosy =tany), = P tan(\ — #) tan(@ — y,) tan(@ — 6), [since tan {(y — p) + (0- W) + ( — 0)} = tano = 0}, = Px 3° 1; or f(aBy) + f(Bya) + f(yaf) = 1. TRIGONOMETRICAL PROBLEMS. 169 1 8. (a) &=— 1 I ee a ee gh da ae at a 4 v ee ee eee kc)! 1 Q Ga. 2497190, “720 1 1 P1 ye xv x” a? x' a” 1--+——-—4+ — - —,, &¢. 2° 6 24 180 720 av a” au v =14+-+4+— —-—+4+0.2# eeavce sabe, 2 12 720 P2 1 g oe 1 a les & < n° -—-+— ee Cc. 4 eS 9 12 720° 6 360 7 Cig tat) ee aot 45 aQ— —-— = ’ ced eee. fe bP STO 121270) Ps 1 5 ey an, at te e i ee x & —_-— —+ — + —_—_ Cc. eG og oh aR ae Cc. 3 18 270 12(270) Gr Odes 1O80n Vv a“ =$3+—-+—+0.4', &c. 2 20 1 “we 9 Da =3+2H(-+—+0.a@, &.J=3+ —, 4 40 Ps 1 A Qu x Ps = = = 4 in ae + me > &e —-+ ek &e 1 + ces, &c 4 40 10 170 TRIGONOMETRICAL PROBLEMS. I 2 v = = = 5 i+ &c ie 1 wv v ; Tey erring, SREY ©} &e 1 —- Lie &ec 5 50 ] = 5 1. &c. = + peo, &e. 1 xv Xv G2 pk Hence e* = SS Cc. La) Ls Qi 3 4 es i ead the successive denominators being taken -—1, 2, —3, + 4, — 5, + 6, Xe. 1 na (3) log, +2) =e - pe one &e. (1 oe. a? a Sh aot 3g +- — — C.) = @ = 2s NS at ee ie LD. : oe & & pe ss op,=1—-pp=- —-— + &e Dp, 5) P Pr 8 9 q Lh Ole tae & CL 8 Balls 2 cal aemt laae atl Rn 2 ceaadh emetcas ea. 4, fa tei rg Aad fa yee) 4 pi vps & 20 3a A =Q+ CPs=p,—2p.= — — —.+ &c D> Pp.” SO ea Ween 42 8 seis 6 ; 1 Qu Ba? Aw’ re pede fy Wsahiag mats on rr a + XC. Dera Wa ik ethan ita Mean ta 2 3 4 Ds QU, Qu A Ow 8x S34 8a pe 3 + —-— +Wc. Ds ke PGs 4 cnn ee AG eee d 1 Qu 8a’ Ay? & TMT Gat Oe se + &e fd ioe a pk a8) eh amen eee: 2s TRIGONOMETRICAL PROBLEMS. YE 2 2 oe Si Ra 4 Bar & or v 5 = 3h Z Wet ree ara oe es Cc. ae ee tO RBG. 7 q 3a 6a" e OC aera ere eae OIC, Tee Dar ret car mas Li laa Ps Ps ” Fe 3x 0 a 184° Pe OY sD. = _ f = os = = Cc, Bee a Oar a eg) Ge anal ai tard 1 8& 6 ax & or p,= —— —- ——— + =i Sc POS ae Beep.) OG TRS Ps SUP, 3a 6.2" =o 2 . Ok = —2?n,.= ae + &e Ps noe ML cle Gan 8 6. aT e z 1 A @ OF ,= + &c Eaten BE Ou Tae se. CORB Le gle Kap P: ce wv «v 2u 2H 8x 3a “. log,(1 + #) =— — — — — — — +e. 1+2+34+2+5+ 24+ 7 For a complete investigation of the law of these two series, see the Cambridge Mathematical Journal, Vol. 1v. p. 237. (y) When 2 is even, n*.(n? — 2°) , | . ¢ ( J fund eee iyear— tained cos 29 = 1 — —— sin’ + Le | Gs ee (Hymers’ Triconometry, Art. 134). Let cos”n@=0, then the values @ are vin . a nm —-1)7 172 TRIGONOMETRICAL PROBLEMS. and all the values of sin 8 will be By .O beg Ns ag b (n _ 1) ~=sin—, ~+siIn—.,,. + sin : Qn Q2n Qn and the values of cosec’ @ will be : 38a (n-1)7 cosec? —-, cosec? —,.., cosec? —————., n but the equation becomes = -- dass fd (cosec’ 0)’ — aie (cosec* 0)” eet (cosec? 6)? —&c.4+(—1)* 2"-' = Now the coefficient of the second term = — (the sum of the roots), Rass 937 (1 —1)7 n .“. cosec* — + cosec’ — +... + cosec? —————- = — . Qn Qn Qn 2 K 2 cos’ @ -— 1 ; (Oo) cos 26 cosec? 20 = — ; — = dS. cosec” 8 — cosec’ 20, 4sin® @ cos? @ 2 cos 40 cosec’ 40 = cosec” 20 — 2 cosec’ 40, 4 cos 8@ cosec” 80 N 2 cosec? 49 — 4 cosec® 80, &c. = &c. 2"-! cos 2” 8 cosec® 2” 8 = 2”-*cosec” 2”—! 8 — 2"-! cosec’ 2” 8, . by addition, 4S = 3 cosec’ @ — 2"~' cosec® 2"8. 9. Let AP, BO (fig. 57) meet in Q: draw PM, PN perpendicular to AO, BO respectively ; then ZAPB= ZAQB- 2PBQ; 2ZAOB=2AQB- ZOAP; TRIGONOMETRICAL PROBLEMS, f 173A PN / PM . £AOB- 2APB= P- 2 OAP= —_¢ = : é Z ZQB 4 PB) AP nearly, hsin(p+C) . . , sin(C +A) % = e —-hsing. Tey \ | \ wp h ; : : i Sapte noo Poin 4 sin p cos 4) A sinC.sin (4 - @)_ ha sin A and this reduced to seconds will be h sin C.sin(4-@) es wade sin C. sin (A-¢) b sin A b sin A 206265 minutes. ‘a ] 1 1 10. > (=) 2 ka art (- Se - | 2 ; (a+b+ $d Mle ar eee a Fs n * 9 T * 9 37 * 9 (2 age 1) i? ip = — + cosec® — +... + cosec — Nn An An An iE 1 T 3a (2n—1)7 | (2n)* = | — + — {cos — +cos — +... + cos ———_—_ sere ett Jae 2 2n 2n 2n 2 See Ex. 8, n : : ee ete (Te 1). When n = 3, a=sin’15, b=sin’45, c=sin*75=cos"15; a sin?15 sin* 15 sin? 45 sin’ 45 cos*15 ~=— cos" 15 =—5—+—- +5 + +35 +55 b sin? 45 cos?15 sin?15 cos?15 = sin?15 ~~ sin?45 sin? 15 + cos*15 (cot 15 +t 15)e2 9 4 sin? 45 sin? 45 sin? 15 cos*15 = $2 (cosec 30)}? + 24 = 3 (3-1). sin® 30 174 TRIGONOMETRICAL PROBLEMS. 11. 2 cosd ¢ n ra] n—2 1 3°72 = 8 0 n—4 = |2cos—|] —m™{2 cos — + - = 2 69s — — &e. n n Pee n (See Hymers’ TriconomeTry, Art. 135). n(n — 3) then 2 cos 0 = (2a@)"—n. (2a)""* + — ie (2”)"-*— &e., OF an, Oe Dr 8 — RC. et ee Oe ee eg where p,, p,, &e. are all independent of @ excepting p,, and 1 Pra am a, 5 6086. where a is constant; .. p, is of the form a + 3 cos@ where a and (3 are constant. Now equation (1) has m roots, and since cos 0 = cos (247 + 0) =... = cos $2 (m - 1) 7 +0} the 2 roots of the equation are 247 +0 2(n-1)7+86 ae ee COS ° cos —, COS n n n If S;,, 8S, be the sums of the m™ and p™ powers of the roots of equation (1), = (cos . (cos ey = SnSp — Snap VW Vt (Hymers’ Tueory or Eauarions, Arts. 161,151), and §,=—p,; also S, can be obtained in terms of p,, p.; Ss in terms of p,, Pe» P33... and §,_,1n terms of p,, p....p,_13 hence §,, Sp... 4 will be independent of 0. TRIGONOMETRICAL PROBLEMS. 175 Again, S,=—(p,8,,-1+poS,-o++--+ pp) Si +2 p,) of which every term is constant except np,; .. S, is of the form A, + B, cos@. Also, Sin oe MSm-1 + ove =e DaSin—m — O 3 oe Stak 5 Py eo Sz,-1 can be expressed in the form a + (6 cos@; and Sn + P,Son-1t+ .-. + p,S, = 03 hence JS, involves p,S,, and therefore is of the form a + $3 cos@ + +y cos’@, or it may be expressed in the form a + B cos + ¥ cos 20. Similarly, 3, involves cos@, cos26, cos30, and by fol- lowing the same reasoning it will appear that S,, = Ay + A; cos 0 + A, cos 20 +... + A, cos gO, . mM where g is the greatest integer contained in —; n and S;, = B, + B, cos 0 + B, cos 20 +... + B, cos 80, where s is the greatest integer contained in ae n Sn+p = Co + Ci cosO + ... + C, cos rd, mM + p bd) where r is the greatest integer contained in and S'S, = Dy) + D, cos@ +... + Dy, cos (¢ + 8) 0. m—m Di pe where m’ and p’ are less than n; m+p—(m+p Q+s)=e——* ‘ Ly and is therefore not greater than 7; 176 TRIGONOMETRICAL PROBLEMS. hence ; ep m S, = £, + £, cos6 + ... + KE, cos 78, where E,, £,... HE, are independent of 0. (12. Let C (fig. 58) be the centre of the circles"A the point of intersection of the two rectangular tangents AB, AD; AP p, one of the cutting lines: so that z BAP, = = = 0; n draw CO, perpendicular to 4P,p,: then since the tangents at P, and p, are perpendicular to the radii CP,, Cp, respectively ; the product of the perpendiculars from 4 upon the tangents at P, and p, = (AP, cos CP,p,) (Ap, cos Cp, P,) = AP,. Ap, .cos*.CP,p, = ARB’ cos’ CP, p, = cos CP,p, = POP =? —COr; now CA 277%, i and z CAO, = 45-86; “. CO, = 7 4/2.sin (45 — 0) = r (cos @ — sin 8) ; hence CO,’ = r* (1 — sin20), and 7* — CO,’ = r* sin 20. Similarly, r° — CO,’ = 1° sin 40, &c. and the product of all the perpendiculars =r sin20.7° sin 40...... r* sin $2 (2n — 1) O} : ee . (2n -1 = (°)2"-1 . sin —. sin ree ae Ge jar 2n 2n 2n Now, 5 : ; j + . 2art+ . (2n-1 sin @ = pbo-1 ey ee EEN oe PO Se ae Qn Q2n 2n Qn therefore if gp = 0; Meche 27) «2A I1) ae 1 sin ® 2n n Se esi. oI Peewee TC 7. egy EMP | 2n 2n — 2n pa Nia) oD g2n-1 — g2n—2 sin —— 2n il ~_ ~ = | Bret) Lawl fet Sees ee rN) hence the product of all the perpendiculars TRIGONOMETRICAL PROBLEMS. 177 13. Let O (fig. 59) be the centre of the circle whose radius = 7: join O4A,, O4,,......OA m > then if PO =a: andr’? +a’ =’, z POA, = 0, 27 Z POA, =— + 0, &c.; m . PA’ = 1° + a” — 2ar cos@ = Bb? — 2arcos@; : 2 PA; = 6* — 2ar cos (= +6) ; m g —] PA’, = b’-—2ar con |=) 44 Mm Hence if S,, S,, $3, &c. represent the sum, the sum of the squares, cubes, &c. of the m quantities 2 4 2 — 1 cos 0, cos (= +9), cos (= +9), Oat ee cos Ue ) Cae oe m m m we have (PAY °OSCP ALY" +. ee. +(P4,,)°" | n—1 ; | = mb" —n.(2ar).b°"?S, +n. (2aryb"-* Si — &e. Zz este os + (- 1)" Gar)’ S,: Bnaeour 14x. XI. when vis <3 S75. Ss:..-..8, are in- dependent of (0); .. =(P4,)*" is independent of @ when is < ™. EE sin? 6 sin? @ ( + sin :) y 2 -. NC ae fe t(a}aonan.¢ = ToRPaT 1—sin?@ 1+sina 1+ sina sin’ 9 (: sin’ @ + sin “) be sin 6. ha 12 178 TRIGONOMETRICAL PROBLEMS. Pe sa : sin? @ sin’ @ + sina tan’@ = a S62 ee 1+sina 1+ sina I sin’ @ sin’ @ + sina ( sin’ 0 sin’ @ + sin 3 ; “6) SCOR ETI a are tienen | eetecamanbel iby fo ae ib OE ayy 1+sina 1+sina \1+sin8 1+sinf sin® @ sin’'@ sin? 9 + sina ee eh eee he ie an ee 1+sina 1+sina 1+sinf sin’ @ + sina sin’ @ + sin 3 ? . ——__~—. tan’ 0 1+sina 1+ sin 3 sin’ @ sin? @ = sin? @ + sina ; aN a tee : 1+sina 1+sina 14+snf sin?@ sin? 9 + sina sin’ @ + sin. 3 1+sina 1+4+snf 1 + siny wy ons -ETIR- NE Gl taking the differential coefficient of the logarithm of this equation, 8a re 8.v mw —(2u)? (3x)? — (2) : a ee @\? wv? T Qa 3a .. taking the differential coefficient of the logarithm, tan v = + &¢.; 1 Qu 2H cotw~=- - &c , 1 ] "la — Ga)” Gn) — way * *} tanw cot tan’s Sa» «2a A TRIGONOMETRICAL PROBLEMS. 179 oe OE oe ar +0 . (m-1)7+0 (y) Sin @ = 2”7! sin—.sin mets ) set m m m and taking the differential coefficient of the logarithm, verano a+ (m-1)7r+0 cot @ = —. | cot — + cot—— + asda + cot : We) Noe 21 m 7 + let ty be substituted for 6; *, m cot junds . + 3 + 2 — 1 se RU i ata a NO A “an geared DS: 2m 2m Mr mt next, let oe) + Le be substituted for 6; m Oe —— ih ae + 2 — | = tan [Oe Bones mee noe Pipa Binns o 2m Q2m ‘ m tan = P ak, hence — = = ta NASEIN SS eat A 1 T co) Aw m cot ——= 15. Let A’ (fig. 60) be the eye of the observer 54 feet above A: O the earth’s centre, 2 A’OB = ¢, 0 the horizontal refraction, 7 the earth’s radius in feet, w the height of B in feet: then since B is raised by refraction through an z (6), the true depression of B from A’ = 49'.14” + @; ZOA'B = 90 — (49'. 14” + 6), A'O = 1919 + 5447 = 19248 + 74, and OBe=r+a, 12—2 180 TRIGONOMETRICAL PROBLEMS. i 19245 +7 sin {p + 90 — (49'. 14" + 8)} cos }p — (49". 14” + Oy} ‘ bos ore cos (49'. 14” + 0) CRD get mien zi .. since 7 is large when compared with a, 19241 — x . Ras + ear ay -1-£ 4g (49.14 + 0), 3849 Fars: 2H ” ° or —-—-_ = 2. (49.14 + 8) — pe r The second observation is taken from B’ 54 feet above B: and since 4 is raised through an Z (0) by refraction, the true elevation of 4 = 31'.31" —@; .«. 2 OB’A =90 + 81'.31" — 94; OA=1919+7r; OB =r+a+ 54; r+ev+5h — sin(p + 90+ 31’. 31” — 0) "1016 4-72) asini(O0RR 6.317670) cos (p + 31’. 31” — 6) 5 Aces Gl iD ae a ae (31'.31” — 6), a 19133 Pp ‘ " Le i eo Lol eee or ls a ; ~ ( 3827 — 2x 4 *. ——_——— = 26(31.31 — 2, i p ( d+ dv are ot) (a0 ae gas a x (49'. 14” + 0) — 6 6 3 42 Q aid wo = 2. (80.45") = 2h x circular measure of 4845 seconds; _ | 9690 121028 x 9690 ”. 7676 — 44 = 1 LELOZE 9090 es re a Hye comad)) 206265 hence 4v = 1990, and w = 4071 feet nearly. ) 2. TRIGONOMETRICAL PROBLEMS. 181 In like manner it may be proved that, if the depression of B and the elevation of A be a, 8 seconds respectively, and d be the distance between the two stations, the difference of the ae : nearly; which is independent both of the height of the instrument and of the horizontal refraction. altitudes of 4 and B= 16. See Simms’ Treatise on the Mathematical Instru- ments employed in Surveying, pp. 3, 59. 109 and 116. 17. Draw Aa, Bb vertical (fig. 61); then ZACa=}, BCbh=p; aCh=a; AC=AB=d; and if CB =a, we have | Ca=dcosrA, Ch=waxcosp, Aa=dsind\, Bbh=asinun; ll or (ab) = d’ cos’ + x cos* pn — 2da@ cos % COs X Cosa; and AB’ = (ab)* + (Bb - Aa)’, “. (ab)? + (@sin » — dsin))? = d?; hence d* + v* — 2dx(cosX cosu cosa + sind sing) = d?; a Pan and w=2d {cos 4 cosaA (cos? S — sin* =) 2 2 PON re AON | + sin au sin A {| cos* — + sin* — 2 2} f oA Hg =2d cos (x — A) cos” = — cos(u + A) sin ale 18. Let AB (fig. 62) be the diameter of the given semi- circle whose radius is R, and centre 0; C the centre of the semicircle whose radius is r,; P,_,, P, the centres of two consecutive circles touching each other and the two circles whose radii are 7, and R; Pu-1> p» their respective radii; draw P,_,M,_,, P,,M, perpendicular to AB; and let MM = Oy tpg OU Gan ee AM, = Uns PM, = Yn 182 TRIGONOMETRICAL PROBLEMS. then (OP 754 = (v,~4 = ry + pies — (7; + paal) 5 (OP,,- 1)” = (v,,_1 as R)* 7s Tye = (R Ti Puli) 3 ee by subtraction, 2(k = 11) By 4 a 2H + 1) Pu 3 Pl Sabie i fats ak oe Ty) Pr-1 R cee Tr) or and is constant for all the circles inscribed between the two given circles ; ale eo, NONE Cake Ce Pr -1 Pr Also UP a). .r, (@,_) “ia @,)° ae (Yu-1 s y.)° a (fei a Pde a et) ee = (p,-1 + 11)" (x, _ r,)° 4 Yy", Pa n Vy, \ oy eid ee \ Pa Vn R + Fy Pn = ——- > 2 3 and —_ = ; pees, 1} = No5 An? + (=) 1 Pn hes ry Pn 2R A one tT, Rr, RES Serre ee See, oper os : , KT. (tls + 7) ANS + 4:— ae Rr 1 1 I MCAS e he ate ah Ke 0+ &c. = ; Tes pte rt be gtk Pe ts A a he, SER ae oC) earae: Ue) Now sin@ = 0 f _ (=) | f - (--) | Lae. 7, , 1 2% I 26 20 &e AR t = RS B >> De pM es (2h 0 184 TRIGONOMETRICAL PROBLEMS. 1 cot @ i i 1 we r fetes | aaa ae . 5 26? 20 0 6’? —-@ (x)? - O oH ee -+- —— — = Sr CT ee 57 ony . 20%, 20,/-1(e9V=1 — e- 8-1) 0? (20)-0 Let @ = — n’s’; 1 1 ems as e-m J E se er ONE Are moral ae aR Ae Ga Sy r(1 oe s) ar” (2? as s”) Qa s(e7s cs ego) 2 a” 5” 1 1 1 or ; rae tat, ee 1° +58 2 4+s° 3+ aw(e?™> + 1) 1 ws — 1 1 = = 4 — BSA le mes 25° s(e?7> — 1) Rr, {ws -1 T and (26, => 4 i Ts 2 s° s(e?™>— 1) ST JOHN’S COLLEGE. Dec. 1844. (No. XVI.) 1. Iran angle receive a given small increment, find the corresponding increment of the tangent of the angle. Shew in what case it is disadvantageous to determine an angle im- mediately from the given value of its tangent, and investigate a formula for finding the angle in that case. 2. Having given sin(#+y)=tan(«—y)=4(tanw-—tany), find the values of w and y. 3. Shew that if 4+ B= 90°, sin(4 — B) = —cos2 A, sin2.4 + sin2B =2cos (A — B), and that if 4+ B+ C = 90°, cos 2A + cos2B+cos2C=1+44sin Asin B sin C. TRIGONOMETRICAL PROBLEMS. 185 4. ‘The hypothenuse (c¢) of a right-angled triangle 4BC is trisected in the points D, &: prove that if CD, CE be joined, the sum of the squares of the sides of the triangle 2¢° CDE = —. « 5. Find the values of # in the foliowing equations : Sec v = 2sin’2@ + cosv; sin 2v = sin’3a. 6. If R, r be the radii of the circles described about and in the triangle ABC, the area of the triangle = Rr (sin A + sin B + sin C). 7. The side BC of the triangle ABC is bisected by the straight line 4D; prove that cot BAD — cot CAD = cot B —- cot C. 8. On the sides of an equilateral triangle three squares are described. Compare the area of the triangle formed by joining the centres of these squares with the area of the equi- lateral triangle. 9. If «w, y, x be the lengths of the straight lines bisecting the angles of a triangle, and terminated by the opposite sides a, b, c; prove that 2 2 3? Ais 2 ov Aas ba : (b +c) 5 + e+) Pam Ulas ) 7 (a+b+c)*, 10. Shew how to express (3 in a series proceeding accord- ing to sines of multiples of a, when sin 3 = m sin (a + (3). If 2 cos a, = «7, + —3; 2coSa@,= 7, + —3...... wv; Vy 1 260s a, = #, + anda; + apt... + 0, = 27 5 On there, a5:.55..0, = 01. 186 TRIGONOMETRICAL PROBLEMS. 11. The sides AB, BC, CD, DA of a quadrilateral figure inscribed in a circle are in a geometrical progression, of which the common ratio is 7; prove that Soe = aso tn BCD 7 +1 12. Eliminate @ from the equations a sin@ + bcos @ =e, a b “ft = dd, sin@ cos@ 13. Shew that if @ be small, 0 = 4 $2 (sin 26 — sin@) + (tan20-— tan @){ nearly, and prove that Tv T j T tan (2"-' + tan —— + 2tan — +...4+ 2"-* tan —] = 1. \ grt) gn g3 gut 14. If the tangents of the angles of a triangle be as the numbers 1, 2, 3, find them; and shew that if a, b, ¢ be the sides of the triangle, and p,, p., p; the perpendiculars on the sides drawn from the opposite angles, then 5 p, p. p, = 3abe. 15. On the sides of a triangle ABC, three circles are described intersecting in P, Q, #; determine the sides of the triangle formed by joining P, Q, #; and shew that if P, p be the perimeters of the triangles ABC, PQR, CA GB outa p = 4P sin — sin — sin—. 2 2 2 16. If C be the middle point of a semicircular are whose diameter is 4B, and the angle ACB be divided into nm + 1 equal angles by lines terminated by the diameter, prove that the sum of the squares of the reciprocals of the dividing ht 1 lines = — 4m + cot —— 29° ——.}, where r is the radius of the 2(n + 1) circle. TRIGONOMETRICAL PROBLEMS, 187 SOLUTIONS TO (No. XVI.) 1. (a) See Hymers’ Triconomerry, Art. 79. (3) Wuen @ is nearly 90°, the change in the tangent is not proportional to the change in the value of 6; but tan @ — 1 tan@ +1 known; and since 9 — 45° is nearly = 45°, its value may be found correctly from the given value of the tangent; and thus @ may be determined. tan (0 — 45°) = , which is known when tan @ is tan w — tan y tan w — tan y 2. OE AE SE MA at , which equation is sa- 1 + tan # tany 2 tisfied by making tan v—tany=0, (a) or 1+tana@ tany=2, (3). Using equation (a), w=m7+y; and sin(v+y)=tan (w—-y)=0; TW T ._@+y=nn, &-Y=mMn, or Sal) a> y=(n-m) —, 2 where 2 and m are any integers, positive or negative. Next, using equation (9); tanatany=1; .°. tan v=coty, ~ 7 : . us x cad hence oy RED * Tv tan (v— y)=(- 1)", al myth Cote ,e+y=mr + ma T . 2a = (mtn) a+ f24(-1)"} A nS sell U8 2y=(m—n) w+ }2—-(-1)"} -; T or v7 = (m+ n) ~ as fa bgt 1)"} cae y) ~—m TT : y= (m —- n) ea + SQ + (— iyi 7 where m and 7 are any integers. 188 TRIGONOMETRICAL PROBLEMS. 3. (a) Sin (4—- B) =sin } 4 — (90°- A)} = sin (24 — 90°) = — cos2J. (8) Sin24+sin2B = 2sin (A + B) cos (A — B) = 2cos (A — B), since sin (4 + B) = 1. (y) 4sin/4 sin B sinC=2sinC {cos(4—- B) -—cos(4+ B)} = sin(B+C-—A)+sin(4+C-B)+sin(4+ B-C)-sin(4+B+C) = cos24 + cos2B + cos2C —-1; * cos 2A +cos2B+cos2?C=1+4snAsinB sinC. 4, CD? (fig. 63) = AC* + AD’ -2AC. AD cos A be b Poe = 6°?4+—-—- —-2—.- =—+-5 9 Hah ss Mgt : ; r iE Done CE” sues ean ee adaamenawtel a A DE’ e 9 oe, Gita CC Ce CD? + CE? + DE? = —— +— =—. 8 3 8 Lae sin? x ; 5. (a) Sec # — cose = 2 sin’; = 2sin’ av; COS &@ hence sinw = 0, or cosw = f. When sin # = 0, «© =m, where m is any integer. WT vin When cosw=4, v= 2ma7+—=(6m+£1)—, where m 3 3 is any integer. > ~ a ae e os (6) Sin2w# = sin’3a; 1 — sin2a@ = cos’3a, and by extracting the square root, cos 7 — sina = + cos 3a. TRIGONOMETRICAL PROBLEMS. 189 Using the upper sign, sina = cos a — cos 3a = 2sinw sin2e; “. sing = 0, or singe = 4, When sin vw = 0, © = mr. ; T T When sin2v= 4, 2t= 2mm +; or a eTeN ae TT T or 2@ = (4m +1)— +-; 2 3 Tv “yy Pe 4m + 1 — + — ° ( ) A eh The equation cos # — sin w = — cos 3w will determine the remaining values of #; but these can only be found by the solution of a cubic equation. b 6. 9 fe cote gE sin 4 sinB sinC and the area of the triangle Meet O 0) z r(R# sin 4+ R sin B+ # sin C) = fr (sin A + sin B + sin C). i Let 2. BAD (fig. 64) = 0, 2 CAD=¢@; sn0? BD, CD \ sing fhente et te ee Sin eee er A Le sin. © or ic: “eawat and sin (0 + @) = sin A =sin(B+C); sin?@ sin’ B d sin*@ — sin?@ — sin? B — sin’ C uti ae ae oY an TaD = cee sin’ @ sin? C’ sin* @ sin? C d sin’@ —sin’d sin’ — sin’ C sin 9 sin ~ sin B sinC ” or 190 TRIGONOMETRICAL PROBLEMS. _ sin(@— dp) sin(O+@) _ sin(B-C) sin (B+ C)- sin @ sin ~ “s sin B sin C ; sin(O-) _ sin(B—C) sin@ sin @ ~ sin BsinG hence , or cot @—cot d = cot B— cot C. 8. Let c, b (fig. 65) be the centres of the squares de- scribed on AB, AC, respectively ; join Ac, Ab, bc; then be is a side of the new triangle which is evidently equilateral ; ~ a and Ac = yi. ZbAc = 90° + 60° = 150°; ib i 0. . side be = 2Ac. sin 75° = a 4/2 cos 15°; Aabe (side be)’ Been ee ee de = 2cos° 15 = 1+ cos 30 =1 4 A ABC a” and Ayi8 ee 9. Let Aa, Bb, Ce (fig. 35) be the three straight lines bisecting the angles 4, B, C of the triangle ABC, then da=a, and ACda+ ABAa= AABC, =ibe sin A ; ro | & A or 1 ba2 sin — +1 2 sin 9 t x o fo) ACU ‘ A . (64 c) « = 2be COa te de A and (b + ¢)? — = 4be cos? — = 48 (8 — a); be 2 i 48'S’ — d), 2 similarly, (@ + ¢)” a ae ‘ oe wo and (a + ier 48 GS’ — c), hence by addition, 2 D) x” 2 y” 5 & 2 9 (b +c)? — + (a +c) — + (a4 6)? — = 48" = (a +b +c)’. be ac ab TRIGONOMETRICAL PROBLEMS. 191 10. (a) «eb Y=T— @-BV=T = m feloH8) VA — e-(eBV=Y . (1 — me*V-1) e28V-1 = 1 — me-@8F1, or 284/—1= log, (1 — me-*¥-1) — log, (1 -— me*Y-1) =m (e* rt Pe a e-#N=1) te 4m’ (e2@ Neher Enten =r) +4m (e84v-1 — gute Vath, &c. hence 6 =m sina + 4 m sin2a+ é m’ sin 8a + &e. (3) wy=cosa;+1/—18iNa,, &%=Ccosar+\/ —1 SiN de,... v, = COS a, + eel sin a, 3 ", @)@y...U, = (cosa, + \/— 1 sina,) (cos a. + \/ — 1 sin Gyre s (cosa, + \/—1 sina,) = cos (a, + a +... + a,) + —1sin (a, + a+... + a,), (See Hymers’ Trigonometry, Art. 130), = cos2a +\/—1 sin2x =1. This result is only true when all the values of #,, v...., are supposed to be of the same form as Vm = COS An + A/e 1 sin. a, or all of the same form as Cm = COS Ay, — Vee 1 sin Amn: If some be taken of the first form, and others of the secend form, the equation will be no longer true, as may be easily seen from a simple case, where there are only two quantities Q,5 Ae- Let a= cosa, +4/—1 sing, @ = COS dy — \/ —1 sin a, = Cos (— as) + 1 — 1 sin (— a); 192 TRIGONOMETRICAL PROBLEMS. v, Vz = COS. (a, a te) + VAR 1 sin (a, _ U2)5 which is not = 1 unless a, — a, 0 or a multiple of 27. SH A Let a, b, c, d be the sides 4B, BC, CD, DA (fig 66) of the quadrilateral figure taken in order ; £ABC=0, 4 BCD=9; then the area of the figure = AABC + AACD = 4 (ab + ed) sinO; similarly, the area of the figure = 4 (ad + be) sing; . (ab + cd) sin @ = (ad + be) sin @. Also, a’ + b?—2ab cos@ = AC’ = cc’ + d’ + 2cd cos 8, or — 2 (ab + cd) cos 8 = c* + d’ — (a’ + b”), and — 2 (ad + be) cos@ = a’? + d’ — (6° +c’); v. fe+ ad — (a? + b’)} tan @ = a’ + d’ — (b+ c*)} tang, tan0d d’—b’-(c’- a’) r = : tan ® (ce -@)(r-1) r-1 ae re ne aN, ey Cae 1) a a | if the sides be in geometric progression. 12. asin0d+bcos@=c, acos@ + bsin 9 =d sin @ cos 9 By multiplying these two equations together, (a? + b*) sin @ cos 0 + ab = ed sin @ cos 8, or Sed — (a? + 6’)} sin 20 = 2ab. Also, by adding the squares of the two first equations d? sin a+6?+2ab sn20@=c’+— 720 e 9 193 TRIGONOMETRICAL PROBLEMS. *) = sin 20 (d’ sin2@ — 8ab), 2ab 2abd* Ca i, cd — (a + b*) | — (a? + b’) : sae gs 44 (a? + b*) + d?- 4cd} hence a’? +b’— c?= (ab)’ G+ ed) AL ike es 6 Q 16 0° Td) (2) tan = @ + — + ——__— + &e.; Swe Ione ate Aa 5 *, 2 (sin 26 — sin) + tan 20 — tan@ 3 5 3 i =2}(20- = ae Cie a i ir pes > 16 (26)? 6° 66° (26)' LOMA EI ah gg tts, 5 OU a aa Sane 2 8 cde & +26 + —— 3 20 Ws An 8 18.2?>-—1 ‘Es = 304+" = 364 = = 38 nearly, when @ is small. (3) tan @ = cot 6 — 2 cot 280, 2 tan 20 = 2 cot 20 — 2? cot 2760, o8—* tan 2°7" @' = 2?-) cot 2"-' 6 —. 2" cot 2° ¢: * tan? + 2 tan 20+... + 2""' tan 2"-'@ = cot 0 — 2" cot 2”0. v Let : Sere = Tv T oo 2" tan — + 2°7) tan — 2 9? T gnt+l Tv cot — = cot 2 grt) eee —_— n—2 > | hence tan are (tan — gat + 2tan— ~ +... + 2"~* tan a 4 Qn ) 1. 13 194 TRIGONOMETRICAL PROBLEMS. 14. TanB=2tanA, tanC = 3tan JA, and tan A + tan B + tanC = tan A tan B tan C, since tan(4 + B+C)=0; “. 6tan 4 =6tan?A, or tand=1, tanB=2, tanC=3; also a=p,(cot B + cotC) = 2p,, b = p,(cot A + cot C) = 4p,, c = p;(cot A + cot B) = 2p,, hence abe = 3p, pop. 15. Let ARPC (fig. 67) be the semicircle described on AC; then since 4 ARC isa right angle, 2 BRC is a right angle, or the semicircle described on BC will pass through R,; and the intersection R of the two semicircles lies in 4B; similarly, the points P, Q are in the sides BC, AC respec- tively. Also the angle subtended by PA at the centre of the circle ARPC =22PAR, and the radius of the circle b =~ Ci PR = 2- sin PAR = b cos B, since Z APB in a semicircle is a right angle. Similarly, PQ =ccosC, and QR = acos 4. Hence p=acos dA + bcos B + ccosC Psin A cos A + Psin B cos B + P sin C cos C a sin 4 + sin B + sin C P sm24+sin2@B+sin2C P 4sinJ4 sin B sinC Te Wain Asie mein Coa a Ope 4 cos — Sis oe C = 4Psin — sin— sin —. 23 2 Q TRIGONOMETRICAL PROBLEMS. 195 16. Let CP,, CP,... CP, (fig. 68) be the dividing lines, Opes Ga 7 OP. = 20, Ke. then AC =14/2, £CAB= Se P Tis sin — OW crue ey aS = aes ee is cos @ + sin @’ sin (= 436 1 1 ge aE PP 4): GPs saul ) Similan ee se eta Cr sin <0) Imiarly, ——- = — (1 + sin 5 ” CP, 29 : 1 in2nQ@Q); ae ae sin27@); CE ap | f ) 1 1 1 > OP OP tf Cp. 1 : ° : Let S = sin 20 + sin4@ +... + sin2n@; *. 2S'cos 20 = sin4@ + sin60 + ... + sin(2” + 2)0 + sin20 + sin4@ +... + sin(2n — 2)0 = S — sin20 + sin(2n + 2)04+ S — sin2né@; . 2,1 — cos 26) = sin 20 — $sin(2n + 2)0 — sin2n@} 2 sin 0 }cos@ — cos(2n + 1)0}, 13—2 196 TRIGONOMETRICAL PROBLEMS. gun agrees Oars eoe (eit ott) CAF COR 2 Ere CO aAaaie eee een 2sin @ 2 sin @ since (2n+1)0=7—-80; 1 1 7 hence > = — (n + cot 0) = — jn + cot ———_—__>. 2" 29" 2(n + 1) ST JOHN’S COLLEGE. Dec. 1845. (No. XVII.) 1. DEFINE the cosine of an angle, and trace the variations of its sign and magnitude through the four quadrants: find ee 2 the values of cos7! , m being an integer. 2. Prove the formule: 1— 2cos2a a tan(a + 30°). tan(a — 30°) = ——_——__—__.. (2) (a ) (a ) 1 + 2cos2a (b) cot = — tan = = 2(tana + 2 cot 2a). a oy cot — + cot — i 2. 2 sn hoes * c ee Ww Qa = 7. (c) iC) Y sin qa’ Y es cot — + cot — 2 2 3. Given two sides a, 6, of a triangle and the included angle C, which is very obtuse, find the value of the third side in a form adapted to logarithms. 4. Determine v and y from the equations 1 1 tan“! a — tan-*y = tan“! ek (hus ee y 2H 12 TRIGONOMETRICAL PROBLEMS, 197 and prove geometrically that 2 tan A tan 24 = ———__—__.. W— tan: A acos(@ + a) + bsin@ a'sin(@ + a) + b’cos@ all values of 6, then aa’ — bb'= (ab — b'a) sina: prove this, and shew that if the sides of a triangle be in harmonical pro- gression, 5. If the fraction be the same for sin 4 sinC cos A + cos C" 6. Two circles have a common radius (r) and a circle is described touching this radius and the two circles: prove that : : ; 3r the radius of the circle which touches the three = Ease 7. At noon a column in the E.S.E. cast upon the ground a shadow, the extremity of which was in the direction N.E. : the angle of elevation of the column being a°, and the distance of the extremity of the shadow from the column (a) feet, deter- mine the height of the column. 8. Squares are described upon the sides of a triangle ABC, and the centres of the squares are joined forming the area LM N area ABC the sum of the cotangents of angles of the triangle ABC. triangle LMN: prove that the 1+ : » p being 9. From the four angular points of a quadrilateral figure, which may be inscribed in a circle, straight lines are drawn to the diagonals, making equal angles with the diagonals: shew that the product of two of these lines taken alternately is equal to the product of the remaining two. 10. Eliminate @ from the equations a tan? + bsecO0 =e, acot@ + bcos@ = d. 198 TRIGONOMETRICAL PROBLEMS. 11. Find the sum of » terms of the series a a tan sa+ tans = ee ta —— + &c.; 1+1.2a° t Goo ote . and prove that Q cot — i j lo 2 cos @ , cos’@ ~=cos* @ an é « =a + C. 8 sin @. 1 7) 3 12. Solve the equation 8#* — 364° + 42% -—-13=0 by Trigonometry, and resolve sin v + cos # into quadratic factors. 13. Circles are inscribed in triangles whose bases are the sides of a regular polygon of m sides and whose vertices lie in one of the angular points: shew that the sum of the areas of the circles the radius of the circle circumscribing the polygon being considered unity. SOLUTIONS TO (No. XVII.) 1. (a) See Hymers’ Trigonometry. Arts. 12, 26. (3) Cosma +\/—1sinma = (cos + \/ — 1 sin Ww)" or cosma = (— 1)”, and cos (m2+0) = cos im cos@ —sin m7 sin 8 = (— 1)" cos@ ; “. cos~* $(— 1)” cos 6} = mm + 0. — TRIGONOMETRICAL PROBLEMS. Let cos @ = 435 38 = 20m & , 3 ae 1 me and yi rece = (Qr+m)7 cP 2. (a) Tan (a + 30) tan (a — 30) oe sin (a + 30) sin(a — 30) cos60 — cos 2a a (b) cot = — tan = = 2cota = 2(tana + 2 a+ y, a (ec) cot— + cot 2 2 a = cot-— + tan 2 a a pe REET, COS — COS + sin — sin 9 9 199 1, 1 — 2c0os 2a I; + 2 cos2a_ cos (a + 30) cos(a — 30) cos 60 + cos 2a .s cot 2a). cos — “ 2 2 2 evant oH 3 ° os [3 sin -- COs sin —.cos 2 2 “ cos — Similarly, cot — + cot+ = - Ys 9 i 3 a+ B 3 sin — cos 9 cot — + cot (9) 3. aG 446 cos? — ~ let sin’? Q ae EP ear array i (a + by’ 9 » 9 2 ‘ C c= a’ +h? -2abcosC = (a + 6) — 4ab cos’ et 4 200 TRIGONOMETRICAL PROBLEMS. C ; then since C' is very obtuse, cos = may be determined cor- rectly, and sin@ is small; so that @ may be determined correctly from the tables, and ¢ = (a + 6) cos@ is known. 1 1 i Ler ay a 1 1l4+4ay Avy hence hence w I] 6-3/3 41/71 — 3864/3 4 9 84/8 =-6 41/71 — 3864/3 4 (8B) Let ABC (fig 69) be a triangle having the angle C a right angle; make 4 ABD = 4 BAC; «. 2 BDC =22 BAC =24; and 4D? = DB’ = DC’? + BC, or 4C* -2A4C.DC+ DC’ =DC+ BC: TRIGONOMETRICAL PROBLEMS, 201 + 240 0DC = AC? — BC; 2BC ee i 2 BC ce A Cag, pat hy, ons 2 tA DC. AC - BC es 2° 1 — tan’ A AC a cos (0+ a) + bsin@ a sin (0 + a) + 6’ cos@ oy ea Ce) a cosacos@ + (b — asina) sind ~ (a sina + 0’) cos@ +a’ cosasin@’ and this will be the same for all values of 0, if acosa a sina+b eens PTT Te deed ee PE ee a | b—asina a cosa or aa’ cos’a = bb’ + (a’'b — Wa) sina — aa sin’ a; . aa — bb =(a@b — Wa) sina. (3) Since the sides are proportional to the sines of the opposite angles, 1 1 g sin A + sinC 1 ° ay 1 Coat erage gly KO San aerurruraaenamsoai ol) Game mamra-commenenanseuerie sind sinC snB sin A sin C poate B’ A+C AC: ‘sn Asin C 2 sin - cos ———- = ——_— Bed sO: sin — cos— 2 2 : B A-C sin A sin C or 2 cos — cos ——— = 2 2 Het deh GR cos cos — 2 ve 202 TRIGONOMETRICAL PROBLEMS. bees. sin A sin C sin A sin C’ . cos = = ——. = : 2 A-C A+C cosd+cosC’ 2 cos — COs 2 B i sin A sin C_ and cos — = aa ee 2 cos A + cos C 6. Let AB (fig. 70) be the common radius of the two circles BC, AC whose centres are A, B respectively ; O the centre of the circle which touches the two circles, and the common radius AB; join AO and produce it to meet the circle BPC in P: draw OM perpendicular to AB; then if OM=p; AO=r-p, and AM ~~; , ae 4" wf — p) =p +3. OF fF — 27 p= — a. 0 era) arte Peet Cham Again, let O' be the centre of the circle which touches the three circles, p’ its radius ; . 2 o| 8 then 40 =r-p', MO =2p+p, AM = 2 ‘ 7?" 2 | Dy) f 3) “ (- py =Cptp)? + Oh Tis 2r0 =4p" + 4p + 7? Qi 3) on AS. an tae Aah (2r + 4p) p = oe ate 7. Let A (fig. 71) be the place of the observer; B the foot of the column, AF the direction of the east, D the ex- tremity of the shadow; then 2 BAKE = 221°; 4 EAD = 45°, and BD is perpendicular to AX, because the sun is in the TRIGONOMETRICAL PROBLEMS. 203 south, and therefore D to the north of B. Also if A be the height of the column, . 45 AB =hcota = 2asin—, since 4 ABD = 674° = 2 BAD; fee hcota = a\/2- 4/2, orh=\/2-/2atana. 8. Let L, M, N (fig. 72) be the centres of the squares described on BC, AC, AB respectively ; join LC, CM, MA, AN, NB, BL; then hexagon LCMANB = A ABC+ ALBC + A AMC+ A ANB a+ bec = AAD CE yas also, hexagon ILCMANB = ALMN + AMAN+ANBL + ALCM, A LMN +4(MA.AN.sin MAN + NB.BL.sin NBL + LC.CM.sin LCM) li il b ALMN + 4( Os = COS c) Gee Cid yi Crue A B a 2 wise “OVERSEE MOVERVE A LMN +4 (be cos A + ac cos B + ab cosC); . ALMN Il : ~ +b +e A ABC-4(be cos A+ace cos B+abcosC) + ; ex (Lh); li ll A ABC — 4 (be cos A + ac cos B + ab cos C) + 4 (be cos d + ae cos B + ab cos C) A ABC + 4 (be cos A + ae cos B + ab cos C) li ‘be . se ab. = AABC + 3(> sind cot A+ — sin B cot B+ — sin C cot c) ‘ = A ABC }(1 + 4 (cot A + cot B + cot Cine (1 + 7) A ABC. 204 TRIGONOMETRICAL PROBLEMS. Since 2(be cos A + ac cos B -+ ab cos C) =(P4+C-@)4+ (C7 4+C-H)4+ (C+ RP -C)=04+0 +0, we have from equation (1) Oe Act eet ye ALMN = AABC - 7 Ceo ee Ot Eo ee eI 8 7 8 ; = AABC + and L, M, N are the middle points of the arcs of the semi- circles described on BC, AC, AB respectively. 9. Let da, Ce (fig. 66) be drawn, making a given angle a with the diagonal BD of the quadrilateral figure ABCD, and Bb, Dd, making the same angle a with the diagonal AC; AE.sin FE CE .sin E ————; Ce= then Aa = 5 9 . sin a sin a hee WEES Dio ee sin a sina Aa.Ce AE.CE ; [2B ED aeRE OD rae since the figure 4BCD can be inscribed in a circle ¢ Aare (3b saa: 10. Multiplying the two equations together, a’ +b’ + ab(sin @ + cos 0) = ed; *. ab (sin 8 + cos 0) = ed — (a’ + b*), and a sin@ + b =e cos 0, Cd =i “. {ab + bc) cos8 = cd — a*, or Gos = ——_— b (a +- Cc) TRIGONOMETRICAL PROBLEMS. 205 Cad —ao-aby— We e cos0 — b a ab(a+c) : and sin@ = a hence (ab)? (a +c) = (c'd — a®c — ab? — b*c)? + @ (cd - a)’. 11. (a) S = tan7'a + (tan 2a — tan“'a) +(tan-'3a—tan-'2a)+...4 }tan7'na—tan~!(m—1)a}=tan-'na, cou= cos — 2 2 rgd 6 2 sin? — cos — 2 2 1 = l _— lc — z _ saan os 27 log, (1 — cos 8) cos’?@ cos?@ + &e. = cos @ + 12. (a) Let w=y + 8, and the equation is reduced to 2y°—-38y-1=0. Again, let y =a cos@; a’? (3 cos@+cos 30) . : a® cos 30 o .p OG and if —- — 3a=0, we have 3 “. @=4/2, and cos3@ = i aot 29 hence 30 = 45, or 27 4453; and @=15 or ates *, the three values of y are /2 cos 15, 4/2 cos 135, and 4/2 cos 105, 206 TRIGONOMETRICAL PROBLEMS, or 4/2 cos (45 — 30), — 4/2 sin 45, and — 4/2 sin (45 — 30), which by substitution become A/ 3 +1 et a/ 8 ee : TEL eS er | ec, as respectively ; and w#=y + 3; *. the three values of #& are é , 4 — 3 EAS Oe ae Pied eee T (3) Let y= a+ i , 3 T , *. sina + cos v = 4/2 sin (« ne “) =a/2 siny -vorb-(}b-IV- (2) or sin#” + cose 7) | Aw+a\? Av+a\? 40+ \? Ave w+ 7] 1-— 1— a oes 4} \ Aor 8 ar 12a 13. Let ABCDEF (fig. 3) be the polygon; 7,, 7, 73, &e. the radii of the circles inscribed in the triangles BAC, CAD, DAE, &c. whose vertical angles at 4 are each = 6 2 3 and the angles ACB, ADC, AED, &c. = gee cee hil &e. nn n respectively. Hence if (a) be a side of the polygon, ABC ACB vial (cot ; + cot ? )=« TRIGONOMETRICAL PROBLEMS, 207 ————— 2 (= a sin 3 eee ee ee Anco ACB : sin ——— sin 9 BAC ARCS ACE or 7, COS =a sin sin * = 2 Tv . (n-22)r . hence 27, cos — ‘= 2a sin -————— sin 2n 2n 2n (n — 3)7 (n-1)7 (si 370 =) = a@ < cos ——————_ — cos ——————- 7? = @ | SIN — — SIN —] 3 2n 2n 2n 2n similarly, ate (n-3)m . 27 We T 27r, COS — = 2a sIn ————— =a sin | — — sin —}]; 2n 271. 2n 2n 6 . (7-4) ar). 3a ar Sa 2 — = Game Ma — — —]; 1, COS 3 tae 2a sin aS sin a {sin F — sin 3 n n n 7 . (2n-3)7 aes. COS —— =a 7, y is negative, and when w=a7+h, y= : however small may be the value of 4; or the curve approxi- mates to, and ultimately coincides with the right lines Aa, Ya s ve vr “tr th) the e > ° aa,aa,aa@,aa , &c. repeated indefinitely where i {1% AA =, AA’ =27, &e., and Aa, A’a’, Aa’, &e. = : w es Hence from # = 0 to#=7, y= re and is positive: when ae ° . e>q7, and <27, y= —, and is negative; or as # passes ; 4 4 al , fet through 7, the value of y is suddenly changed from + — to ‘ 4 ae A ‘ . — —, and is discontinuous. 4 oP o@trace the change in the value of 2 a” S;, = @ COs @ — — cos $v + &e., vo as w changes from 0 to 2q. L5 226 TRIGONOMETRICAL PROBLEMS. a’ Let Ue Sn Pie Trucs 8u + &O5 *, 2y = tan-!ae’¥-14 tan“) ae7* V4 4 (GON + Ge _, 24a cos av = tan Bnesce way. 69 Yao } re (AGG 1-—a 1-a fats 2a COS & or tan2y = ————— y 1-@ 2a ; sie Soe When wv = 0, tan2y = 53 as & increases y diminishes, —a wT : and when v = ae 0, or the curve cuts the axis of a. T vs Pee Also f ( + x) =e f (= a v), and the curve is bisected by the axis of w, having two similar and equal portions from 7 7 ; ve=0tov= a and from «# = . to v = 7, respectively above and below the axis. 3 If AB= 2 AA =7, AB= oa AA” = 2a, the curve is that represented by fig. 83. As (a) approaches to unity, tan2y is very large unless : 7 T , © is nearly = 57 gs nearly = 7 for all values of # except T : ° (2r + 1)—; and the curve changes to the form given in fig. 84. 2 When a = 1, tan 2y is infinite for all values of a less than T T Tv T ° . ae or when w< PIs Fie and when # = 5° ¥ 3s indeter- minate; in this case the curve is changed into the series of terminated right lines ab, bBb’, b'b", b’b”, bb" (fig. 85), the lines parallel to the axis of w being all = 7, and those parallel to the axis of y all equal to =: ao = —_— a) %. TRIGONOMETRICAL PROBLEMS. 22 4. ‘To trace the change in the value of S, =acose — a’cos2a + a’ cos3v — &c. as & changes from 0 to 27. Let y=acosx —a’cos2u + &c., PQepende "Ny gerev-l = grterenal — ghe-N=1 4 &e. ae? Na deme \—1 2a(a + cos x) Se oo NI att On cok a” 1+ ae 1+ae ne 1+a°+2acos@# a(1 — a’) sin 2 and d,y=-— as *! (1 + a@ + 2acos 2a)’ hence y is a maximum when sinw = 0, or when w=0, 7a, a(1 + @) a Cn a ee Also when # =0, y= When cosv=-—-a, y=0, and the curve cuts the axis of 2. a | When w=7, y= - : ; and the curve assumes the —a form (fig. 86). Wh h : ges. en (a) approaches to unity, y = 1 — WIT Cea! and yis nearly =1 for all values of «(2m—1)r<(Qm+1)7; os and as w increases and passes through 77, S$ instantaneously pete 5 changes from (— 1)' a5 to (- 1)’ a WT : When «>ra<(r + 1)7, Sy = (- 1)’ re and as w Increases C : ff and passes through rar, S, is suddenly changed from (—1)’~’ : Tt to (—1)"—. (<1 ape Wwe. oa When a> (2r -1)> < (27 +41) — 5 S3 (- 1)’ a and as , 8, is suddenly a increases and passes through (2r — 1) ~° 139 changed from (— 1)'7! * to (-— 1)’ - ae 4 4 As a approaches to unity, S, approaches indefinitely near to unity, except when a is an odd multiple of a, in which case S', becomes indefinitely large and is negative. In the same manner we can trace the change in the values of asina + da singe ++ta°sinsw + &e.; and acos a + a’ cos2a 4- &c. when @ approaches to unity, as @ Increases from 0 to 27. MS Ae EE: vee Metcal#e & Patmer, Luthoy”* Metcalfe k Patmer, Lithoy”* v4 ce . R, 4 eo j A tal - & a Se oS N | | = ‘ y / oe —ooy 7 § Y : N \ ee g No mn P&S % ye 7; | 8 S | s qo a . = i x S © s S br) as ‘ Sat | a! Mi a eheralty x sata THE SOLUTIONS OF GEOMETRICAL PROBLEMS, CONSISTING CHIEFLY OF EXAMPLES IN PLANE CO-ORDINATE GEOMETRY, PROPOSED AT ST JOHN’S COLLEGE, CAMBRIDGE, FROM Dec 1880 TO Dec. 1846. WITH AN APPENDIX, CONTAINING SEVERAL GENERAL PROPERTIES OF CURVES OF THE SECOND ORDER, AND THE DETERMINATION OF THE MAGNITUDE AND POSITION OF THE AXES OF THE CONIC SECTION REPRESENTED BY THE GENERAL EQUATION OF THE SECOND DEGREE, BY THOMAS GASKIN, M.A., LATE FELLOW AND TUTOR OF JESUS COLLEGE, CAMBRIDGE. CAMBRIDGE: J. & J. J. DEIGHTON; LONDON: JOHN W. PARKER, M DCCC. XLVIL. Cambridae : Printed at the Gnthersity Press. TO WILLIAM CRACKANTHORPE, Esa., THIS BOOK IS INSCRIBED AS A TESTIMONY OF THE HIGHEST RESPECT AND ESTEEM, AND AS A GRATEFUL ACKNOWLEDGMENT OF NUMEROUS AND EXTENSIVE OBLIGATIONS CONFERRED UPON THE AUTHOR. “a al . ae PREFACE. Tue Examination Papers which the Author has select-_ ed for solution in the present Treatise have been proposed in the several years from 1830 to 1846 to the students of St John’s College at the end of their fourth term of resi- dence, and according to the plan which he adopted in the solution of the Trigonometrical Problems, he has endea- voured to place them before the reader in a proper form © for the inspection of the eines ~The problems are sufficiently varied in their character to exercise the student in the ordinary properties of the straight line, circle, and conic sections; they have been proposed by some of the most distinguished members of the society ; the generality of the results are remarkable for their neatness and simplicity; and except in one instance it is needless here to make any further comment. J In Question 6, Dec. 1833. (No. IV), it is required “To inscribe in a circle a triangle whose sides or sides produced shall pass through three given points in the same plane.” This problem has been solved analytically in a most ingenious and elegant treatise, entitled “ Researches on Curves of the second order” lately published by Mr Hearn, vi PREFACE, who remarks that it was proposed by M. Cramer to M. de Castillon, and that Lagrange has given a purely analytical solution which may be found in the memoirs of the Academy of Berlin (1776). The problem then being one of acknowledged difficulty, the author hopes that the first Appendix in which an analytical solution has been given when a conic section is substituted for the circle, will not be entirely devoid of interest. The case of the three different erst sections has been separately considered, and the author has afterwards still further generalized the problem, by inscribing in a given conic section, a polygon whose m sides taken in order shall pass through 7 fixed points. A simple and concise geometrical solution of M. Cramer’s problem has been extracted from “The Liver- pool Apollonius by J. H. Swale,” and inserted in the third Appendix; and when the triangle is to be inscribed in a conic section, the problem has been reduced to that of inscribing in a circle a triangle whose three sides shall pass through three fixed points, so as to afford a com- paratively simple geometrical solution in the more general case. Some apology may be considered necessary for the introduction into the second appendix of two different methods of determining the magnitude and position of the conic section represented by the general equation of the second degree. PREFACE. Vil By transferring the equation to the focus the author has endeavoured to point out the change which takes place when the curve approaches to a parabola; and on that account he has deduced the latus rectum and the co-ordinates of the vertex of the parabola from those of the ellipse in its transition state. The reduction of the equation to the focus led to the forms which have been determined for the elements of the curve; and it was afterwards found that most of them might be obtained more briefly by the polar equation from the centre. In the case of oblique co-ordinates the expressions are remarkably symmetrical, and are placed in such a form that the reader may perceive at once their con- nexion with the corresponding results when the co-ordi- _ nates are rectangular. In the latter part of the second appendix the author has endeavoured to trace a conic section geometrically as far as it appeared practicable, by the determination of a successive series of points when five points of the curve can in any manner be found. A remarkably simple construction for a tangent at one of the five given points has been obtained (Art. 89); a tangent has also been drawn from a given point without the curve; and the position of a point in the conic section in any proposed direction has been determined by its points of intersection with a given straight line. vill PREFACE, In a subject which for ages has exercised the skill and ingenuity of the most profound mathematicians, little ean be expected which is really original; but as only a very small portion of the matter in the appendices has been met with by the author elsewhere, even if he may have been anticipated in any or all the properties which are inserted, it is extremely probable that the proofs now given will be widely different from any which have been hitherto published. A reference has been made in several places to an edition of Euclid lately published by Mr Potts, in which will be found much valuable information; it is well de- © serving the attention of every one who wishes to study Geometry. T. G. CAMBRIDGE, Nov. 1847. GEOMETRICAL PROBLEMS, ST JOHN’S COLLEGE. Dec. 1830. (No. I.) 1. PaRaLLELocRAMs upon the same base and between the same parallels are equal to one another. 2. Of unequal magnitudes, the greater has a greater ratio to the same than the less. 3. If the diameter of a circle be one of the equal sides of an isosceles triangle, the base will be bisected by the cir- cumference. 4, The line joining the centres of the inscribed and cir- cumscribed circles of a triangle subtends at any one of the angular points an angle equal to the semidifference of the other two angles. 5. Find a point without a given circle, such that the sum of the two lines drawn from it touching the circle, shall be equal to the line drawn from it through the centre to meet the circumference. 6. If a circle roll within another of twice its size, any point in its circumference will trace out a diameter of the first. 7. If from any point in the circumference of a circle, a chord and tangent be drawn, the perpendiculars dropped upon them from the middle point of the subtended arc, are equal to one another. 8. If a, 6, y represent the distances of the angles of a triangle from the centre of the inscribed circle, and a, b, c the sides respectively opposite to them, then aa + 3°b + y’c = abe. 9. Describe a circle through a given point and touching a given straight line, so that the chord joining the given point > GEOMETRICAL PROBLEMS. NO. I. DEC. 1830. and point of contact, may cut off a segment capable of a given angle. 10. Shew that the perimeter of the triangle formed by joining the feet of the perpendiculars dropped from the angles upon the opposite sides of a triangle, is less than the perimeter of any other triangle whose angular points are on the sides of the first. 11. Explain what is meant by the equation to a curve; find the equation to a straight line, and state clearly the meaning of the constants involved. 12. Trace the circle whose equation is aw +y)+b (w~+y)=0; draw the lines represented by the equations y — 2Hy secat a2” =0, and shew that the angle between them is a. 13. The portion of a straight line intercepted by two rectangular axes, and the perpendicular upon it from their intersection are each of given length; what is the equation to the line P 14. Find the equation to an ellipse, and deduce that to | the parabola from it. 15. Find the co-ordinates of the point from which if three lines be drawn to the angles of a triangle, its area is trisected. 16. In the last question, the (distance)? from the angle A ; 2 : of the required point = 5 (0 +c — =) : 17. If the centre of the inscribed circle of a triangle be fixed, and a, 2, y represent the distances of its angles from any fixed point in space; then whatever position the triangle assume, the expression a’a + (3°h + °c is invariable. GEOMETRICAL PROBLEMS. NO. I. DEC. 1830. 2 SOLUTIONS TO (No. IL) He uciip,; Prop. 85, Book rT. 2. Kuclid, Prop. 8. Book v. 3. Let AB, AC (fig. 1) be the equal sides of an isosceles triangle; upon AB describe a semicircle cutting the base BC in D; join AD: then 2 ADB is a right angle = 2 ADC; also ZABD=2Z ACD, and AD is common to the two tri- angles ABD, ADC, .. BD = DC. 4, Let ABC (fig. 2) be a triangle, d, D the centres of the inscribed and circumscribing circles; draw DE, de perpen- dicular to 4B, and join 4D, Ad: then £ADE=142 ADB =2C; or ee Gea ie a a ee J 2 2 also Wades & LDA oe d to In like manner it may be proved by joining DB, dB, DCG, dC that A-C A-B ZDBd= 3 and ZDCd= ETN. ~ If 2 B be less than 2 C, 2 DAd will be negative, which shews that AD would in that case lie below Ad; and so of the rest. 5. Let D (fig. 3) be the required point in any diameter ACB produced; DE a tangent drawn from D; then DE? = DA. DB, and DA = 2DE, . De =4DE? =4DA.DB or DA=4DB; AB hence AB=3BD, and BD= aes oO which determines the position of the point D. A2 4 _ GEOMETRICAL PROBLEMS. NO. I. DEC. 1830, 6. Let C (fig. 4) be the centre of the larger circle; 4 the original point of contact of the two circles; let the inner circle roll until P becomes the point of contact; join CP and bisect it in O; then CP is the diameter, and O the centre of the inner circle PQC when P becomes the point of contact. Let the circle PQC cut AC in Q, and join OQ; then 4POQ =22PCQ, or oO ae 2 *- arc PQ = a are PA = arc PA; .. Q is the point which originally coincided with A, and it is in the radius AC; hence the locus of the point Q is the diameter which passes through A. 7. Let AB (fig. 5) be a chord and AT’ a tangent at the point 4 of a circle; C the middle point of the arc AB; join AC, CB and draw CM, CN perpendicular to 4B, AT’ re- spectively ; then CN=CAsinzCAT=CA.sinzCBA=CB.sinzCBA=CM. 8. Let A, B, C be the angles of the triangle respectively opposite to the sides a, b, c, and r the radius of the inscribed circle; then . 8B r 7 a= = —— = eh Shel 2? on’ Co slIn — sin — sin — 2 Q a b Cc Qr" ( i Las a b ae = ° $y Weer pam | © 6) sin 4 2 snB 2 % sin C Ws 2ar ( ot 4 2B ae , = - cot — cot — — sin A ous gt oO | aes [ae eee eee GEOMETRICAL PROBLEMS. -_NO. I. DEC. 1830. 5 A B C but rol = 5 — 4; Ot 5 =O, reo = S—e, a+b+e where Sere aaaepmet X A B C “5 r (cot = + cot > + cot =} = JS; 2 2 2 ~ x area of A ABC 2a or a’a + 3°b + ye = ——rS = — B Y sin A sin A Ga. (be. = i= sin 4) = abe. sin 4 \ 2 9. Let P (fig. 6) be the given point, and 4B the given straight line; draw PC parallel to AB, make z CPD = the given angle, and let PD meet AB in D; bisect DP in E, and draw DF, EF respectively perpendicular to 4B, DP meeting each other in F'; with centre F’ and distance FD describe a circle; this will pass through the point P and touch the straight line 4B; and 2 PDB = the angle in the alternate segment cut off by DP = 2 CPD = the given angle. 10. Let b, ¢ (fig. 7) be two of the angular points of the triangle which is required to be of the least possible perimeter ; a the remaining angular point in the side BC; then ba+ca will be least when 2Cab=2ZBac. (Ports? Evc rip, p. 293.) Now if a be the angular point properly determined, and } the angular point in the side AC, then ac + cb is least when ZBca=ZAcb: similarly, if a, c be two angular points of the triangle, ab + bc will be least when 2 4be =ZCba: hence the perimeter of the triangle abe will be the least possible when ZCab=2Z Bac; Z2Cbha=2Z Abe, and 2 Acb=Z Bea. Now if da, Bb, Ce be drawn from the angles A, B, C per- pendicular to the opposite sides respectively, the circle described on the diameter AC will pass through the points c, a, because the angles AcC, AaC are right angles ; hence LAac=4ACo=~~ A, and Bac=—- dac=A. 6 GEOMETRICAL PROBLEMS. NO. I. DEC. 1830. Similarly, the circle described on the diameter AB will pass through the voints 6b, a; and 2 dab=2 ABb= = —A; : £Cab=~ - dab= 4, or ZBac=2Cab; similarly, 2 Cha=Z Abe, and 4 Ach=2Z Bea; therefore the triangle abc has the least possible perimeter. 11. See Hymers’ Contc Sections. Art. 13. a bFa 12s 1) © ee? 4 be 2 = b* ( 4) ‘ (y Qa 2a” the equation to a circle the co-ordinates of whose centre are 6? b* b* —-~-, —-——, and whose radius = ——-. Hence if Aa, Ay 2a 2 a. \/ 2 a b? (fig. 8) be the co-ordinate axes, take AB, BC each = op ee a and with centre C and radius = —— describe a circle AED; 2a it will pass through A and meet the axis of # in the point D such that DB = BA; bad LACB=4 DCB ="; ri LACD ==; hence AED is a quadrant of the circle having the chord pro- duced for the axis of a. (2) Since y? — 2wy seca + w*® = 0, we have 2 a (2) — 2seca = + (sec a)? = (tana)’; YU he a = seca + tan a = tan (45 = 5), > & GEOMETRICAL PROBLEMS. NO. I. DEC. 18380. 7 which are the equations to two straight lines passing through Pa ; e a the origin and inclined at angles 45 + ; and 45 — ; respectively to the axis of #; hence the angle between them 13. Let O (fig. 9) be the origin; AB the straight line meeting the axes of # and y in the points A, B respectively ; draw OP perpendicular to AB and let 4B=a, OP =}, OA=m, OB =n; then equation to the line AB is ve ak where m? +n? =a’, and mn =2A AOB = AB.OP=ab; rs m+n=+\/a? + 2ab, m—-n=+\/a —2ab: “or 2m = & (\/a? + 2ab + \/a? — 2ab), and 2” = + (\/a? + 2ab = \/a? — 2ab); therefore if Ja? +2ab+/ a —2ab = 2a, and \/a?+2ab —+/ a —2ab =2, we have m= +a, or + (3, and the corresponding values of n are nad Gy +a; hence aie pe bee OF; ed ahs hing each of eG B a which equations will give the equations to four straight lines by the four combinations of the double sign +. If O4, OA’, OA", OA” be taken each equal to a, and OB, OB’, OB’, OB" each equal to 3; then AB, BA”, A” B", B’A; AB’, BA", AB", B’A’ will be the straight lines required. 14. See Hymers’ Conic Secrions. Art. 104. The equation to the ellipse is 2 y = (1 - e’) (2 w= at) epi +e)a-(1-e) ats and when e =1, y’ = 4px, which is the equation to the para- bola. 8 GEOMETRICAL PROBLEMS. NO. I. DEC. 1830. 15. Let P (fig. 10) be the required point within the triangle ABC; join AP, BP, CP and produce them to meet the opposite sides in a, b, ¢ respectively: then A APB = 4 AB. Pec.sn PcB = +A ACB = 4(44B.Cc.sin PcB), *. Pe= similarly, Pb = ad and Pa = ae ; 3 Join ac, then PC =2Pc, PA =2 Pa, and..Pa 2: (Po %: BP Aer er, or ac is parallel to AC, and Bes BA ::'ca 2: -CA™=: Posters “. BA=2Be, and the side BA is bisected in C; and in like manner it may be proved that a, 6 are the middle points of BC, CA respectively. Draw PM parallel to AC, then if AB, AC be taken for the co-ordinate axes, and wv, y be the co- ordinates of P; PMS PAC Po tC Camor PM =34AC; b Ve Bee ,and AM: MP: de: ca: 2: 2; = 3 3 w 16. Let PA =a, then Aa=—: and since BC is bi- sected in a, 9 2 9a” a” D 9Aa’+2Ba? = AB* + AC’, or Saati ROD spe ~ NS) 9 2 a” ._a=- ee). eae 17. First let P (fig. 11) be a point in the plane ABC, O the centre of the inscribed circle whose radius is 7; join PA, PB, PC, PO, AO, BO, CO; then if P4 =a, PB =, PC =r, 4 AOP =9, we have GEOMETRICAL PROBLEMS. NO. I. DEC. 1830. 9 aat+ Bb+y'?c = a(PO’ + AO? —-2P0.A0.cos POA) + 6(PO? + BO? - 2P0.BO.cos POB) + c(PO? + CO? -2P0.CO.cos POC) =(a+b+c)PO0'+a.A40°+b.BO0?+e¢.CO —-2P0O(a.A0.cosPOA +b.BO.cosPOB+¢.CO.cos POC) =(a+b+c)PO’+ abe ° A ° B ° s1n— s1n— siIn— 2 2 : A+B f A -aP 0}, ae : fake os CRC 2 (from Quest. 8) =(a+b+c)PO' 4+ abe -4P0O. A 3 Joos 3 cos — yi sin (+6) + pane (o- =) 2 2 2 a af 2 | a sin 4 2 \ =(a+6+c) PO’ + abe G B —-4P0. a {cos = cos 8 - (sin cos = + cos = sin >) cos | sin A 2 2 Oy 2 Q , A nie G = (a+b +0) PO’ + abe -4PO “(cos 5 — sin i ) cos 0 sin 4 2 =(a+b+c) PO’ + abe, ; : +C iA since sin = SOE ee. Next let P be without the plane 4BC; draw PQ perpen- dicular to the plane, and join Q4, QB, QC, QO, then PA = QP? + QA’, PO’ = QP? + QO’; and a?a+B°b-+y%c = a(QP’+ Q4’) +0(QP?+ QB’) +e(QP?+ QC) =(a+b6+c) QP’? +a.Q4* + b.QB’ +e.QC’ =(a+b4+c)QP?+(a+b+0) QO° + abe (by the first case) = (a+6+4c) PO’ + abe, and is therefore invariable, since P and O are two fixed points. 10 GEOMETRICAL PROBLEMS. NO. Il. DEC, 1831. ST JOHN’S COLLEGE. Dec. 1831. (No. II.) 1. Is four magnitudes of the same kind are propor- tionals, the greatest and least of them together are greater than the other two together. 2. If two straight lines meeting one another be parallel to two others that meet one another, and are not in the same plane with the first two; the first two and the other two shall contain equal angles. 3. The sum of the perpendiculars from any point in the base of an isosceles triangle is equal to a line of fixed length. 4. To find a point in the side or side produced of any parallelogram, such that the angle it makes with the line joining the point and one extremity of the opposite side, may be bisected by the line joining it with the other extremity. 5. The lines which bisect the vertical angles of all triangles on the same base and with the same vertical angle, all intersect in one point. 6. If a semicircle be described on the hypothenuse 4B of a right-angled triangle ABC, and from the centre F& the radius ED be drawn at right angles to 4B, shew that the difference of the segments on the two sides equal twice the sector CHD. "The locus of the centres of the circles which are inscribed in all right-angled triangles on the same hypothenuse is the quadrant described on the hypothenuse. 8. Of all the angles which a straight line makes with any straight lines drawn in a given plane to meet it, the least is that which measures the inclination of the line to the plane. 9. Find the sine of the inclination to each other of two straight lines whose equations are given. i GEOMETRICAL PROBLEMS. NO. Il. DEC. 1831. 11 10. Find the length of the perpendicular from the origin of co-ordinates upon the line whose equation is a(w—a)+b(y—-b) =0, and the part of the line intercepted between the co-ordinate aAXes, 11. The equation to a circle is y? + # =a (y +x); what is the equation to that diameter which passes through the origin of co-ordinates ? 12. A side of a triangle being assumed as the axis of a, the equations to the equations to the other sides are y=av+), and y= aa; determine the sides and angles of the triangle. 13. If through any point of a quadrant whose radius is R, two circles be drawn touching the bounding radii of the quadrant, and 7,7 be the radii of these circles, shew that rr = R’, FY GEOMETRICAL PROBLEMS. NO. II. DEC. 1831. SOLUTIONS TO (No. II.) 1. Evctip, Prop. 25. Book v. 2. Euclid, Prop. 10. Book x1. 3. Let D (fig. 12) be a point in the base AB of the isosceles triangle ABC; draw DE, DF perpendicular to AC, BC respectively ; then DE + DF =AD.sn A+ DB.sin B= (AD + DB) sin A = AB.sin A equal the perpendicular from B upon the side AC, and is therefore constant wherever D be taken in AC. 4. Let ABCD (fig. 13) be the parallelogram ; with centre B and radius BA describe a circle cutting DC in the points E, F; jon AF, BF; then BF = BA and LBFA= ZLBAF = 2 AFD, or ZDFB is bisected by the straight line 4F. Similarly if AE, BE be joined, 2 DEB is bisected by AE. Also if with centre 4 and radius AB a circle be described cutting DC in the points G, H, the angles CGA, CHA will be bisected by the straight lines BG, BH respectively. 5. Let ACB (fig. 14) be one of the triangles; about it describe the circle ACBD; then since the vertical angle is constant, the vertices of all the triangles will be in the circum- ference ACB. Bisect the arc AB in D, and join CD; then since arc AD =arc DB, 2 ACD =z BCD, and CD bisects the angle ACB; hence the line which bisects the angle ACB will always pass through the fixed point D. 6. Let ABC (fig. 15) be the right-angled triangle, 4DCB the semicircle described upon the hypothenuse 4B; then since AR = EB, AAEC= ABEC; and segment on AC —segment on BC'= (segment on AC + A AEC) — (segment on BC+ A BEC) = sector AEC - sector BEC = (sector AED + sector DEC) — (sector BED — sector DEC) = 2 sector DEC since quadrant 4D = quadrant BED. GEOMETRICAL PROBLEMS. NO. II.. DEC. 1831. 13 7. Let O (fig. 16) be the centre of the circle inscribed in the A ACB; join OA, OB; then 4 AOB=-7 - (OAB + OBA) ee r-C wr C TT — 2 2 2 2 and is constant. Hence the locus of the point O is a segment of a circle containing an angle == TLL e >] up Em 20m ee and the 2 ADB in the alternate segment br CC men 2 2 2 2 therefore the angle subtended by AOB at the centre E =a —C. When the A 4CB is right-angled, £C=o; and £AEB = - C="; therefore AOB is the quadrant described upon the hypothe- nuse 4B. | 8. Let P (fig. 17) be any point in the line 4P which meets the plane in the point 4; draw PM perpendicular to the plane; join 4M, and through 4 draw any other line 4Q in the plane; draw PQ perpendicular to AQ, and join QM; then if 0, 6’, be the inclinations of AP to AM, AQ respec- tively, PM Rr pee G) sin 9 = AP’ and sin @ =p? but PQ’ = PM’ + MQ? since 2 PMQ is a right angle, or PQ is greater than PM; therefore sin @ is less than sin @’ and @ is less than 9’. Now @ measures the inclination of the line AP to the plane, which is therefore less than the inclination of the line 4P to any other line 4Q drawn to meet it in that plane. 9. Let y=anv+b, y=a a+b’ be the equations to the two straight lines; then if 0, 0 be the angles which they respectively make with the axis of x, tand=a, tan@’ =a, 14 GEOMETRICAL PROBLEMS. NO. II. DEC, 183]. a =a / (1 +a”) (144) which is the sine of the inclination of the two-straight lines to each other. 10. Let O (fig. 9) be the origin, and 4, B the points in which the given straight line meets the axes of w and y respec- tively; then the equation to the line OP drawn through O and sin (6’—@) =cos0@cos@’ (tan 6’ — tan 0) = b: ; : perpendicular to 4B is y= “ ; and if a’, y be the co-ordi- nates of the point of intersection P, we have , bw \ oe b Sas Ai 5 = 0, a (a —a)+ es ) , , Sa, or w =a, and y =—=),; a therefore the length of the perpendicular OP =f x? +4 y%=/S/a +B. Again in the equation to the given straight line, make . : a4 Bb y = 0, the corresponding value of # is OA = ; and by 2 b? making w = 0, the value of y is BO = By ; ; See (a? “ b°)3 -, 4B =/ AO? + BO Seat ll. # -avr+y’-—ay=0, s) ( = a (7-5 SECA oy Te the equation to a circle, the co-ordinates of whose centre are a a g° 2 Let y = ma be the equation to the diameter which passes through the origin, then since this straight line passes through Be ihen 7 a a ; a point >» E+ we have ren vars or m=1; and y= is the equation required. . GEOMETRICAL PROBLEMS. NO. Il. DEC. 1831. 15 12. Let ABC (fig. 18) be the triangle, A the origin, and BA produced the axis of w, the equations to AC, BC are y=aa, and y=aun+b; *, tan (CA) = tan(w — A) =a’ or tan d= —a, and tanCBe =tanB=a; ’ a—-a hence tan C = tan(CAw — CBx) = t+ aa’ Draw CM perpendicular to AB, and in the equation to BC make y =0, then the corresponding value of v = — AB b b , , . = --or AB=-: let x, y be the co-ordinates of M, then a a ve b a & aware or i ee May BT RES ; ; (eee) ! aE a ra , ab Se Te ON ea y = CM =a'e' =——;_ ». ACH VW? 4 y? = “A ; b ba’ also ii) ip wae eae a Gai a (an aye ; cB ae eVv 1 ate a(a —a) 18. Let the bounding radii AB, AC (fig. 19) of the quadrant be taken for the axes of # and y respectively ; bisect the 2 BAC by the straight line AD, and from any point D in AD draw DE perpendicular to AB, then a circle described with centre D and radius DE will touch both the bounding radii 4B, AC: andif AE = ED= pp, the radius will also be = ps and the equation to the circle is (w —p)’+ (y—p)’ =p or # +y°-2p(w@+y) +p? =0. Let this circle cut the quadrant in the point P whose co-ordinates are h, k; wh? +kh?-2p(h+k)+p?=0, or R-2(h+kh)pt+ p?=93 from this equation the two values r, 7” of p may be deter- mined, and rv’ = R?, r+7r =2(h +k). 16 GEOMETRICAL PROBLEMS. NO. III. JAN. 1833. ST JOHN’S COLLEGE. Jan. 1833. (No. IIL) 1. Maenrrupes which have the same ratio to the same magnitude are equal to one another; and those to which the same magnitude has the same ratio are equal to one another. 2. If a solid angle be contained by three plane angles, any two of them are greater than the third, 3. If a straight line be divided into any two parts; to find a point without the line at which the segments shall contain equal angles. 4. Given the base and sum of the sides containing the vertical angle, and the line drawn from one extremity of the base perpendicular to it to meet a side or side produced ; to construct the triangle. 5. In a triangle ABC, if CD be drawn bisecting the vertical angle C, and meeting the base in D; and DE, DF be drawn respectively parallel to the sides AC, BC and meeting them in FE and F’; prove that DE = DF. 6. If two circles be described about the same centre, the radius of one being double that of the other, and a point be taken within the inner circle; to draw from this point to the outer circumference a straight line which shall be bisected by the inner circumference. 7. ACB is a segment of a circle, and any chord AC is produced to a point P, so that AC : CP in a given ratio. Required to find the locus of P. 8. If FACB be a line passing through the centre C of a circle, CD a radius perpendicular to the diameter ACB; DEF, DGH any two lines cutting the circle in KE, H, and the straight line FACB in F, G; shew that a circle may be made to pass through the four points /’, E, G, ads GEOMETRICAL PROBLEMS. NO. III. JAN. 1833. 17 9. If through any point O within a triangle, three straight lines be drawn from the angles 4, B, C to meet the opposite sides in a, b, ¢ respectively ; prove that Ac. Ba.Cb=cB.aC.baA. 10. Find the equation to a straight line which cuts a given circle, when the lines drawn from the points of inter- section to the centre contain a right angle, and one of them is inclined to the axis of # at a given angle. 11. If from a point without a given circle, two straight lines be drawn touching the circle; find the equation to the line joining the points of contact. 12. If a,, a, be the sides of a right-angled triangle, and 6,, 0» the diameters of the circles inscribed in the triangles formed by joining the vertex and the middle point of the hypothenuse ; shew that 13. Find the equation to the straight line drawn from a given point to bisect a given equilateral triangle. 18 GEOMETRICAL PROBLEMS. NO. lf. JAN. 1833. SOLUTIONS TO (No. III). 1. Evctiip, Prop. 9. Book v. 2. Euclid, Prop. 20, Book xt. 3. Let AB (fig. 20) be divided into two parts in the point C; E the required point such that 2 AEC = Z BEC; then if AC=c, BC=€c, CE=p; 4ECB=®, 2, AEC =2BEC =6, ae ae ee sin 0 Cc sin @ eat ke _ sin (p + @) - sin(@ - 8) © : (5 -=)e= ane = 2 cos d, or p= ; cos, which is the polar equation to a circle Codh ee Paes , is in the direction CB. F 26¢ whose diameter = Produce CB to D, take CD =~ cee and upon CD —¢ _ describe a circle; from any point # in ee circumference draw AE, EC, EB, then will 2 AEC = z BEC. 4, Let AB (fig. 21) be the given base; draw AC per- pendicular to the base, and of the given length, then BC will be the direction of one of the sides of the triangle. In BC take BD =the sum of the sides, join 4D, and make 2 DAE =Z ADB, then AEB will be the triangle required; for ZADE=2DAE, «. AE=DE, and AE + EB = DB the sum of the sides. 5. DECF (fig. 22) is evidently a parallelogram, and WODE = 2DCKF = 2DCE: -\ DE= CE = DF, 6. Let A (fig. 23) be the common centre of the two circles, B the given point; BPQ a line drawn through B cutting the circles in P and Q; join AP, AQ and let BP=p, BQ=p, ZABQ=0, AP=a, AQ=2a, AB=c; *. p +e —2cpcos0 =a’; p> +e - 2cp cos 0 = 4a’; GEOMETRICAL PROBLEMS. NO. IIl. JAN. 1833. 19 and if p!=2p, we have 4p° + c? — 4cpcos@ = 4a’, but 4° + 4c? — 8ep cos@ = 4a’; *. 4ep cos @ = 3c or p cos 9 = — A Hence in 4B take BM = 3 AB or AM = ae draw MP perpendicular to 4B, meeting the inner circle in P; join BPQ, and BP will be equal to PQ. 7. Let O (fig. 24) be the centre, CAO =0, AO=a; then CA=2acos@, and CP=n (CA); AP =(n+1)CA=2(n +1) acos8, or the polar equation to the locus of P is p =2(n + 1) acos 0, which is the equation to a circle whose centre is in the line AO, and radius = ( + 1) a4. 8. Draw DK (fig. 25) touching the circle at D; then ZLDFG=2KDE=2 EMG: hence a circle may be described through the four points F, E, G, H since EFG and EAG will be angles in the same segment and upon the same base EG, and are proved to be equal to one another. Ac sindAOc Be sin BOc Dome Eta. 80) AO sindcO’ BO sin BcO’ Ac AO sin AOc ’ Be BO sin BOc’ ae Ba BO sin BOa Similarly Ca. GO’ sin COa Cb CO sin COb Ab AO’ sin AOb’ hence by multiplication Ac. Ba.Cb a _B .Cbh= Be. AU: et 1, or Ac. Ba e.Ca 10. Let PQ (fig. 27) be the straight line whose equation is required, C the centre, 7 the radius of the given circle; CA B2 20 GEOMETRICAL PROBLEMS. NO. lll. JAN. 1833. the axis of 7; 2 PCA=a, .. ZQCA = 90+ a; and the co- ordinates of the points P, Q are acosa, asina; and —a sin a; acosa respectively ; therefore the equation to PQ is acosa — asina y —asina (@ — a cosa) —asina —acosa sina — cosa ; (vw — a cosa) ; sina + cosa sina — cosa a y= __ & + -—__. sina + cosa sina +cosa | a hence y = tan (a — 45) v + —=sec (a — 45) af 2 is the equation required. 11. Let A, & be the co-ordinates of the given point without the circle; a’, y', 2”, y” the co-ordinates of the two points of contact, then the equations to the two tangents are av+yy=a’, and wv’ at+yy=a’; and since they both pass through a point whose co-ordinates are h, k, webeove he’ + ky =, he’ +ky =a’; and the equation to the line joining the points of contact is yl" A yf y” ss y! As (@ — a’); but Son rT y-Y= : = & / h 3 / , “Y-Y= pg ee. orhat+ky=hw +ky =a the equation required. 12. Let ACB (fig. 28) be the right-angled triangle, D the middle point of the hypothenuse 4B; AC =a, BC =a., AB =c; a A B Petite 3 = tane = =a tan—: | 65 =-estan — 1 9 9 1 a. 2 2 ae. cot — cot — cot — cot — 2 2 1 a 2 or ~— — y= - — = -\- Se O; Oe ay Os ec \sinB sinA GEOMETRICAL PROBLEMS. NO. Ill. JAN. 1833. 2.cos’ —.— 2. cos* — g 2 1 (cos 4 — cos B) Cc sin Asin B c¢ sin Asin B e(cos4—cosB) a, — a, (csin A)esinB aa, ” Pepe kes age Ort 31 21 13. Let a, 8 be the co-ordinates of the given point P (fig. 29), referred to the two sides AB, AC as axes; DPE the required straight line meeting AB, AC in the points D, E; then if 4D =m, AE =n, the equation to DPE is — + ~ =1, m nr and since this passes through the point P whose co-ordinates a are a, (3, we Hea ee yp mon 2 and A EAD = asin d= $a CAB =4 (5 sin A) . mn=—, hencean+ Pm=mn=—, vi and an — Bm = £\/(an+ Bm) — 4aBmn >t , datas a a setae Ear 4: B 2 Re *-. Zan E (ra a/ a 2 8a8 2 ig 2 2ama© (1m / tenet 2 a” yi 1 1 a fs 1 ; 8af3 . ao se —f{ 1+ 1— > m al V/ a° n mee hte a and the equation to DE to DPE becomes 1 2a “I 8) py GEOMETRICAL PROBLEMS. NO. IV. DEC. 1833. ST. JOHN’S COLLEGE. Dec. 1833. (No. IV.) 1. Fuinp the centre of a given circle, and state where Euclid’s demonstration is imperfect. What is the meaning of a given circle in Analytical Geometry ? 2. Ifa straight line be at right angles to a plane, every plane in which the straight line lies shall be at right angles to that plane. 3. Inscribe the greatest quadrilateral figure in a given circle. Can a circle always be inscribed in a proposed quad- rilateral figure, or described about one? 4. Given a polygon traced upon a plane, describe the triangle that shall have an equivalent area. 5. ACB is an isosceles triangle having a right angle C; with centre C and distance CA describe a circle; if from a point Q in the circumference of the circle, QRr be drawn parallel to AB meeting AC, BC in R&R, r respectively, prove that QR? + Qr? = AB’. 6. Inscribe in a circle a triangle whose sides or sides produced shall pass through three given points in the same plane. 7. ABC is a triangle inscribed in a circle; AB, AC, BC produced cut a line given in position in the points m, ”, p respectively. If ¢,,, t,, t, be the lengths of the lines drawn from m, n, p touching the circle, shew that tnt,t, = An.Bm.Cp = Am. Bp.Cn. 8. An indefinite number of straight lines (not in the same plane) are situated so that from a fixed point perpendiculars can be drawn to them which are all equal to one another: required the locus of the intersections of these perpendiculars with the given lines. 9. Given a circle traced upon a plane, describe another whose area is exactly twice as great as the former. GEOMETRICAL PROBLEMS. NO. IV. DEC. 1833. Te 10. Investigate the line or lines represented by the equation ye +(e —a)y? + (v’-—@)y+ (a - a) (w~+a) =0. 11. Taking the requisite data to fix a parallelogram in a plane by equations to its sides; prove that the diagonals bisect each other. 12. Given the equation to a circle and the chord of the circle; shew that a perpendicular let fall from the centre of the circle upon the chord, bisects the chord. 13. One of the vertices of a triangle being taken for the origin of rectangular co-ordinates, and a’, y’, v”, y” the co- ordinates of the other two; prove that the area of the triangle =$ (a'y” — y'e”). 24 GEOMETRICAL PROBLEMS. NO. IV. DEC. 1833, SOLUTIONS TO (No. IV.) 1. Evcurp, Prop. 1, Book 111. See Potts’ Euclid, Page 107. In Analytical Geometry if the equation to a circle is given, the position of the centre and radius may be deter- mined, and the circle is therefore determined in magnitude and position. 2. Euclid, Prop. 18, Book xt. 3. (a) Let ABCD (fig. 30) be a quadrilateral figure; draw the diagonals AC, BD intersecting in E; then if a be the angle between the diagonals, AABC=4 AC. BE. sina: Also AADC =44AC.DE.sina; therefore by addition, the area of the quadrilateral ABCD =4 (AC. BD.sina): this will be the greatest when AC, BD and sina are respectively greatest. Now if the quadrilateral be inscribed in a circle, AC, BD are the greatest possible when they are two diameters; and sina is greatest, when AC, BD are at right angles. Hence if two diameters AC, BD be drawn at right angles, and 4B, BC, CD, DA be joined, the square ABCD will be the greatest possible quadrilateral figure which can be inscribed in the circle. (3) When a quadrilateral figure is inscribed in a circle, the opposite angles are together equal to two right angles; if this condition be not satisfied aye Geen tareedt figure cannot be inscribed in a circle. (y) If a circle can be inscribed in ABCD (fig. 31) touching the sides AB, BC, CD, DA in the points a, b, ce, d respectively, then Aa= Ad, Ba= Bb, De = Dd, Ce= Cb, - 406+ Ba+De+ Ce=Ad+ Dd + Bb + Co, or 4B+ CD=BC+DA4; hence the sum of two opposite sides is equal to the sum of the two remaining sides. 4. Let the parallelogram ABCD (fig. 32) be described GEOMETRICAL PROBLEMS. NO. Iv. DEC. 1833. 25 equal to the given rectilineal figure ; produce 4B to E making BE = AB; join AC, CE; then AACE =2AACB = 0 ABCD equal the given rectilineal figure. 5. Draw CDN (fig. 33) perpendicular to the base, meet- ing AB, Rrin Dand N respectively; then since Rr is bisected in VN, RQ’ + Qr’? = 2(RN? + QN’) =2 (CN? + NQ’) (since 2 CRN = 4a right angle = 2 RCN) = 2C A? = CA? + CB = AB’. 6. See Appendix 1. Cor, Art. 1. 7. Let 2 Amp (fig. 34) = 0, 2Anp=9, 4 Bpm=\, ia Jae eo Meal Bp iy mnie Syme An sin@° Bm sin’ Cp sing Am. Bp.Cn | therefore by multiplication, 6 ERE b) also ,= Am. Bm, t,=Cn.An, t= Bp.Cp; Ee 2 a aa ‘ “ t,.t,. =(Am.Bp. Cn) (An. Bm.Cp) = (Am. Bp. Cn)’, or tn. t,.t, = Am. Bp.Cn= An. Bm.Cp. 8. Let the fixed point 4 be taken for the origin, P any point in the locus; then since AP is constant, it is evident that the locus of P is a sphere round the centre A, and radius = AP. 9. Find C (fig. 35) the centre of the circle; draw any radius CA, and AB perpendicular to C'A and equal to it; join CB and with centre C and radius CB describe a circle, this will be the circle required. For CB’ = CA?+ AB? =2C A’, and circles are to one another as the squares of their radii; therefore the circle whose radius is CB’: circle whose radius is CA :: CB? : C42 : 2:1. 26 GEOMETRICAL PROBLEMS. NO. IV. DEC. 1833. 10. The equation may be reduced to the form (y +x-a) (y+ a°- a) =0, which is satisfied by making separately y+uv—a=0, and y’+a2°-a=0, If CA, CB (fig. 36) be taken for the co-ordinate axes, the former equation represents the straight line 4B, where CA = CB =a, and the latter the circle ABD whose centre is C and radius CA. Hence the proposed equation represents the circle ABD, and the chord of the quadrant AB. 11. Let the two sides 4B, AD (fig. 37) of the parallel- ogram ABCD be taken for the co-ordinate axes; join 4C, BD, intersecting each other in HL, draw EF parallel to AD, and let AB=a, AD=b; then the equations to the diagonals AC’, BD b } t are y = ute , and ~ + = 1, and if a’, y' be the co-ordinates of a their point of intersection E, ba’ av’ / gf eons Nene eae a a b 20 , a 24 : b | 21, or a = AF =>; and =1, Ot eee paces Ak AF AE AC AG: ABs Pi es apie a BE OEF BD da = 99 BE = —; NBD PA Dee, 2 and the diagonals of the parallelogram bisects each other. 12. Let the equation to the circle referred to the centre C (fig. 38) be a + y? =a’, and the equation to the chord PQ, y=mau+c; then the equation to CR drawn through C per- : : 1 ; pendicular to PQ is y = — PL and if .X, Y be the co-ordinates of the point &, em Cc Y = —m’Y +c, or Y =———/and X¥ = — ; 1+m? 1 an” GEOMETRICAL PROBLEMS. NO. IV. DEC. 1833. 27 and to find the co-ordinates of the points of intersection of PQ with the circle, we have uv? + (mex +c)’ =a’, or (1 +m’) wv + 2mceu + (c?— a’) =0; therefore if a’, x” be the two values of x ace are the abscissze of the points P, Q 2mc Cee = a ee 2X, 1+m? or & is the middle point of PQ. 13. Let ABC (fig. 39) be the triangle, At, Ay the co- ordinate axes; 4 BAx= 0, CAa=0", AB=p, AC =p then the area of the triangle A BC = 1 AB. AC. sin 2 BAC= 1 p'p’ sin (6" - 0) ae) ” =i (p” sin Q’’. p cos Q’ ae p- cos Q”’. p. sin 0’) id dyke L(y" a — wy’). 28 GEOMETRICAL PROBLEMS. NO. Vv. DEC. 1834. ST JOHN’S COLLEGE. Dec. 1834. (No. V.) 1. Wuar objections have been urged against the doctrine of parallel straight lines as it is laid down by Euclid ? Where does the difficulty originate, and what has been suggested to remove it ? 2. Magnitudes have the same ratio to one another which their equimultiples have. When is the first of four magni- tudes said to have to the second the same ratio which the third has to the fourth; and when a greater ratio ? Do the definitions and theorems of Book v. include in- commensurable magnitudes ? 3. If a solid angle be contained by three plane angles, any two are together greater than the third. Define the inclination of a plane to a plane, and shew that it is equal to the inclination of their normals. 4. If there be two concentric circles, and any chord of the greater circle cut the less in any point, this point will divide the chord into two segments whose rectangle is in- variable. 5. Divide algebraically a given line (a) into two parts such that the rectangle contained by the whole and one part may be equal to the square of the other. Deduce Euclid’s construction from one solution, and explain the other. 6. Find a straight line, which shall have to a given straight line the ratio of 1 :A/ 5. 7. ACB is a triangle whose base AB is divided in E and produced to F, so that dE: EB and also AF’: FB as AC: CB. Join CE, CF and shew that 2 ECF is a right angle. 8. The point C is the centre of a given circle, and £ is any point in the radius; find that point in the circumference where CE subtends the greatest angle. GEOMETRICAL PROBLEMS. NO. V. DEC. 1834. 29 9. Two points are taken in the diameter of a circle at any equal distances from the centre; through one of these draw any chord, and join its extremities and the other point. The triangle so formed has the sum of its sides invariable. 10. If p,, po, p; be perpendiculars from any point within a triangle on the sides; P,, P., P; perpendiculars from the angular points on the same sides respectively, prove that Pi Pe Ps sla st iE ale Pit Pier 11. MANP is a parallelogram having a given angle at A, and also its perimeter a given quantity. Find the locus of P for all such parallelograms and construct it. 12. Find the locus of a point such that if straight lines oe drawn from it to the four corners of a given square, the sum of their squares shall be invariable. 13. Given the equations to two straight lines passing through two given points, find the locus of their point of concourse when the straight lines intersect each other at a given angle. 14. In any parallelopiped, the sum of the squares of the four diagonals is equal to the sum of the squares of the twelve edges. 15. Ifa triangular pyramid have one of its solid angles a right angle, i. e. contained by three plane right angles, the square of the face subtending the right angle is equal to the squares of the three faces which contain it. 30 GEOMETRICAL PROBLEMS. NO. V. DEC. 1834. SOLUTIONS TO (No. V.) | 1.” See Potts’ Euclid, p. 50. 2. Euclid, Prop. 15. Book v. and Definitions 5 and 7 Book v.; and Potts’ Euclid, note to Definition 5. Book. v. page 162. 3. Euclid, Prop. 20. Book x1. and Definition 6, Book x1. Let A (fig. 40) be a point in the common section AB of the two planes, from which draw AC, AD at right angles to AB in the two planes; draw also 4H, EF' perpendicular to the two planes BAC, BAD respectively; then AE is per- pendicular to the plane BAC and therefore to 4B. Hence AB is perpendicular to AE, AC, AD which are consequently in the same plane. Similarly 4 may be proved to be in the same plane with AC and AD; and right 2 EAC = right LFAD; .. LEAF = 2DAC. 4. Let C (fig. 41) be the common centre of the two circles; PpqQ a chord cutting the inner circle in p and q; draw CD perpendicular to PQ, it will bisect PQ and pq in the point D; join CP, Cp; then Qp.Pp =PD*- p D*=(PC*- CD’) - (pC? - CD*) = CP*- Cp’, and is therefore invariable. 5. See Potts’ Euclid, p. 73. Taking Euclid’s figure, Ay Ae cae AH = AF = soak VAR : ae . EF-~. a= /o+ = / Ais AB = BE, which gives Euclid’s construction. GEOMETRICAL PROBLEMS. NO. Vv. DEC. 1834. at 6. Let AB (fig. 43) be the given straight lin {draw BC at right angles to AB and =2AB, join AC, and-draw BD perpendicular to AC; then AD will be the line required. For AC =\/AB + BO =\/s. AB, \ end eA) AB PA OAC 1% cee 7. Produce AC (fig. 43) to G; then CH, CF bisect the angles ACB, BCG respectively ; LECF = 2BCF+ 2 BCE=4(4BCG+ 2 BCA) equal a right angle. 8. Draw EP (fig. 44) perpendicular to EC meeting the circle in P: on CP as a diameter describe a semicircle which will touch the circle in P and pass through E, because 4 PEC is a right angle. Draw any line CQ’Q meeting the circles in Q’, Q; join EQ, EQ’; then zCPE=LCQE is greater than ZCQE, or ZCPE is greater than any other angle subtended by CE at a point in the circumference of the given circle. 9. Let C (fig. 45) be the centre of the circle, D, FE two points in the diameter AB at equal distances from C; PDQ any chord through D; join CP, CQ, EP, EQ; then PD + PE? =2CP? +2CD’=2AC’ +2CD’; DQ@ + QE? = 2C0Q? + 2CD’ =2AC°+2CD'; 2DQ.DP=2AD.DB = 2AC* -—-2CD’; therefore by addition PQ + PE’ + QE? =6AC? + 2CD’, and is invariable for every position of the chord PQ passing through D. 10. Let P (fig. 46) be the point within the triangle mabC; jon AP, BP, CP; then 32 GEOMETRICAL PROBLEMS.. NO. V. DEC. 1834. similarly A APC = e A ABC; 2 and AAPB= = A ABC: therefore by addition Pi eee ps” WT bY ed AAPC +AAPB = pak Rieder Bye TOE! 33 C+ + aABC=(F4+5 +5 ‘11. Let AM, AN (fig. 47) be taken for the axes of # and yw; and w, y the co-ordinates of P; then w+y= S the perimeter of the parallelogram -f ; or the locus of P is a straight line cutting the axes of v and y at distances AB, AC =* from A. 12. Let ABCD (fig. 48) be the given square whose side AB ='a; P a point in the required locus, draw PM perpen- dicular to AB, and let AM =a, PM =y; then AP =att+y', BP? =(a- a) +9 DP? =«°+(a-y), CP=(a- a) + (@-y)s , AP? + BP? + DP? + CP? =2§a° +(a-) +y°+(a-y)*} = : J ce — 40° ee ee A eye LY ae ONE a\? @& —2a and (» -¢ +{y-= = , 2 3 2 4 the equation to a circle the co-ordinates of whose centre are a : 2 > and radius = 3 \/c? — 2a’. wo | & GEOMETRICAL PROBLEMS. NO. Vv: DEC. 1834. 3S Hence in order that the problem may be possible c? must not be less than 2a’. 13. Let the straight line joining the two points 4, B (fig. 49) be taken for the axis of w; C, the middle point of AB, the origin; AC =c; then the equations to the two lines AP, BP are y=m(w+c), y=m(«-—Cc) respectively. And if (a) be the angle between them ; HW, m—m U-—-C &+C 2cy pL ee ret Oye ee ee ae ogee ao Tg 5 1+mm y it Ua GC -F ea Ce . 2 +y?—c =2ccotay, and x + (y — cota)? = (ccoseca)’; which is the equation to a circle whose centre is in the axis of y at a distance c cot a from the origin, and radius = ¢ coseca. 14. Let AB, BC, CD, DA (fig. 50) be the four sides of the base of the parallelopiped ; ab, bc, cd, da the correspond- ing sides of the top; then ACea will form a parallelogram whose sides are two diagonals AC, ca of the base and top, and the two sides 4a, Ce of the parallelopiped ; and the diagonals Ac, Ca of this parallelogram will be two diagonals of the parallelopiped. Similarly, BDdb will form a parallelogram having its two diagonals Bd, Db the two remaining diagonals of the parallelopiped. Now if dB=a, AD=b, zBAD=a; BD=d, AC =d, we have d’ =a’ +b? —2abcosa, and d* =a’ +b’ + 2abcosa; od +d” =2(a’ +0’), or in any parallelogram, the sum of the squares of the diagonals is double the sum of the squares of the sides. Hence Ac + Ca® = 2(AC’ + Cc’), Bd + Db =2(BD + Bb’) = 2(BD + Ce’); - Ae? + Ca + Bd + DP =2(AC? + BD’) + 4Ce’ = 2{2(AB’ + BC’)} + 4Cc* = 4(AB’ + BC + Ce’) = the com of the squares of the twelve edges. C 34 GEOMETRICAL PROBLEMS. NO. V. DEC. 1834. 15. Let O (fig. 51) be the vertex of the pyramid; OA, OB, OC the three sides at right angles to one another = a, b, e respectively: then AB =a? +B, AC=Vae+e, BCH=a VU +e, 2 acme 2 2 and Be IOC ee 2A4B.AC / (a? + b’) (a? + 0) - sin Z BAC= Vabi+ ae + be i Cte VJ (a? + b%) (a? + c*)” and AABC=}AB.AC.sin 2 BAC =4/ 00 + ac? + Be’; or (AABC)? = (=) : (ar 2 ea = (A AOB)? + (A AOC)? + (A BOC)*. GEOMETRICAL PROBLEMS. NO. VI. DEC. 1835. ao ST JOHN’S COLLEGE. Dec. 1835. (No. VI.) 1. Iw some treatises on Geometry it is laid down as an axiom more evident than Euclid’s 12th, that two straight lines which cut one another, cannot both be parailel to the same straight line. Shew that this is only a disguise of Euclid’s axiom. Give an instance to shew how some of the fundamental theorems of Geometry may be proved a priori from considera- tions purely analytical. Two solid angles may be unequal, which are contained by the same number of equal angles in the same order. 2. If four magnitudes be proportionals, they shall also be proportionals when taken alternately. Prove by taking equi- multiples according to Euclid’s definition, that the magnitudes 4, 5, 7, 9 are not proportional. 3. Similar triangles are to one another in the duplicate ratio of their homologous sides, How does it appear from Euclid that the duplicate ratio of two magnitudes is the same as that of their squares ? 4, ) each other in the hyperbola whose equation is y’ = and whose axes are 0, 20. 2 o As 6 diminishes, the hyperbola y’ = ne approaches to 40 GEOMETRICAL PROBLEMS. NO. VI. DEC. 1835. 2 ae the two straight lines represented by the equations y° = tt or y= ata ; hence the parabolas ultimately intersect in the two straight lines y = BS and y = — Bt The straight lines mani- festly touch all the parabolas represented by the equation y? = a (w — a) when every possible value is assigned to a. . b? Fi b hee rah PP ear yee a*b hence CP =a’ 4+y’? = a+b /a+b; CD = CP -a@ +B = 0? + ba? +0'; a’ b? CD hence PY? = CP? — CY’ =a'?+06? or PY=CS. 10. or @ =a? + CY? = PF = "7 2b PsP; 11. Taking the centre of the hyperbola for the origin of co-ordinates, the equation to the circle whose centre is H and radius 6 is (w+ ae)’?+y* = 6’; and the equation to a tangent at the point wv, y is , e+ ae ; Boe ohh - (a — x), or yy + (a#+ae)a’ =y’ + w (wt ae) =6?-aex -—-ae=—-a’-aeuv; now if this passes through a point h, k, . ky +(a@+ae)h=—a’ -aex, or ky+(h+ae)v=-a@- aeh; (1) which is the equation to the straight line joining the two points GEOMETRICAL PROBLEMS. NO. VI. DEC. 1835. 41 of contact, when tangents are drawn to the circle from the point h, kh. The equation to the circle described on the transverse axis is v +y?= a’; and the equation to a tangent at the point v,, y, is wxv,+yy,= a’; and in order that this may coincide with equation (1), we must have My h+ae a, h+ae —> = LL a* a +aeh a a+eh k k and vA Ty wa Oe ; a a*+aeh a ateh KP + (h+ ae)’ | (2 (a + eh)’ “ =| + go eS als es ris) i pom or kh? —(e—-1)h? +a’ (e®-1)=0; hb? Ke? ; , : hence — — ja = 1, which shews that the point h, k is a point a in the hyperbola. 12. Let tangents be drawn from a point whose co- ordinates measured from the centre are h, k; then the equation to the straight line joining the points of contact is he ky : Be winb) Aa 4 let this pass through the focus; therefore when w = ae, y = 0; he alee, : : hence a i 1, or 4 =-: which shews that the point h, & lies e in the directrix. 13. Let PK, DK’ be two normals at the points P, D whose co-ordinates are #, y and «’, y’ respectively ; then 4 bt PR*=y'+—a°; DK? =y? + — 2"; a Gat: and va; poh 42 GEOMETRICAL PROBLEMS. NO. VI. DEC. 1835. b? b? b? ve PR? DK? x (1 +3) y +- = (1 += )a? a a a b? 2 2 b? = 6 (1 +5) (f+ =) = (a° +b"), b? a’ a” and is constant. 14, Let C (fig. 55) be the centre, P,, P,,...P>, the points of division of the circumference ; T 27 LACP, =0@; one Ce Cir Oe AC Be Uae ce &e. and SP}=SC’+ CP!-28C.CP,cos0 = SC? + CA? -2.SC.CAcos6; SP? SP? SP?, = SC? + CA? -2,§8C.CAcos (047) : 2 = SC? + CA? —2,SC.CAcos (0+="); 2n—I1 = SC+ CA -28C.CAcos) 94" - (SP? + SP2+...4+ SP?) =2n(SC? + CA’) - 280.CA foos 0 + cos (0 +=) +... + cos (0+ env) {sin (2r+0-74) -sin(o-2)| nm 2n 2n “ =2n(SC2+ C4’) —SC.CA . T sin — Qn = 2n (SC* + CA’) = n (AS? + SB’) since AB is bisected in C. 15. Let V be the volume of the pyramid; S,, S82, S83, S; the areas of the four triangular faces opposite to the angular points 4, B, C, D respectively ; then . 3 GEOMETRICAL PROBLEMS. NO. VI. DEC, 1835.. 43 r Bag Sic So + Ss + Si) 5 ? V= 3 (Se + Ss + Ss — 51) 5 ? V= = (Si + Ss +S — 52) 5 Ue V= — (S, + S, + S, — S35); oO "4 V= g 081 + Sat Ss — 54) 5 1 1 1 1 2. hence V(—+— +4) =3 (S454 548) = . 4 “aor, 4:4, GEOMETRICAL PROBLEMS. NO. VII. DEC. 1836. ST JOHN’S COLLEGE. Dec. 1836. (No. VIT.) 1. Iw equal circles, angles whether at the centres or circumferences have the same ratio as the ares which subtend them. 2. Every solid angle is contained by plane angles, which together are less than four right angles, 3. If two pairs of common tangents be drawn to two un- equal circles, and 2a, 2a’ be the angles which the two of each pair make with each other ; then sina R-r sin a’ R+r 4. A, B, C, D are four points in order in a straight line, find a point FE between B and C such that 4F.EB=ED.EC by a geometrical construction. 5. If from the centre of a rectangular hyperbola a line be drawn through the point of intersection of two tangents ; and if @ and @ be the angles which this line and the chord joining the points of contact, respectively make with the real axis; then will tang. tan ¢’ = 1. 6. There are any number of ellipses having a common centre, and their axes majores in the same position. Shew analytically that if all the ellipses be twisted through the same angle @ in the same direction, the loci of the intersections of each ellipse with its original position, are two straight lines whose equations are =wtan—, and y= —zcot-. i] Q° Y 9 7. ‘Two given unequal circles touch each other externally ; shew that the locus of the centre of the circle which always touches the other two is a hyperbola. Find the axes and eccentricity, and shew what the figure becomes when the given circles are equal. GEOMETRICAL PROBLEMS. NO. VII. DEC. 1836. 45 8. A vessel whose outward figure is a paraboloid of revolution, is required to be of equal thickness throughout ; find the figure of the interior surface. 9. In an ellipse, if through the foci S and H, chords PSP’, and QHQ’ be drawn parallel to any pair of conjugate diameters, shew that SP.SP’+ HQ.HQ =02+2? where b and 7 are respectively the semi-axis minor, and semi-latus rectum. 10. In any circle draw a chord AB: from the middle point of the lesser segment draw any line cutting AB in C and meeting the circumference in D; join AD and take AP = AC; find the locus of P. 11. Round a given ellipse circumscribe a rhombus; about this rhombus circumscribe a second ellipse, and so on for m times; prove that all the ellipses are similar, and find the sum of the areas of the 7 ellipses. 12. An ellipse has a square described touching it at the extremities of the minor axis: an ellipse upon the same axis- major circumscribes the square. ‘This ellipse is dealt with in the same manner as before, and the operation is continued till there are altogether mn + 1 ellipses; prove that if the original eccentricity = ———— the last ellipse becomes a circle. S/n +i 13. Given the equation Ay’ + Bay + Ca’ + D=0 to be the equation to the hyperbola; find the position of the asymp- totes, and the equation to the hyperbola referred to them as axes. 14. Find the axes and position of the curve represented by the equation y —20y + 3a’ 4+ 2y-—4a-3=0. 46 GEOMETRICAL PROBLEMS. NO. VII. DEC. 1836. SOLUTIONS TO (No. VII.) . 1. Euctip, Prop. 33, Book v1. 2. Euclid, Prop. 11, Book x1. 3. If C, C’ (fig. 56) be the centres of the two circles; DDT, ETE’ common tangents to the circles, meeting CC’ in J, T" respectively; join CD, C’D', CE, C’E’; then if LCTD=a, LE'T'C =a’, we have YM pubic LAr es sin a= 2 ey CC’ sina R-r sina = sing’ Rir 4. Take any point F (fig. 57) not in AB; about the triangle ABF describe a circle 4BF'G, and about the triangle DCF describe a circle DCFG. Let the circles intersect each other in G; join GF and produce it to meet 4D in E; then EA. ERB=EF .EG = EC. ED. 5. Let h, k& be the co-ordinates of the point of inter- section of the two tangents; then tan p= z° and the equation to the line joining the points of contact, when tangents are drawn to the hyperbola from the point h, k, is hex _ ky ‘a oe 2 bh b 2 tan p= — 7 or tan g tan p= —; and if the hyperbola be rectangular, tan @ tang’ = 6. Taking the centre for the origin, the polar equation b? to the ellipse is p? = ————_——_ ; and the polar equation to GEOMETRICAL PROBLEMS. NO. VII. DEC. 1836. 47 the same ellipse when the axis-major is twisted through an b° 1 — e’ cos’ (fh — 8)” points of intersection of the ellipses in the two different positions, cos’ d = cos” (fd — 9), - P=O0-g, or T+O—-; le. P= both which values are independent of the eccentricity and magnitude of the axes. Hence every corresponding pair of ellipses will intersect each other in two straight lines passing angle @ becomes p" == therefore at the Q zr 0 =, OL — =; Q° Bae oie 7 0 through the origin and inclined at angles > and 4 + : to the axis-major; or the equations to the two lines will be 0 = w tan— =-—w#cot—-. Y 9? y 9 7. Let S, A (fig. 58) be the centres of the two circles whose radii are 7, 7’; P the centre of a circle touching both circles: then SP — HP = SQ-— HR =r -— 7’, and is constant, or the locus of P is a hyperbola whose foci are S, H. If 2a, - 2b be the axes of the hyperbola, 2a=r—7r, 20e=SH=ri?r'; ? r+r os ae e= aan 2b=2aVfer—1=2\/rr’. —f? Le When r=7", SP — HP = 0, or P lies in the straight line which is drawn perpendicular to SH bisecting it; therefore the locus of P in this case becomes the common tangent. 8. Let 0 be the angle which the normal PG (fig. 59) to the parabola makes with the axis AG; XY, Y the co-ordinates of P; in PG take PQ = 65, then the locus of Q will be a curve which by its revolution round AQ will form the inner surface of the vessel. ye Since the subnormal = 2a, tan 9 = aes a * Y=2atan@, X =a tan’ @; 48 GEOMETRICAL PROBLEMS. NO. VII. DEC. 1836. hence if aw, y be the co-ordinates of Q, v= X +bcos@ =a tan’@ + bcos 6, y= Y -bsin@ = 2a tan@ — bsin@; 1 — 400 = - ae + 6? (1 — cos’ @), or 6? cos? @ + (y? — 4av — b’) cos8 + 4ab=0; (1) and bcos*@—(w+a)co’O0+a=0. (2) Multiply (2) by 6 and subtract from (1) ; . b (a +a) cos’ 0 + (y® — 4aa@ — 6’) cos0 + 3ab=0. (3) Multiply (2) by 4b and subtract (1), or 3b’ cos*@ — 4b (a + a) cos0 — (y?- 4ax-—b*)=0. (4) Multiply (3) by 3b, and (4) by (# + a) and subtract ; . {3b (y? — 4a@ — b’) + 4b (a + @)*} cosO + 9ab’+ (w@+a)(y’ — 4av — 0b’) =0. Multiply (3) by y’ — 4a@ — 6’, and (4) by 3ab and add’; . §b(w + a) (y’® — 4a” — 0”) + 9ab*} cosé + (y? —4au — 0’)? - 12ab? (@ + a) = 0. Hence Soab’ + (@ + a) (y? — 44a - b*)}? ={3(y’-4aa—b’) +4(w+ a)*} 3 (y?— 44a —b*)? - 12ab*(w+a)}, which is the equation required. 9. If Aa (fig. 60) be the axis-major, and a’ be the semi- diameter parallel to PSP’, we have PS.SP’ : SA.Sa:: a? : @; b2 a’ , a* Or (ePStSP) = b? Similarly, QH.HQ’= at b?; b? pe PS .SP’ + QH. HQ’ = = (a? os. b’) 2} b? = 3 + b*) = b? at ft GEOMETRICAL PROBLEMS. No. VII. DEC. 1836. 49 10. Let # (fig. 61) be the middle point of the circum- ference AB; join AH, EB; and let LEAB=a, LEAD=$9; : thn LEDA=LCHBA=a, LAED=7-(P+a), LECB=LEAC+LAECHr-; «. LACE=90; and if dE =a, AP=AC=p,, ag oe 8) sin b the equation to the locus of P. 11. If a’, b’ be the equal semidiameters of an ellipse, and a the angle between them, 2ab a’ —b’ 5; oF Cosa = 2a" = a? +67, and, sin a*s= ——— es 08 : a” + D* abs and the tangents at the extremities of these diameters will form a rhombus whose side = 2a’, and whose diagonals are , e a , a 4a Des and 4a oe . Also the diagonals of a rhombus are at right angles to one _ another; and if an ellipse be described upon the diagonals as axes it will circumscribe the rhombus; let a,, 6, be the semiaxes, then 1—cosa & Qa Steals 1s ee = 4 2 oe Ze re 9 2? I1+cosa a’ a, or the ellipse will be similar to the original ellipse; hence if A, A, be the areas of the two ellipses, ofthe 3 a aa A, =74,b, = 47a” sin a deat 27a’ sina =27rab = 2A. Similarly, if A,, 43,....4, be the areas of the second, third, and nm" ellipses described in the same manner, A, =2A, = 274, A, = 2A, = 2°A, &e. and 4,+ 4,+...+ A, =(24+27+274+...4+ 2°.) A fe (2°t} a 2) A= (255; _ 2} a ab, D 50 GEOMETRICAL PROBLEMS. NO. VII. DEC. 1836. In this problem it will be necessary to assume, that the ellipses are all described upon the diagonals of the successive rhombuses as axes. For if ABEF (fig. 62) be a rhombus circumscribing an ellipse, 4H, BF its diagonals intersecting one another in C, 2 b? AC=6, BC=6; then CP? = CD’ = — : 8° 4.8? = AB’ =4CD? =2 (a? +B’), 60 =2A\ ACB =40 ABEF =2ab; 2 ao muy Yy .Oo= \/2a,0= \/2. an A ag ae + oR will be the equation to an ellipse em haere the rhombus, where m may have any magnitude less than unity. Let e’ be the eccentricity of this ellipse; then 1 1 Bare f 20 20° fi coy. 1 LA? m\?° ssa; ay. 2—e” 2—¢ A omen Sa /e! + 4m? (1 — e’) 2 e f e hence —,—1 is never greater than —— 1; or e’ is never less e e than e, but may have any value greater than e. 12. Let P (fig. 63) be one of the angular points of the square; then if 6, be the semi-axis minor of the ellipse circum- scribing the square, the co-ordinates of P are 6, b, e b° 1 1 1 pei? or — —-— =~—; ees hileye slouch be the semi-axes minor of the 2™%, gtd, ellipses ; —— eee GEOMETRICAL PROBLEMS. NO. VII. DEC. 1836.: 51 J 1 a ae 1 i n Bee : ‘ : 1 m+ and if b,=a, the n™ ellipse becomes a circle, or = = —,—, b a b? ri and ,=1-é= 5 Pe AY/ eee a Tarai 1 n+l 13. Ay’?+ Bay+Cv=-D; D AG ert) se ere v y buyers. D hence eae fipe Aa? and the equations to the asymptotes are _-B+VB- 4dC 2A Pi If a, a’ be the inclinations of the two asymptotes to the axis of v7; , B , tana + tana = ——; tanatana men Fe A and if a’, y’ be the co-ordinates of any point referred to the asymptotes as axes, @©=# cosaty cosa, y= sina+y sing’; +, (Asin? a + Bsinacosa + Ccos* a) &” + (Asin’a’ + Bsina’ cos a’ + C cos’ a’) y” ‘+ 2A sinasina + Bsin(a + a) + 2C cosacosa’) ay’ + D=0. D2 52 GEOMETRICAL PROBLEMS. NO. VII. DEC. 1836. Now the two first terms vanish ; , A , B , ya .. 2C cosacosa qtanatana +5 (tana + tana’) +1 vy+D=0, 2 24C ° , / Nowe ota) Bo cos(asta) ati Cy or 2C cosacosa’ (2 - )a'y + D =o. cos a cosa’ A’ cosacosa’ Ati 1 _V(4A- C+ Be “ cosacosa’ A : ay A D VASO . ey =—____, ——___ = —_________.. D cosacosa B* —4AC B* —4AC 14. The curve is an ellipse since hee eee Wags Let a, ( be the co-ordinates of the centre, and let the origin be transferred to the centre by making v=a'+a,y=y +P; “ (y'+B)°—2(@' +a) (y'+ B)+3(a' +a)*+2(y' +B) -4(e' +a)-3=0; hence 28 —-2a+2=0, —-28+6a-4=0; ° a=4,8=-4; and the equation to the ellipse becomes 2 Poe? lo 9 Y —-2LRY +3" — ar 0. Let @ be the inclination of the axis of the ellipse to the ay? axis of a, and let 2 + a 1 be the equation to the curve referred to its principal diameters ; transform the axes through an angle 0, then (x cos@ +y' sin 0)? (y’ cos@ — a’ sin 8)? 2 ns sie? a b* cos’@ sin?@ 6 - be pase NO. VII. DEC. 1836. 53 GEOMETRICAL PROBLEMS. sin?Q@ cos*@ 2 1 ; ee 53 5h and sin 20 (5 -;)- ci a Gs 8 1 1 4: Pa? and cos 20 (= -3) a a 1 hence — + — i eee or tan 20 = —1, and 20=135; ee Ry eh ies ere Poa 9. as 9’ 2 8-4/2 ‘ 9 9(2-+4/2) DE ings », Ora = == — ’ a 9 4 — 2/2 4 and b2 =" (2-2); T a. “LF hence a= 8c0s—, Las 54 GEOMETRICAL PROBLEMS. NO, VIII. DEC. 1837. ST JOHN’S COLLEGE. Dec. 1837. (No. VIII.) 1. Iv two triangles which have two sides of the one pro- portional to two sides of the other be joined at one angle so as to have their homologous sides parallel to one another; the remaining sides shall be in a straight line. 2. If a solid angle be contained by three plane angles, any two of them are greater than the third. 3. If from any point in the diagonal of a parallelogram, straight lines be drawn to the angles, the parallelogram will be divided into two pairs of equal triangles. 4, Shew how to find the focus of a traced conic section. 5. From three given centres describe three circles touching one anothier. 6. SY, HZ are perpendiculars from the foci on the tangent at P to an ellipse whose centre is C; SP, HP cut CY, CZ in Q, R; shew that CQPR is a parallelogram. 7. Let the two circles, radii R, r, which touch first, the three sides of a triangle ABC, and secondly one side BC and the other two produced, touch 4B in D,, D,, AC in E,, E.,; shew that BD,.BD, = CE,.CE, = Rr. 8. The side of an equilateral hexagon inscribed in an ellipse, eccentricity e, with two sides parallel to the axis major : side of one inscribed in the circle on the axis-major 2: 4—-2e : 4—e?, 9. If one of the co-ordinates of the centre of the curve GEOMETRICAL PROBLEMS. NO. VIII. DEC. 1837. 55 0 ay’ + bey+cau°+dy+exuv+f =0, assume the form — , shew that the equation becomes ba+d 1 cathe Aten. =- + —/d? — 4 Y 2a SEA af, and explain the meaning of it. 10. AP is a parabola, vertex A, focus §; T' the point where the axis intersects the directrix; join PY’ and produce it to meet the latus rectum in NV; draw SPQ to meet NQ, which is parallel to S'7’ in Q; and shew that the locus of Q is a circle. 11. If O be a point in the directrix of a parabola; and OA=a, OB =b, tangents at dA and B; shew that the equation to the parabola referred to OA, OB as axes, assumes the form Ree es a b 12. Shew that fe + we = 1, is the equation to a a 4a? b? parabola whose latus rectum = ae 13. If in (11) any tangent to the parabola cut OA, OB in P and Q; shew that OP OR es CN Od TORS Ae where OP or OQ is considered as negative, if P or Q lies in AO or BO produced backwards. 14. Two parallel planes revolve in their own planes about fixed points 4, B, in the same direction with equal angular velocities; shew that the curve traced upon the first by a pencil P fixed perpendicular to the plane of the second 56 GEOMETRICAL PROBLEMS. NO. VIII. DEC. 1837. is a circle: or if the planes revolve in opposite directions the equation to the curve is a” — 2arcos0+7" = ( C where 4 is the origin, BP =a, AB =c, and the prime radius is the line originally in the position AB. 15. ABCD is any quadrilateral. Bisect 4C, BD in E and F': EF is the locus of the centres of all the inscribed ellipses. 16. Shew that all lines drawn from an external point to touch a sphere are equal to one another; and thence prove that if a tetrahedron can have a sphere inscribed in it, touching its six edges, the sum of every two opposite edges is the same. ! GEOMETRICAL PROBLEMS. NO. VIII. DEC, 1837. 57 SOLUTIONS TO (No. VIII.) 1. Evucurp, Prop. 32. Book v1. 2. Euclid, Prop. 20. Book x1. 3. Let ABCD (fig. 64) be the parallelogram whose diagonal is AC; E any point in it; join DE, EB; then since AACD = A ABC, the perpendicular front D on AC equal the perpendicular from B on AC; hence the altitudes of the triangles ADE, ABE are equal; and they are upon the same base, therefore AADE = AABE. Similarly A DEC = ABEC. 4. Find C the centre of the ellipse (fig. 65) by joining the points of bisection of two parallel chords; take any point D in the curve, and with centre C and radius CD describe a circle cutting the ellipse in the four points D, E, F, G; through C draw Ad’, BB’ parallel to DE, EF respectively ; these will be the two axes; and with centre B and radius = AC describe a circle cutting 4A’ in S, H; these will be the two foci required. 5. Let A, B, C (fig. 66) be the three given centres; find O the centre of the circle inscribed in the triangle ABC; draw Oa, Ob, Oc perpendicular to the three sides BC, AC, AB respectively; then 4b=Ac, Ch=Ca, Ba=Bc; and the three circles described with centres 4, B, C and radii Ac, Be, Ca, respectively, will touch one another in the points a, 0, ¢. 6. Produce HP, SY (fig. 67) to meet in V; then since ZSPY = 2YPV, SY=YV, and HC =CS;; therefore HP is parallel to CY; similarly SP is- parallel to CZ; or CQPR is — a parallelogram. 7. BD,=S-b; BD,=S-c; CE,=S-—c; CE,=S—-); Bal (SSW, 58 GEOMETRICAL PROBLEMS. NO. VIII. DEC. 1837. S'.1S —b — fesolne/f Serko oF or BD,. BD, = CE,. CE; = Rr. Rr = (S- 6) (8-0). 8. Let APQa (fig. 68) be half the hexagon, Aa being one of its diagonals; then if x, y be the co-ordinates of P measured from the centre C, PQ = 22a, AP=22; and (a-«2)’?+y?= AP’ = 42°, or 30° + 2a —a’ = (1 — e’) (a — 2”); 2 — e - Ca) ct li ee ey AOL a = a3 —e 4 — 2¢° j : ; ; and 2v = Satine and the side of the hexagon inscribed in —e the circle on the axis major =a; therefore the side of the hexagon inscribed in the ellipse: the side of the hexagon in- scribed in the circle on the axis-major 2 4—2e:4— e?, 9. Let a, 8 be the co-ordinates of the centre; transform the origin to that point by making e=U+a y=y +B; aly + BY’ +b (a +a) (y+ B) +0 (e' + a)? +d(y+B)+e(a'+a)+f=o; hence 2a8+ba+d=0; b8+2ca+e=0; _ 2ae—bd b* — 4ac 2ae—-bd=0; &—4ac=0. 0) a ; and when a assumes the form -, O In this case the equation to the curve becomes (2ay + bw)’ +2d ay+bez) +4af =0, "1 bx+d ie ( 2a ey ee +— Vd -40f (1) _which represents the equations to two parallel straight lines ; GEOMETRICAL PROBLEMS. NO. VIL. DEC. 1837. 59 and any point in the straight line whose equation is yo - (“= **) 20 will be equidistant from the two straight lines represented by equation (1); therefore any line drawn through that point to meet the two lines will be bisected in the same point. In this case the centre is not limited to a single point ; be+d ., will but any point whatever in the straight line y = — bisect every line passing through it and terminated by the two straight lines represented by the given equation; and will therefore satisfy the definition which has been assumed for the centre. 10. Let AM (fig. 69) =2, MP=y; draw QN’ per- pendicular to 77'S, and let SN’ =a’, NQ=y’'; 20a. then y= SN = or AP aG+wxv SM a- 2X Ca a, N= “ — and & UP Q F y ee a-@ G+" 2a beet lend Usha Ys and sO ONSEN lado Uoge-Y y y y 2ay 2a Hence by multiplication 4a? — aw? 12 My re a =1 or 7° +y" =40. The equation to a circle whose centre is 8’ and radius =2a= ST’. 1] and 12. See Appendix, 1. Art. 19. wv 13. Let ve + se = 1 be the equation to the parabola ; a & 1 A : é and — + ce 1, the equation to the tangent ; then, at the point m n in which the tangent meets the curve, 60 GEOMETRICAL PROBLEMS. NO. VII. DEC. 1837. @ v eat y=b(1-2 +o )an 1——]; | aa m and since w has only one value, the quadratic equation (G+ 2) 0-9 as @-n) m0 am /a must have its two roots equal ; b? bY in BG bn bn n? —=(6-n)(-+-—)=— —_ -— -—; a a em a m a m b b n mn or —=-+4-; —+—-=1. mam WR iad 14. First, let the plane of the paper which represents one of the given planes revolve round the point 4 (fig. 70) ; and let the other plane revolve round a point whose projection upon the plane of the paper is B. Let the point P describe the angle PBP' round B; then if the plane of the paper revolve in the same direction round A, the straight line AB will move into the position AB’, so that BAB = 2PBP = dQ; since the angular velocities are equal; and if AP =r, 4 P'AR' = 0, the relation between 7 and @ will be the polar equation to the curve required. Let AB=c, PB = a, LABP=a, then PBA = ota, and AP B=7- S(p + a) +(9-¢)} =7-~(0+a); =a’ +r + 2ar cos (0+ a), which is the equation to a circle, Secondly, suppose the planes to move in opposite direc- tions, then AB will move into the position 4B,, and if 4 PAB, = 0, we have LPAB=9 +0, and LP BA=p+ a; GEOMETRICAL PROBLEMS. NO. VIII. DEC. 1837. 61 *. rsin(@ + 0) — asin (Pp + a) =0, and 7 cos(f + 0) + acos(@ +a) =c; or 7 cos (@ + 0) — acos(P+ a) =c — 2acos(P + a) ; and by adding the squares of the two last equations r? —2ar cos (9 -a) +. a° = $c — 2a cos (p +a)? e+e rr? a? — r2 el Ck ee : Cc Cc 15. Let A (fig. 71) be a point without a sphere whose centre is O; AP any straight line drawn from A touching the sphere in P; let the plane 4PO cut the sphere, the section will be a circle, and AP will be a tangent to this circle ; .. AP? = AO’ — OP’, and is the same for every position of P. Next let A (fig. 72) be the vertex, and BDC the base of the tetrahedron ; and let the sphere touch AB, AC, DC, DB in the points c’, b’, b, c respectively ; then , Ac = Ab’*, Cb’=Cb, Db=De, Bc=Be'; “ De+cB+ Abl'+ Cl’ = Db+ Be + Ac’ + Cb, or BD+ AC =AB+CD; and in like manner it may be proved that AB +CD= BC + AD. 62 GEOMETRICAL PROBLEMS. NO. IX. DEC. 1838. - ST JOHN’S COLLEGE. Dec. 1838. (No. IX.) 1. Mewnrion the principal methods that have been pro- posed for establishing the theory of parallel straight lines ; and shew that the following principle will suffice. Through any point within an angle, a straight line may be supposed to pass which shall cut the two straight lines that contain the angle. 2. Give Euclid’s definition of proportion ; and apply it to shew that in equal circles, angles at the centres have the same ratio which the circumferences on which they stand have to one another. 3. Every solid angle is contained by plane angles which together are less than four right angles. 4. If from the point where the common tangent to two circles meets the line joining their centres any line be drawn cutting the circles, it will cut off similar segments. 5. Of the two squares which can be inscribed in a right- angled triangle, which is the greatest ? 6. Construct all right-angled triangles whose sides shall be rational, upon a given straight line as their base. 7. In the sides 4B, AC of a given triangle ABC, take two points M, N; and in the line joining them, take a point MB AN MP P such that Wham NG = pi? prove that if PB, PC be joined, the triangle PBC is twice the triangle AMN. 8. In No. 7 the circle described about the triangle d4MN will always pass through a fixed point. 9. Draw the straight lines represented by the equation (Qy-a+c)(3y+u-—c)=0; _and determine where they intersect, and at what angle. GEOMETRICAL PROBLEMS. NO. IX. DEc. 1838. 63 10. Find the equation to the line which is equidistant from the two lines represented by the equation y=me+cex#e. 11. The circles represented by the equation (2 +1) (a+ y’) = ax + bny, when ” assumes various values, will have a common chord. 12. Find the locus of a point at which the base of a triangle subtends an angle equal to the sum of the angles at the base. 13. In No. 7 find the locus of P. 14. If the vertex and nearer focus of an ellipse remain fixed, whilst the centre moves in the line joining them to an infinite distance, the curve will become a parabola; but if it move in the opposite direction, the curve will become succes- sively a circle, a straight line, hyperbola, and parabola. 15. An ellipse and hyperbola that have the same foci and centre will cut one another at right angles. 16. In No. 15 if from any point in the circumference of the circle which passes through the points of intersection of the ellipse and hyperbola, tangents be drawn to those curves, they will be at right angles. 17. Trace the curve whose equation is y=Lv424\/— a — 3x +10. 18. If the length of the axis of an oblique cone be equal to the radius of the base, every section perpendicular to the axis will be a circle. 19. In the general equation of the second order ay’ + bey+cu’+dy+ex+ f=0; shew what the curve which it represents becomes when be —4ac= +o. 20. Find the conditions in order that two given equations of the second order may represent similar and similarly situated curves. 64 GEOMETRICAL PROBLEMS. NO. 1X. DEC. 1838. SOLUTIONS TO (No. IX.) 1. Sex Potts’ Euclid, p. 50. 2. Euclid, Def. 5, Book v. and Prop. 33, Book v1. 3. Euclid, Prop. 21, Book x1. 4. Let C (fig. 73) be the point in which the common tangent CDE meets the straight line CBA joining the centres of the two circles; CFGHK a straight line cutting both circles; join BG, BD, BF, DG, DF, AK, AE, AH, EK, EH; then because the angles CDB, CEA are right angles, the triangles CDB, CEA are similar ; eCOD CE: CB: CA: BD SEAg UBF : AH Ort fe hence the triangle CDF’, CEH are similar, and DF is parallel to EH. Similarly, DG is parallel to HK, «. 2F DG =2zKEH, and the segments PF DG, HEK are similar. 5. Let # be the side of the square abcd (fig. 74) inscribed in the triangle ABC, and having one of its sides ed on the hypothenuse 4B; then Ca=wusn A, Ba=«xsec A, or aw (sin 4 + sec A) = BC =a; acos A x 1 + sin 4 cos 4 Next, let w’ be the side of the square Ca'c'b’ (fig. 75) which has two of its sides coincident with the two sides CA, CB of the triangle; then CB = Ba + Ca’, or aw tan4A +a’ =a; . acos A sin 4 + cos 4° Now (1 — sin 4) (1 — cos A) is positive ; “. 1—sin 4 — cos A + sin Ad cos 4 >0, or 1 + sin Acos 4>sin A + cos 4; hence w is less than a; and the square which has one of. its angular points in the hypothenuse is the greatest. GEOMETRICAL PROBLEMS. NO. IX. DEC. 1838. 65 6. Let 2 and mw be the two sides; then 1 2 x (1 +m’) = (hypothenuse)? = 2” (m + -| : n gh teas 2m Ee Ue ee hs. OPT Fm =m + hes. = 3 n n Qn and if # = 2ma, the sides of the triangle are 2na, (n’ — 1)a;3 and the hypothenuse = (n? + 1)a. Hence if AB (fig. 76) be the base, divide 4B into 27” equal parts, and let 4D, DB contain n +1 and nm —1 parts re- spectively, and BE one part; then if BC be a fourth pro- portional between BE, BD and AD, BC =(n?—-1)a, and AC = (n* +1) a; or ABC will be the triangle required. 7. Let 4M (fig. 77)=a, MB=na; .. AB =(n 4 1)a; NC=6b, AN=nb, AC = (n+ 1) 6; PNe=c, PM =2=nce, MN =(n +1) ¢; “ AABC=1 AB. AC.sin A =4 (n+ 1)’ absin A; AAMN=44AM.ANsin A=4n ab sin A, 1)? ON BC a A AMN ; PN.NC 1 OR AN Me oe ln AEN? Seo aR AEN. AN n.(n +1) PM.MB n? ia aeaperrane ee A = M. 3 APMB ward 24 Ae oe N - A PBC = 6 ABC— AAMN —- APNC ~ a PMB 2 1 2 -{e2% BG [ene ee - "| 6 AMIN = 24MIN, n m(n+1) nm+1 Sct 1h Cena ee ah ty OB n+l n+ and the equation to the circle passing through 4, M, JN, re- ferred to 4B, AC as axes becomes v+y —-2eycosA=a,x + by, E 66 GEOMETRICAL PROBLEMS. NO. IX. DEC. 1838, but when y 0, 7=AM=a,; cms a, = AL: n similarly, when # =0, y =b, = AN, or b, = : n and the equation to the circle is v+y?—2vy cosA = gue y a wie | Hence if aw = Py; a’ +y’-2eycosA=axv= By; from which equations two values of # and y may be found in- dependent of n; therefore all the circles will pass through the two points so determined; hence they have a common chord a(3” Ba Spare? 5 = 7 pme & (BC)? (BC) the co-ordinates of the common point through which they all pass. whose equation is aw = Sy; and w= re 9. The equations to the two straight lines are 2y—e+e=0, and 8y+u-c=0; and by addition 5y=0, or y=0, and wa =c; hence the two straight lines intersect in the axis of w at a distance c from the origin. Let m, m’ be the tangents of the angles which the two straight lines make with the axis of w; and @ the angle between them; .. m=4, m'=—-1, é 1 1 m—m ha Nieto 2 and tan Qisp 322 i Se ie: pO aa 1+mm 1-4 10. Let ecosat+ysina=p, (1) wcosatysina=p, (2) xcosatysina=p, (3) be the equations to three parallel straight lines, whose perpen- dicular distances from the origin are p, p’, p” respectively ; then if the second straight line be equidistant from the first and third, DEC. 1838. 67 GEOMETRICAL PROBLEMS. NO. IX. and if cotana =—m, the equations are , a” Pp : yY=met+——, Yomu+——, Yume +—-; sin a sina sna or the second equation is ‘ Ad a i{ P Pp } Se Po oma te ke Y z (* a sina Hence in the proposed example the equation becomes, y=maetdaf(ctc)t+(c-—c)i=mare. 11. Equating the coefficients of n, we have e+y=by, and a +y=a2; hence the straight line whose equation is av = by will meet all the circles in the same point, or the circles have a common chord whose equation is aw = by. Let C be the vertical angle, and Z2C=24-2DRD= 7 —C ; ZC = 90°, and the locus of C is a circle described on the 12. . diameter AB. 13. Draw PV (fig. 77) parallel to AC meeting AB in V; and let 4B, AC be taken for the axes of w and y respectively ; then if AV=a, VP =y, ] AM=(n+1)a@;3 Ay = "+ ) : AB =(n+1)AM=(n+4+1)v=a; 1 ri? 4c ="** 4n = ("=") aye n n 7 os d y 1 n re ror — = ———_ ame [ys V i b Petes a SY which is the equation to a parabola referred to two tangents AAC. | E 2 68 GEOMETRICAL PROBLEMS. NO. IX. DEC. 1838. 14. Let A be the vertex, § the focus, Cc AS =c=a(1-—e); *” Lte=2--, a and y? (1 - e*) aa — x’) (1 +e) }2a(1-e)v-(1 -e) a? Cc Ce (2 ~ =) (20a = a2): a a Cc e e e e If a>c, 2-—- is positive; and as a increases the curve a continues to be an ellipse, having the axis of w for its axis- major until a@ = ¢, when y’ = 4c, and the curve approaches to a parabola. If a diminishes until a =c; the equation becomes y? = 2cxu — 2’, or the curve becomes a circle when C moves up to S. c : RT : Ifa 5) the curve becomes an ellipse having its axis- “~ major in the direction of the axis of w. C e s e e When ae: y=0, and the ellipse coincides with the axis of 2. c C c When Geeiee y= (< ~ 2) (< v= 200) , and the curve a a becomes a hyperbola. 2 , Cc When a=0, y’? = (<) xv’, or «=0, and the curve co- incides with the axis of y. (This result is not mentioned in the problem.) When a is negative, y? = (< = 2) (- a + zen), which is a a still the equation to a hyperbola. When a = — ©, y’=4cw, and the curve again becomes a parabola. GEOMETRICAL PROBLEMS. NO. IX. DEC. 1838. 69 2 2 2 2 y v Yy ° 15. Let ate 1, and oT TOL be the equations to the ellipse and hyperbola; then CS® = a? — b? = a? +b”, since the curves have the same foci and centre; and at the points of intersection —_ — a PY by —_— 295 — ° aa MANS aE ; ee GE DER aa? Now if 0, 6 be the angles which the tangents to the two curves at the points of intersection make with the axis of a, or en ery a—a? ba bh? wv tan 9 = — =aee) tan 0’ = ars ay ay b°b? a? *. tan@ tan0’ = —- — ==-1; 2 y or the tangents at the points of intersection of the two curves are at right angles. 16. Let a tangent be drawn to an ellipse from a point h, k, and let m be the tangent of the angle which the tangent makes with the axis of x; then its equation is y - ma =/b? + am; and since it passes through the point h, k, k—-mh=/0 + am’. Similarly, if a tangent be drawn from the point h, & to an hyperbola whose semiaxes are a’, b’, and m, be the tangent of ‘its inclination to the axis of a, k — mh =1/a"®m2 — 6”; and when the tangents to the ellipse and hyperbola are at right angles 1 2 Th SS m= — —, oF km +h =\/a™ — b?m; and k — mh = \/b? + a'm’; 70 GEOMETRICAL PROBLEMS. NO. IX. DEC. 1838. therefore adding the squares of these two equations, (1 + m*) (h? + kh’) = 0 4+ a? + (a? — b”) m? = (14+ m’) (BP +4”); hence h?+#? =0? +a”. (1) But if w, y be the co-ordinates of the point of intersection of the ellipse and hyperbola, x ae q'? x? qa’? y? y’ a 2 z= 3 ig Rares) Re hare Bog Oa Lf 5; po Net ais a Beeb? b* 6 oy A) OPAL. hence 7 = fyb 5 had y 12 b’?? a’ a) b” date ye a tt a? (BP) +0 (b? + 0”) and @+y = — = a’ ae bh’? a? - hb” ify) 12 2 2 12 4] a*(a* +b*) +6? (a sc 52 which is the equation to a circle passing through the points of intersection of the ellipse and hyperbola: and from equation (1), 4, & are co-ordinates of any point in this circle. 17. y=La+24s/ (2-2) (5+). Let A (fig. 78) be the origin; draw the straight line BCD whose equation is y=inv+2 by making 4B =8, AC=2; then BCD is a diameter to the curve; and if # =2 or y= — 5, the curve will intersect the diameter in the points D, E. Bisect ED in G, then G will be the centre, and the curve will be sym- metrical with respect to ED, having equal portions above and below ED. When #=0, y=2+ 4/10; hence if 4H =24+ 4/10, and AF =4/10 —2, the curve will pass through the points F, H, and it is manifest that # cannot be greater than 2 nor less than — 5, hence the abscissee of D and E are the greatest possible, and the tangents at those points will be parallel to the axis of Y. 18. Let AD (fig. 79) be the axis; BAC a section of GEOMETRICAL PROBLEMS. NO. IX. DEC. 1838. 71 the cone made by a plane through the axis perpendicular to the base; then since AD=DB, 2BAD= LABD=a; Ae AG ee BAC = 2.DC A= (9 hence 4BAC=a+ (3; and 2 (a+) =7, T or at B = 2 9 but the angle 42D which the circular section of the cone makes with AB = ZACB=£; LADF=ZAED + £EAD=B+a=—; or every section perpendicular to the axis will be circular. 19. When 6°-—4ac = ©, we must have one or more of the quantities a, b, c = ©. (1) Let @ be infinite, and 6, ¢ finite; then'y? = 0, and the equation represents the axis of a, (2) Let 6 be infinite, and a, c finite; then ey=0; ..e€=0, ory=0; and the equation represents the axes of y and a. (3) Let ¢ be infinite and a, 6, finite; then «# = 0, and the equation represents the axis of y. (4) Let a, b, ¢ be infinite, and d, e, f finite; | ae a Bp then y? +-ay+-a2 =0, a a rb V(EI-E 20 2 eae Si) a 1 i & 2a 26 a) ( ) which may be finite or infinite, depending upon the values of Cc. and. =. a a rhe GEOMETRICAL PROBLEMS, NO. IX. DEC. 1838. If 6° — 4ac be negative, equation (1) can only be satisfied by supposing ° J ‘ b e e sd e in which case % = — Be which is the equation to a straight @ a line. Hence when 6’ — 4ac = + o, the equation represents one or two straight lines, 20. Let ay? + bey + ca® + dy + ex+f=0; ay” ey: Ba'y’ a Cw? ae: d'y’ fe ea’ +f" pa 0, be the equations to two curves similar and similarly situated ; then if 2 =ma, y’ must = my; samy +b muy + cma + d'my +ema fio must agree with the equation ay’ + bay +ca°+dy+exr+f=0; 3 Ph bide bi ayes cd if Leon fs ye _d er e Fee yards of bf of _ aft _ ef" or m* = = = se af Uf cf df ep y= DIREC NU Tid Ce hah =-—-— COC ee iS eee —— which are the conditions required. GEOMETRICAL PROBLEMS. NO. X. DEC. 1839. 73 ST JOHN’S COLLEGE. Dec. 1839. (No. X.) 1. Descrize a rectilineal figure which shall be similar to one, and equal to another given rectilineal figure. 2. Explain and illustrate the fifth and seventh defini- tions in the fifth book of Euclid; and shew that a magnitude has a greater ratio to the less of two unequal magnitudes than it has to the greater. 3. O is the centre of the circle inscribed in an equilateral triangle ABC; shew that if AD be drawn perpendicular to the base intersecting the circle in HE, AD is divided into three equal parts in O and E. 4. ABC is an equilateral triangle; 47, BE are drawn perpendicular to the sides BC, AC intersecting each other in D; shew that if F'G be drawn to the middle point of AB, it will be a tangent to the circle described about CEDF. 5. If in the figure of Euclid, Book tv. Prop. 10 the straight lines DC, BA be produced to meet the circle again in E, F, and EF be joined, shew that the triangle CLF is to the triangle ABD as 3+ Af 5 : 2, and that the triangle ABD is a mean proportional between CEF and BCD. 6. O is a point within the triangle ABC: D, E, F any points in the sides BC, AC, AB respectively; shew that if OD, OE, OF be joined, and Aa, Bb, Ce be drawn parallel to them from the angles 4, B, C to the opposite sides, then OD OE OF eT, Varaey ab 7. Find the polar co-ordinates of the points of inter- section of the straight lines | p =2asee (9-7), p=asec (9-7), and the angle between them. 74 GEOMETRICAL PROBLEMS. NO. X. DEC. 1839. 8. Prove analytically that the angles in the same segment of a circle are equal to each other. 9. Lp is a normal to a parabola at LZ the extremity of the latus rectum, meeting the parabola again in p; shew that the diameter in which the tangents at L and p intersect, passes through / the other extremity of the latus rectum. 10. Shew that if two equal parabolas have their axes in the same straight line and towards the same parts, the segment of the exterior one cut off by any straight line which touches the interior is invariable so long as the distance between the vertices is unchanged. 11. If SQ be drawn always bisecting the angle PSC in an ellipse, and equal to the mean proportional between §C and S'P, find the eccentricity of the curve which is the locus of Q. 12. CPLD is a parallelogram whose sides CP, CD are semiconjugate diameters of a rectangular hyperbola inclined to one another at an angle of 60°, find the equation to the ellipse which passes through C, P, LZ, D, and cuts the conjugate hyperbola at D at an angle of 15°. 13. Define similar curves; and shew that all curves similar to that whose equation referred to rectangular axes is y= wv are included in the equation k (y — b) cos@ —k (w — a) sin 0 = F jk (y — 6) sin@ + k (w — a) cos 0}. 14. CP, CD are any semiconjugate diameters of an ellipse; jo DP, draw CP’ parallel to DP, and join PP’; then the area of the trapezium CP’PD is to that of the ellipse 1 AS cat 2 > 27. 15. SY, HZ are perpendiculars from the foci of an ellipse upon the tangent at any point; find the locus of the point in which AY, SZ intersect each other. GEOMETRICAL PROBLEMS. NO. X. DEC. 1839. 75 16. AB, AG, AD are three edges of a given parallelo- piped inclined to each other at angles a, 9, y respectively, and to the vertical at angles 0, d, vy respectively ; in the side BECF a point P is taken whose co-ordinates referred to the oblique axes BE, BF are h,k; find the inclination of the straight line AP to the horizon. 17. Three circles are so inscribed ina triangle that each touches the other two and two sides of the triangle; prove that the radius of that which touches the sides 4B, AC is r being the radius of the circle inscribed in the triangle. 76 GEOMETRICAL PROBLEMS. NO. X. DEC. 1839, SOLUTIONS TO (No. X.) 1. Evcutp, Prop. 25, Book v1. 2. See Potts’ Euclid, p. 162, and Euclid, Prop. 8. Book v. AC 3k (Fig. 80) AF = ren = DC: and AD = AC*— CD’ =4CD?— CD? = 8CD* =3 AF? =3AE.AD; “-s AD=8AE, or 4B =; atsn ED = AD - 4p ="* ° >) * on=~, and op =“. 4, Since the angles at KE and F (fig. 81) are right angles, a circle may be described round CEDF; also since AB, BC are bisected in G and F, FG is parallel to AC, and £BFG=60°; ». AFG = 30 = LDCF; hence FG is a tangent to the circle. 5. Let dAB=a; ; NY Bh se AC = BD = 2a sin == V* SY Rae yb): also CRN We a cates 2 “* AC.CF =a’; Ay = Pe 2 and BC=a~ 40 =4(2— 4) - 4 (VO *) Rnd UCL ACD RC LOR (aces "| a’ ; ee a's GEOMETRICAL PROBLEMS. NO. X. DEG. 1839. 77 hence AECF = 4.CE.CF.sin ECF, AABD=%3 AB. BD.sin ABD, A BCD = 4.BC.CD.sin BCD; MEGge Cr ChalCH AB) xs ABD Fae ai heere serra $ CiaD BD CD. BC” BC ABCD: hence (A ECF) (A BCD) = (A ABD)’, and AECF: AABD:: AB: BC ::34/5: 2 OD 6. (Fig. 82) A BOC = Tie A ABC, OC eee ARC. POR URC Bb Ce therefore by addition we oi OF MAHG (—— +a te male A ABC: ODTNOL GS OF OF ae Aa Bb Ce 4. If p be the perpendicular upon the straight line from the origin, and a the angle which the perpendicular makes with the first radius, the equation to the straight line is p =p sec (0 — a); and if p =p’ sec(@—a’) be the equation to another straight line, the angle between them equal the angle between the per- ie =a-—a; hence the angle between the two given At the ae of intersection 7 T 2a sec ( - =) = asec (o- =); 2 6 , 200s (9 - 7) = sind; Odie: 6 Q and p = 2a sec (0-5) = 2a. 78 GEOMETRICAL PROBLEMS. NO. X. DEC. 1839. 8. Let A (fig. 83) be the origin, AB the axis of w; and the equation to the circle a + y? = ax+by; therefore when y=0, «= AB=a; and if 2’, y’ be the co-ordinates of any point P, tan’ PAB 2, “tn PBA = v a— & -. tan APB = — tan (PAB + PBA) Vee’, a a at ay’ ay a Lae y? oe? + y®—ae' by’ ax’ — x? hence ZAP = tan, and is independent of the position of P. 9. Generally let QS'q (fig. 84) be any chord of a para- bola passing through S$, Qp a normal at Q; then since the tangents at Q and q are at right angles, Qp is parallel to the tangent at q; therefore the diameter to Qp passes through q ; but the tangents at Q and p intersect in the diameter to Qp; hence the diameter in which the tangents at Q and p intersect will pass through gq. If Q be one extremity of the latus rectum, q will be the other extremity. 10. Let A, a (fig. 85) be the vertices of the two parabola; QPq a tangent to the inner parabola; draw Pp parallel to the axis meeting the exterior parabola in p; and draw PM, pm perpendicular to the axis; then if Z be the latus rectum, PM? pm’? AM =—— = -3- = am; therefore Pp = Aa; and if a be the angle which Qq makes with the axis, PQ@= a Pp; sin? (or hence area of segment Qpg=4Pp.PQ.sina GEOMETRICAL PROBLEMS. NO. X. DEC. 1839. 79 =4PpV/L. Pp =4\/L (day, and is constant. 11. Let SQ=p; (fig. 86) zCSQ=86; ERC 2 Oe anieoi en 1 — ecos 20 ae (1 — e’) en 27 at : go ome Ge) (5) 1 —ecos2@’ or p° $1 — e(cos’@ — sin’ )} = ae (1 — e’); . @+y—e(v’-y’)=a'e (1 -e’), or (1—e) a’? + (l+e)y=a%e(1—e’); th ae eee ae(l+e) ae(1—-e) 2 hence 5 which is the equation to an ellipse whose centre is 8, semiaxes a\/e(1+e), and ar/e(l —e), and eccentricity / i é lte IOs Since the hyperbola is rectangular, CP = CD (fig. 87); and if CL, DP be joined to meet in FE, 2 DEL is ‘a right angle; and E will be the centre of the ellipse which passes through the points C, P, L, D. Let CA be the semiaxis of the hyperbola = a; , ° 3 CP=a; .. a®sin60 =a’, or aiV3 a, and the equation to the ellipse is k vy “a geese 2 1 ae ae m = LL EI a : where m is a constant to be determined ; 80 GEOMETRICAL PROBLEMS. NO. X. DEC. 1839. and since the tangent at D makes an angle of 15° with DL, its equation is ON ey ; 3 a a Ao? 4—24/3\ )? 4 — 24/3 za fr (AEM) als ma (1-2 FY 85) ot Oa a a and since in this equation w must have only one value, the coefficient of w will vanish ; 4 AS ee lea) a « and the equation becomes 4a 44? 4(4-24/73 eae ( AAI We le ey A et a 13. Similar curves are such that if two lines can be taken for the abscissse, and any two lines equally inclined to them for the ordinates, if the abscissze be taken in a constant ratio, the corresponding ordinates will always be in the same ratio. If y = F'x be the equation to a curve; let the origin be changed to a point whose co-ordinates are —a, — (3, and the axes transformed through an angle @; then the equation becomes (y — B) cos@ — (w- a) sin0 = F $(y— B) sin 8 + (# — a) cos0} ; and the equation to a similar curve is found by putting ka and ky for w and y respectively ; and if a=ka, 8 =kb, the equation becomes k(y—b)cos0 —k(#—a)sin0=F {k(y—b)sin8 +k (w— a) cos}. 14. Let the ellipse be referred to two conjugate diameters CP, CD; (fig. 88) then the equation to PD is ~ re Y eo GEOMETRICAL PROBLEMS. NO. X. DEC. 1839. 81 and the equation to CP’ is v 7 +5 =05 hence for the co- ordinates of P’, y ¥ b Hi yen oe eas and ere sin _ ESS oe : ao. ~Y oS S \/2 ren on b also 4 PCD =ha'b' sina = —; ; 1 + —— b 1 \ 2 .. trapezium CDPP’ = z (14 wis ( v ji \/2 Q9r 15. Draw the normal PG (fig. 89) ;_ then SG SEE Mp8 ¥. PAS QE GH PH PZ HZ” QZ’ therefore Q the point of intersection of SZ, HY is in the normal PG. Tene Me ues G OG. OPaZ yA SH. Ee *. PQ = QG; and PG is bisected in the point Q. Hence if wv, y be the co-ordinates of P; x’, y' the co- ordinates of Q ; Y= CG+iGM=aer+l(1 a) nih Z ? Y= gsGM=er+t(l-e’)ra Rs PM y 2 / f , 4 SS Ss C= = =o 5 aaa Fis sc y=2y Bani yeaa? r a foarte =1;3 {- Cine a b which is the equation to an ellipse whose centre is C, and axes a (1 + e’) and 6 respectively. ¥ 82 GEOMETRICAL PROBLEMS. NO. X. DEC. 1839. 16. Let AP (fig. 90) make angles a’, 8, y’, with the three edges of the parallelopiped AB, AG, AD respectively ; then -, a+t+hcosat+kcosB , acoeath+kcosy COSI Ro tere Se 5 cos [3 = ———_________——__; AP AP ; +h k tan _ acos eae . hence a’, (, yy’ are known. Let a spherical surface (fig. 91) be described with centre A, cutting the three edges and AP, AO in the points a, b,c; p, O respectively ; i 0 =.a, ac =f, be=%¥;3 pia pb=f, pe =r; Oa=0, Ob=¢, Oc= 3 cos 0 — cos 3 cos Wy sin 9 sin Wy 4 ’ ’ cos a’ — cos (3 cos : POR mm doicceny ; . Z£Ocp is known ; sin 3 sin + and cos Op = cos cos y’ + sin yy sin y' cos Ocp ; *-- in A Oac, cos Oca = and cosacp = which determines the cosine of the inclination of Ap to the vertical, or the sine of its inclination to the horizon. 17. Let 7, 25 73 be the radii of the three circles which touch the sides terminated at 4, B, C respectively (fig. 92) ; O,, O», O; their centres; then 40,, BO,, CO; bisect the angles A, B, C; and if O,a,, O;a, be drawn perpendicular to BC, we have 0,03 = T2 + 133 4,42 = VJ (1% + 13) — (12 — 7) = 2\/ P13; B C — * Cee ra col + M3Cot = + 2/7735 B Cc — B Cc or fhe 3 + rcot > + 2/ M75 = 7 saath, BS ag : GEOMETRICAL PROBLEMS. NO. X. DEC. 1839. 83 re A G — A C Similarly, 7, cot 5 + 73 cot = + 24/0, =9 (cot 5 + cot 5) ; at. t— +2 nr = tS + cot =] Teo Yo CO 1)’ =7 | CO 3 oe till a wa 2 2 A C cot — —— Tae COt es 2 2/115 2 a 1 C A A cot — + cot — cot —+cot— cot— + cot — 2 2 2 4 2 2 ~ hence 7, ee ale cot — 7. COL — Q VASE Penh to = $<. + ——________ +. —____"_; C cot — + cot — cot — + cot — cot — + cot — g 2 Q ie g Q cn G : : : cos — sin — sin — sin — cos — sin — aelte VERS 2 BY (0D ae Ly) SSeS Sp r;?3 ———_——_ + 13 a ee B B B cos — cos — cos — zZ Pe wie cos — sin — sIn ge sin —_ cos — sin es +2 VS Pos i : : = 5 ————————qqx“—- Tr: "caer © acaeerricees ee 9". a ° 2 A 2°3 A 3 9 cos — cos — cos — 2 Z a A 7A ed mee A Cae A - 7, cos? — + 24/7,73 sin — cos — + rz cot — sin — cos — 2 2 2 Di ih ao B B Sioa re we! B CoB = 7, COs” — + 2 V/ Po13 Sin — cos — + 7; cot — sin — cos— ; 2 2 2 2 2 2 A : : hence 1, cos” a +2 J 11s sin = cos e + 75 sin® = B a B a 2 7 sexi = 7 COS” — + 2 7.7. SIN — COS — + 7. SIN” — . : 2) Vas 2 2 , 2) et re A a — tte — _ B or Vr cos © + /7, sin © = ¥/7, cos = + 7 8i = Kez 84 GEOMETRICAL PROBLEMS. NO. X. DEC. 1839. ney an OF —,C - B ~.,.B Similarly, V/'7,c08— + Jr; sin5 = / Ty COS pape \/ 7, sin ne B 4/7 ,cos— S+V/75(sinS—sin) = =/ rose +a/ 7; (sing —sin), B ue Gabses, or /m (cos + sin ~ sin 5) = Va, (cos + sin = -sin 5), ae (aici Hee sin. —————_ + So aT A : é , Now cos — + sin — — sin — = 2sin — 2 2 2 sin 4 \ 4 gil w+A =4sin —.sin cos — 3 A. A. _B,arta CY, SMB er Gee eee °, 4sin— sin cos — 4/1, = 4sin —.sin .COS — 4/153 A 4 4 4, 2 1 + tan — 1 fe, A Ue 1 + tan — MANE (1 t A A rr Cc + tan ) hence 7, a er TEs Ot aa A from which by reduction B Cc (1 + tan =] (1 + tan =) A 4, Yr 2) 3 vat a 1+ tan — A and similarly the values of r,, 7; are determined. Next let 4B, AC be produced to D, E, and R, the radius of the escribed circle touching BC; also let three circles be GEOMETRICAL PROBLEMS. NO. X. DEC. 1839. 85 described touching one another, and two of the lines BC, BD, CE; then we must suppose the triangle to have the angles a — B, w—C, and — A; and if pi be the radius of the circle which touches BD, CE, 1 tan ==) (1 tan =“) pn (ite : + tan 2 pi= 1 — tan — A 2R, is B C AN (: + tan 7) (1 + tan = (1 — tan =) A A A, le i i 86 GEOMETRICAL PROBLEMS. NO. XI. DEC. 1840. ST JOHN’S COLLEGE. Dec. 1840. (No. XL) 1. Ir two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by two sides of one of them greater than the angle contained by the two sides equal to them of the other, the base of that which has the greater angle shall be greater than the base of the other. 2. Define duplicate ratio, and prove that similar tri- angles have to one another the duplicate ratio of their ho- mologous sides. 3. Draw a straight line perpendicular to a plane from a given point above it. 4. Prove analytically that the angle in a semicircle is a right angle. 5. If ABC be a triangle where the angle C is a right angle, and if the sides CA, CB be produced, the straight lines bisecting the exterior angles at A and B when produced to meet, include an angle which is half a right angle. 6. If triangles be drawn with two sides coincident in direction, the locus of the centres of the inscribed circles is a straight line; and if the third side be also given in length, the locus of the centres of the circumscribed circles is a circle. 7. Through any fixed point A in the circumference of a circle draw chords AP, AQ at right angles to one another, and join PQ. If O be a point in PQ such that PO = n x PQ, the locus of O is a circle. 8. In two given straight lines drawn from a point O, take points P, Q in one, and P’, Q’ in the other, so that OP, OQ, OP’, OQ are in harmonic progression; find the locus of the intersection of PQ’, P’Q. 9. Considering the tangent as the limiting form of a secant, shew that the equation to the tangent to a parabola is GEOMETRICAL PROBLEMS. NO. XI. DEC. 1840. 87 a the focus being pole, and 9’ being the spiral aie of the ; point of contact. A\ 10. Determine the position and dimensions of the-Conie Section By” — 8ay + xX —21/3ay + 8aur/3 =0, and trace the curve y* a. vy? = a? (x? dul y’). 11. ‘T'wo cones whose vertical angles are supplementary, are placed with their vertices coincident, and their axes per- pendicular ; when a plane cuts them, compare the minor axis of the elliptic section of one, with the conjugate axis of the hyperbolic section of the other. 12. From A, B extremities of the diameter of a circle, draw chords AP, BP. Find the locus of the intersection of circles described on 4P, BP as diameters. 13. A pyramid is constructed on a square base, having all its edges equal in length; find the inclination of two of the triangular faces to one another. 14. Describe two concentric and similarly situated ellipses not intersecting. Draw a tangent to the interior one, and at its intersections with the exterior ellipse, draw tangents to the latter. Find the locus of the intersections of the latter pairs of tangents. 15. Find the locus of the middle points of a system of parallel chords drawn between an hyperbola and the conjugate hyperbola. 16. If a pair of conjugate diameters of an ellipse when produced be asymptotes to an hyperbola, the points of the hyperbola at which a tangent to the hyperbola will also be a tangent to the ellipse lie in an ellipse similar to the given one. 8& GEOMETRICAL PROBLEMS. NO. XI. DEC. 1840, SOLUTIONS TO (No. XI.) 1. Eucxiip, Prop. 24. Book 1. 2. Euclid, Def. 10. Book v. and Prop. 19. Book vt. 3. Euclid, Prop. 11. Book. x1, 4. Let A (fig. 93) be the origin, and the diameter AB the axis of w; then the equation to the circle is y? =2aer—x"; and if y = mw be the equation AP, at the point P 2a 2am ma =2ae-—a; .. &©=——, and y= 5 1 + m* 1+m hence the equation to BP which passes through the points — 2a 2am 20,0: and ; , becomes, 1+m 1+m 2am 1 +m = —___—____(w - 2a y Oa ( ); — 24 1 + m? ‘ Wee | or y= — —(v— 2a) which is perpendicular to AP; m therefore z APB is a right angle. 5. Let the two straight lines meet in D (fig. 94); then L£DAB=%3(r-A); 4DBA=1(r-B); A+B » ADB =a ~ {2(n- A) + bw BY} =~" ==. 6. Let AB, AC (fig. 95) be two sides of the triangle including a given angle A; bisect z BAC by the straight line AO, then the centre of the inscribed circle will be in the line AO, whatever be the lengths of 4B, AC. Next let O' be the centre of the circumscribing circle ; GC then radius AV=4 ida and if BC and 24 be constant, sin AO’ is constant; and the locus of O' is a circle whose centre BC is A. and raditie =: === | 2sin A GEOMETRICAL PROBLEMS NO, XI, DEC. 1840. 7. PQ (fig. 96) passes through the centre ; PO=2na; a5 ee PQ = 2a, and CO = (1-—2n) a which is constant ; hence the locus of O is a circle whose centre is C, and radius = (1—2n)a 89 8. Let OPQ, OPQ’ (fig. 97) be taken for the axes of x and y respectively ; let OP=a, OQ=a, OP’ =b, OQ’ =b then the equation to PQ’ is Sie ae es and the equation to P’Q is Tito ho kee oe th Se = 0; Ga a Oe b J hence the intersection will always be found in the straight line een whose equation is w — y = 0, which bisects the angle POQ 9, secant 3 Let r = psec (9 — 3) be the polar equation to the hence p = rcos(9— (3); and if 6’, 6” be the polar ? angles of the points of intersection of the secant and curve, , tad . . r,7 the corresponding focal distances, we have p=r cos(0 — 2), per’ cos (@”— 2); P 2a ‘ 2a pep mit cos. _ 00s (6’—B). 1+ cosQ” ’ or (1 + cos 0”) (cos 6’ cos 8 + sin @ sin 3) 1 + cos@’ 08 (0’ — 3) 1 + cos @’ = (1 + cos 0’) (cos 6” cos 3 + sin 6” sin 3) hence (cos6’—cos 6”) cos 3 = sin 6” —sin’ + sin (0”—@’){ sin 3 § 90 GEOMETRICAL PROBLEMS. NO. XI. DEC. 1840. . 8 +6" ( 0” + 6 0” - 0 ig S10 a ee + cos — ) tan , ad = 2 cos — cos — tanp; 9 9g Bs / Q” 6 hence tan 3 = 4 (tan % + tan =) : \ ~~ and when 6’, 0” approach to one another, 6’ 0 tan P = tan—, or eee ay @Q’ Q’ Q’ also p = 7" cos (0 — 3) = 1" cos — = a sec? = COS = = a eC > 5 > hence the polar equation to the tangent becomes , / BM MS HS (0-5) 3 2 2 10. The equation is that of a parabola. Transform the origin to a point a, $ in the curve by making # =a +a, y=y +B; . 8 (y! + BY -2V/3(w' +a) (y' +B) + (@' +a)? ~8a(y' +) + 8ar/3 (a +a) =0; hence 3y?-2\/3a'y' +a? +dy' + ea’ =0, where @(a,8)=0; 68 - 2/3a-8a=d, 2a —- 21/384 8avV/3 =e. Again, transform the equation to polar co-ordinates by putting # = pcos@, y' = psing; ie) (4/3 sin 0 — cos 6)’ + (d' sin 0 + e' cos 0) = 0; tia | and if — = —==tand = tan —; e 3 6 4. sin? (0 — 0) p + e’ secd. cos (8 — 6) = 0; which is the equation to a parabola whose axis makes an Zz 30 e’ sec 0 with the axis of a, and the latus rectum = — Tit; GEOMETRICAL PROBLEMS. NO. XI. DEC. 1840. 91 63-2\/3a-—8a 1 Now —_>————= = —= 2a -2»/33+8a\/3 3 rae BV/3-a=2avV/3; 4ar/3 2 (a, B) = $B? -2»/3a3 + a — 808 + 8a\/3.a=0; . 8a(B -/3a) = (3B — a)? = 122’, or 2(8-/3a) =3a; and 63 —2/3a = 12a, *. 48 = 9a; and pa=: = 9a 3a also 2/3a= 28 -8a=—-8a=—; / 3a A, e r] hence a = a, (3 are the co-ordinates of the vertex of the parabola, and the negative sign of the latus rectum shews that it extends in- definitely on the negative side of the origin. (2) To trace the curve y' + 2’y’ = a’ (a — y’), let the equation be transformed to polar co-ordinates by putting w=pcos@, y=psind; .. p= a (cot’@ —1); and when 0=0, p=: also p’ sin’ @ = a? cos 20 = a’: hence the value of the ordinate = +a when @ is infinite, or the curve has two asymptotes parallel to the axis of w at a distance a from it. 92 GEOMETRICAL PROBLEMS. NO. XI. DEC. 1840. : 7 a eA Kero As @ increases from 0 to ic p diminishes from infinity to 0, and the curve passes through the origin. Since by putting — a for w, or —y for y, the equation to the curve is unaltered, it is symmetrical both with respect to the axis of w and y, and has a point of contrary flexure at the origin A. The curve is of the form traced in fig. 98. 11. Let AP (fig. 99) =p be the perpendicular upon the cutting plane PBC; @ the inclination of AP to the axis of the cone; draw BD, CE perpendicular to the axis; and let 2a be the vertical angle of the cone, 26 the minor axis of the section, then 46° = BD.CE = 2p sec (9 — a) sina {2p sec (0 +a) sina} 4° sin’ a 4p” sin’ a cos(9+a)cos(9—a) cos?@ —sin?a Similarly, if 6 be the angle which 4P makes with the axis of the second cone whose vertical angle is 2a’; 2 ne is Pa A sin* a — cos’ @ , , T and when 2a’ = 7 — 2a, and @ pea ee 4b" f Th = cot’a; which is the ratio required. 12. Draw PM (fig. 83) perpendicular to 4B; then since the angles AMP, BMP are right angles, the semicircles de- scribed on AP and BP both pass through M which is their point of intersection; and the locus of the intersection of the circles is the line 4B. The same is equally true when P is any point whatever, and the locus is independent of the circle APB. GEOMETRICAL PROBLEMS. NO. XI. DEC. 1840. 93 13. Let A (fig. 100) be one of the angles of the base ; with centre A describe a spherical surface intersecting the two triangular faces and the base in ab, ae, be; then ab=ac=60°, be = 90: cos 90 — cos 60 . cos 60 *, cos Zbac = : : Sy ay is sin 60. sin 60 Sa and Zbaec which measures the inclination between two tri- angular faces = cos~'(— 4) = 7 — cos"! 4, x” 2 2 2 a ; lav ‘Get x ie a = 1, and ak 5 =1 be the equations to the two ellipses; then the equation to a tangent at a point a’, y’ of the first ellipse is / , Ue yy in ee Let a pair of tangents be drawn from a point XY, Y to the second ellipse, then the equation to the line joining the Yy b? may coincide with equation (1) we must have , , 4. 9 2 ob ples Cs Eine (=) + (£) a1, | a a2? BF 6?’ a es : ff) one a i" a mae which is the equation to an ellipse whose semiaxes are points of contact is a + =1; and in order that this a a? b” The Re 15. Transform the origin to a point a, 3; then the equation to the hyperbola is O29) (A - 94 GEOMETRICAL PROBLEMS. NO. XI. DEC. 1840. and if y=mzw be the equation to a chord passing through the origin, when the chord meets the hyperbola (e+a)? (ma + By? a - (SS) Pls a and the values of w in which the chord meets the conjugate hyperbola will be determined from the equation (a + a)? ee + BY? | A a b " ; but when the origin is the middle point of the chord a’ = — wv; 1 : 2a 2 : (5-F)e+(S- eae Bo ae a? b? a’ b2 1 m a m a’ Y md (j- =) @ 73 Geer ag oe ME a 6? and by addition i 2 g 2 (=F) +S - Eno: hence by eliminating a, ( 1 i) an ——. oe the 5: which gives a relation between a and #, and is the equation required. 6-3) E-8)-« a” b? he he 12 12 16. Let the equation to the ellipse be _ * 7 =1; and a let vw, y be the co-ordinates of a point in the hyperbola at which a tangent to the hyperbola will be also a tangent to the ellipse, then Xu Yy oa b”? aE; ya 13 and mY oe a acl GEOMETRICAL PROBLEMS. NO. XI. DEC. 1840. 95 Face \2@ 2Yy or a”y? +b? a? = 4a°y*, and wy =m’; oh ay? + b'? a = 4m’*, which shews that the point wv, y lies in an ellipse whose semi- 2 2 and —-, or the ellipse is similar a : : 2 conjugate diameters are — to the original one. In this example an ellipse has been found similar to the original one which will pass through the points denoted by x, y; but the two equations ay’ + b’a =4m', and vy =m’ may be combined in every possible way, giving different curves which will intersect the hyperbola in the required points. The curve may be reduced to a right line for a’y? +2a'b vy +b? a? = 4m‘ + 2a’ bm’, or @dy+b a= + m/ 4m? £2ab'; either of which pair of straight lines will intersect the hyperbola in the four points required. 96 GEOMETRICAL PROBLEMS. NO. XII. DEC. 1841. ST JOHN’S COLLEGE. Dec. 1841. (No. XII.) 1. Tue rectangle contained by the diagonals of a quad- rilateral figure inscribed in a circle is equal to the sum of the rectangles contained by its opposite sides. 2. Four circles are drawn, of which each touches one side of a quadrilateral figure and the adjacent sides produced ; shew that the centres of these four circles will all lie in the circumference of a circle. 3. If ABCD be any quadrilateral, M, NV, P, Q the bi- sections of its sides, prove that AC’ + BD* =2(MP* + NQ). 4. Three circles, whose radii form a geometric pro- gression, having 2 as a common ratio, touch each other ; find the angles of the enveloping triangle. 5. The lines joining the angles of a triangle with the points in which the escribed circles touch the opposite sides, meet in a point. Shew also that if the base and the sum of the other sides is constant, the locus of the centre of the escribed circle touching the base is an ellipse. 6. Two semicircles are described on the segments of the diameter of a semicircle whose radius is 7; shew that the locus of the centre of the circle touching these three semi- : , : ; Ar EM circles is an ellipse whose semiaxes are eo and ——. 7. Tangents are drawn to an ellipse so that the product of the trigonometrical tangents of their inclinations to the major axis is constant; prove that the locus of their intersection is a conic section. 8. Through 4 the common vertex of two similar ellipses ABB', ADD’ whose greater axes coincide, draw chords ABD, AB'D and join BB’, DD’; BB’ will be parallel to DD’. 9. If A be the elliptic area contained by two semi- GEOMETRICAL PROBLEMS. NO. XII. DEC. 1841. 97 diameters including an angle a, and B that contained between two semi-diameters at right angles to the first, then a | P , 2A ; 2B 5 sup i ab” ab 10. The straight lines drawn from any point in an equi- lateral hyperbola to the extremities of any diameter, are in- clined at equal angles to the asymptotes. 11. From any fixed point P in an hyperbola, draw PA, PB parallel to its asymptotes, and from another fixed point Q draw any straight line cutting these in a, 6 and the curve inp; then pa « pb. 12. A, B are two fixed points, P, Q any two points in the line 4B or AB produced, such that PA QB QB PA | ey Sp Beas Bf | SE Ey Rg A m 7 7 m find two other fixed points M, N, such that n m QN PM and find the values of K, AM, AN. 13. If any number of quadrilaterals inscribed in a circle, have a common side, and the sides adjacent to this be pro- duced to meet; the lines joining the point of concourse, with the intersection of the diagonals of the quadrilateral shall all meet in the same point. 98 GEOMETRICAL PROBLEMS. NO. XII. DEC. 1841. SOLUTIONS TO (No. XII.) 1. Eocuip, Prop. D. Book vi. 2. Let ABCD (fig. 101) be the quadrilateral figure; O the centre of the circle touching AB and the two sides CB, DA produced ; then OA, OB bisect the exterior angles of the figure, and 2 AOB = 7 — (OAB + OBA) A+B a =7-{3 (7-4) +3(r- B)} = Similarly if O’ be the centre of the circle touching CD and the two sides BC, AD produced, Qa — B B PDO rte os sken ee Te ae = 7 —AOB;: hence 2 AOB+ 2DOC=rz. Also if P, P’ be the centres of the circles touching the sides AD, BC respectively, 2APD+ z2BP’C=7; and the op- posite angles of the quadrilateral figure OPO'P’ are together equal to two right angles; hence a circle may be described about OPO'P’; and the centres lie in the circumference of this circle. 3. MN and PQ (fig. 102) are each parallel to AC, and AC equal to gu also QM and PWN are each parallel to BD, and BD equal to apie hence MNPQ is a parallelogram whose di- agonals are MP, NQ; *. MP’ + NQ?’=2 MN? +2QM’?=44AC) +35 BD’; *. AC? + BD’ = 2(MP’ + NQ’). 4, Let A, B, C (fig. 103) be the centres of the three circles whose radii are as 4, 2, 1 respectively; ab the direction of the common tangent to the circles whose centres are A, B; ac the direction of the common tangent to the circles whose GEOMETRICAL PROBLEMS. NO. XII. DEC. 1841. 99 centres are A, C; bc the common tangent to the circles whose centres are B, C; join AB, AC, BC and produce them to meet ab, ac, be respectively in d, e, f; then the triangle abe will touch the three circles with the sides bc, and the two sides ab, ac produced; but there cannot be any triangle which envelopes the circles and touches them with its three sides. Let 7,, 72, 7; be the radii of the three circles whose centres are 4, B,C; £Cfe=0, 4 Aec=¢@; then LABC—zBfco=a2 - 2abe—ZAdb; | or Zabe=7-ABC=7r - B. Again Zead+LeAd=L Adb+ 4Aec=0+9,; or Lcab=0+@-LCAB; - Lach=a7—i{xn-B+O0+G-At}=A+B-(0+9); ; : ; Y,-7 3 but snO=4; sind= +=; BC=3r,, AC=57,, AB=6r;; M+, +O 2/14 2/14 an/ 14 or sin 4 = ; sn B= : = ; 15 9 ko ‘Z.a= sin714 + sin-!3 — sin iS" 5 15 E g Ze=sin7?} Tene + sin 15 5. (1) Let O (fig. 104) be the centre of the circle which touches the side BC and the other two sides produced ; G2 100 GEOMETRICAL PROBLEMS. NO. XII. DEC, 1841. M, N, P the three points of contact of the escribed circles with the sides; then. BM=S-c; BP=S-a; CN=S-a; and if BC, BA be the co-ordinate axes, the equations to AM, C'P are v y v y +>=1; —+ =1; Sic hee a We isS sik , and the co-ordinates of N are ; (AN), and = (CN), or = (S =e), and = ($ — a); hence the equation to BN is y c(S-—a) a” a(S'—c)’ and for the points of intersection of 4M, BN we have PAS 50) OR = 5 » Y= hye 3 which are the same as the co-ordinates of the point of inter- section of BN, CP; therefore AM, BN, CP meet in the same point. @ (2) Next let BC be the axis of a; BM =x, MO=y; BC=a; BA+AC=s; then -C : 5 cos ] tan — tan — b+e sinB+sin€d “2 S a sin (B + C) B+C Pere cos 1 — tan — tan — 2 2 Ay OA eel tal: A te a an— = and —~ =@c aes 2 LN ey ery GEOMETRICAL PROBLEMS. NO. XII. DEC. 1841. 101 S+a (ax — x’) and y*? = s-a which is the equation to an ellipse having the base for its . e e . S + a axis Minor, and axis major = a s-—a 6. Let AB (fig. 105) be the diameter of the given semi- circle whose centre is O, and radius r; Ca point in AB the centre of another semi-circle whose radius is r’; D the centre and p the radius of any other circle which touches the two former semi-circles; P the centre and p the radius of the circle which touches the three circles whose centres are C, O, D; draw PM perpendicular to AB, and let AM =a, MP=y; x’, y the co-ordinates of D; then (Solutions of Trigonometrical Problems. Ex. 18. No. 15) Xv r+ 7 Y y —= ee see Bat ie p Cigh tore oc 8 and when the point D is in AB, y’ = 0; Oo ntr y >) ep ra and (w-ryt+ye=(ptry, “ @ -2raty =p? +2pr'; anc 5 .. et +y—p=2r (v7 +p) =2r(x#—- 0); Ae y= p' = 27! (w+ p) = 27 (w~p); awe z) or a’ 7 = 2p l(xv—=+]; aN 4s ( y. hence a® — 2rx” + 2—+ ry=0; d : 47 which is the equation to an ellipse whose semi-axes are re 102 GEOMETRICAL PROBLEMS. NO. XII. DEC. 1841. If P be the centre of the »™ circle inscribed between the circles whose centres are C, O, D; «,y, its co-ordinates; we have Ly, +P ¥ : y —=—_; Ap de BEY or = 2n, since y’ = 0; Ee op Rate De Pn and (#, — 7)’ + y;, = (p, +7"); % a, — 27 a, +Y, =p, +27" pn; and a, +, — p, =21" (a, + pn) = 27 (@n— pr) 5 vty — nF On An? Qn 2 (4? — 1) 92 c . at A Laie a Weyl iaaouns An? —1 Qn 2 An? . (vw, - 7)? + +(= > (y-= 5 *) moray 4? wh An? ——__ 5 we Y ha . 3 An? a 1 9 an? — 1 and co-ordinates of the centre 7, and rere ni” — Kooeek he equation to a tangent at the point a’, y’ of an ellipse is esa a = 1; and if tangents be drawn from a a” point , Y without the ellipse, a Sean) YG re Ee a b* Let m be the tangent of the angle which one of the tangents makes with the axis of w; then ba’ Be m=-—, and Y—-mX =— ay yf w am y' amy? y? b t—s- Sears @ tie =~ —— Hs bd ($F) +343 - GEOMETRICAL. PROBLEMS. NO. xu. DEC. 1841. 103 2 Be ae or en AAG TH arm hence Y—-mX =\/0? + am’, Y or m?(X? — a?) -2XY¥m4Y°-68'=0; (1) and the product of the two values of m =a; Vy? — BP . hes 2 aed AS al A ae which is the equation to a conic section whose centre is the centre of the given ellipse. From equation (1) it appears that if the sum of the two 2XY pwr ge Ge which is the values of m be constant = /; equation to a hyperbola. 8. Taking the polar equation from the vertex, if 2 BAC (fig. 106) =90; 2B AC=6', AB=p, AD=>p'; 2a, 2a’ the major-axes, and ¢ the eccentricity ; then 2a (1 — e”) cos 0 , 2a (1 —e’) cosO = ———_—_—.—__, = +; 1 — e’ cos’ 0 1 — e’ cos’ @ eee ay Similarly Feat A p a AD a AB AB AB AD fea Get ahs ABM ADS hence BB’ is parallel to DD’. 9. If x, y, x, y’ be the co-ordinates of P, P’ in the ellipse (fig. 107); then area PCP’ = area ACP — ACP 104 GEOMETRICAL PROBLEMS. NO. XU. DEC. 184d. b a or 4=—(t or ae ot), ay ay 2 be’ ba a@ (tan @ — tan @) 2 Bala Ne RAS PERT I a a0 1 + (5) aes ‘i+ (2) tan 9 tan @’ b L b where 2 PCA = 8, £PCA=6, and &-@O=a; tan @’ — tan @ ee ae et 1 + tan @ tan 0 : a 2 ye (=) tan @ tan @’ a a a 1 + tan @ tan 6’ a 2 1+ [(-] cot @cot@ Similar] ree cot () imilar Cots 2ie6 mat Cas aah Hp J? ab a 1 + cot @ cot 6 a\* : (=) + tan 6 tan @ = )—/cOt |) 2 ee ee a 1 + tan @ tan @ ; 2 24 A ae b ‘ 3 _ a -) ? ete ———_s Co SS ee Oo —. = a ) and 7 = = aan 4 sin” w whence the radius, and co-ordinates of the centre are de- termined. Any line passing through the centre will be a diameter to the circle, and y — 8 = m(# —~a) will be its equation where m is arbitrary. To determine the diameter which passes through the origin, its equation is b-—acosw =—%, or y =———_—_ J a a — bcosw 8. Let C (fig. 126) be the centre of the circle ; draw CA perpendicular to the given straight line 4B; and let AC, AB be taken for the co-ordinate axes; suppose PQ any position of 122 GEOMETRICAL PROBLEMS. NO. XIV. DEC. 1843. the line whose length is 7; draw PM perpendicular to AC, and let AM=2, MP=y, AQ=y'; AC =c, CP=r; P+ y- YP al, Y+(e- ay ar’; and if X, Y be the co-ordinates of the middle point of PQ, i y-274 ay ey ef P aXe: y+y =2Y; or 2y=2V4 Y/R - 4X? =24/P —(e —-2X)’, which is the equation required. 9. (1) Let 4, B (fig. 127) be the two fixed points; P the intersection of the lines PA, PB in any position; then if 4 PAB =0, and 2 PBC = 28, we have 2 APB =0, and PB= AB; hence the locus of P is a circle whose centre is B. (2) If ABP =20, AP =p, AB =a, then asin 20 2a cos 6 2acos@ io = 2S Se ee or hee ar Se ke i sin 30 3 —4sin? @’ p 4cos?@ — 1 . 4a” — (a# + y”) = 2aa, or 3x° -y? = 2a; 9 hence 7? = #}(« — “) - ae 3 9 which is the equation to a hyperbola whose semi-axes are VT ee —, >=, and eccentricity 2. 3 5 V3 9 ne Bis the focus of the hyperbola, and 4 the vertex of the exterior hyperbola. 10. Let PQRS (fig. 128) be a straight line cutting the two circles whose centres are A, B, and intercepting the chords PQ, RS respectively proportional to 4P, BR. Pro- duce BA to meet SRQP produced in 7’; and draw AM, BN perpendicular to PS’, then PQ RS = MP RN UR UBER (AP PsBRe GEOMETRICAL PROBLEMS. NO. XIV. DEC. 1843. 128 therefore AP is parallel to BR, and eet tA) ABS BT BR’ “AR BR - AP’ hence AT is constant; and every line passing through the fixed point 7’ and cutting both circles will intercept chords PQ, RS proportional to their radii. If 7” be between the circles, it may be proved in like Ale AP manner, that —— AB = BRiAP’ and every line drawn through T” will intercept chords proportional to the radii of the circles. 11. Since PQ, RS are always proportional to the radii of the circles, they vanish together ; ; or when FP and Q coincide in P’, R and S will coincide in A’, and TPR’ will be a common tangent; also Z APT isa aig angle; hence if AT be taken a foureh proportional to BR — AP, AP and AB, and upon the diameter AT’ a circle be described cutting the given circle in P’, P’; TP’, TP” will be common tangents. Similarly, if AT’ be taken a fourth proportional to BR + AP, AP and AB, and a circle be described on AT’, this will meet the given circle in two points CUO andl: Che T’ Q” will be common tangents. Also let DT’E', D'T'E be the pair of common tangents ; then if any point be taken within the angles DT'E, or DT'E and exterior to the bases DE, D'E’ it will be impossible to draw a line cutting both circles, but a straight line may be. drawn through a point in any other position so as to cut both circles. 12. Let ASB (fig. 129) be the line bisecting the given angle; AS =a, PS'p any line passing through S; @ the Pidale point of Pp; 2 PAS =a; 4PSB=6; “5Q See analy a sin a asin a SP = ————_. ee ee aS sin (@ — a) OP sin (8 + a) 124 GEOMETRICAL PROBLEMS. NO, XIV. DEC. 1843. asina 1 } “ SQ=p=4 (SP — Sp) =——_ |—_____ - ____ p= ( P) 2 CE Sree _ @sin’a . cos 0 ~ sin? @— sin?a” oy’ — sina (a? 4+ y’) = a sina a, or y’ cos’a — wv’ sin?’ a = asin’a a; e+e 9 Yy oF —_——__ <= er — 9 a (; tan a) 2 which is the equation to a hyperbola whose axes are a, a tan a> and whose centre is the middle point of 4S. wo 13." \Fig./ 180, 4 §dbv Ab = de waee | Be. Bc =Ba.Ba; Ca.Ca =Cb.Cd;; “, (4b.Be.Ca.)(Ab’. Be’.Ca’) = (Ac.Ba.Cb.) Ac’.Ba’.Cb’; but 46.Be.Ca=Ac.Ba.Cb; -. Ab’. Be’. Ca'= Ac’. Ba'.Ct’; or da’, Bb’, Cc’ meet in the same point. 14. (a) Let C be the centre, (fig. 131) S the focus, PT, QT two tangents meeting in 7'; ZPSC=0; 2PSQ=2a; «. 2PST =a; LSPT = 9; £TSC=; ST = p; SP. sinh yi SP /P'sin(p +a) cosa +sinacot p’ -esin R Elie e 6 ang ON Ty cera Pas Ak GEOMETRICAL PROBLEMS. NO. XIV. DEC. 1843. 125 a (1 — e’) ye, (1 — ecos@) cosa+esin@sina’ ks. a (1 — e*) a (1 —e’) pcos a — € COS (O+a) cosa —ecos Wy cos’ a — 2e cosa cos (8 + a) + e’ cos’ (8 + a) ja (1 — e*)}? : and if SP=r, SQ=7r'; pte vate ial INES. CCN tnt: Ul a (1 — e’) (B) Also = = ; at 1 —e {cos +cos(0+2a)} + e* cos’ (8 +a) —sin*a)} Seed 4 hence chk) Near) Sinton Fake eae) Neg l 2 2\2 a Teg a’ (1 — e*) 1 sin’ a : bry’ = ——_— — or = rr be? CUS fa yt sin which shews that in the ellipse p? cannot = r7 unless a =0; but in the parabola 6 is infinite, .. p? = 17". 15. Let the base 4B (fig. 132) be taken for the axis of wv; A the origin; a, y the co-ordinates of the point P from which the perpendiculars Pa, Pb, Pe are drawn upon the three sides of the triangle ABC; then the co-ordinates of the points c, b, a respectively are x, 0; (x cos A +ysin A) cos A, (wcos A + ysin A) sin A; c—{(c—ax)cosB+ysinB} cosB; {(c—«#)cosB+ysin Bi sin B; but when three points 2, Y,3 25 Y2; 3, Y; are in the same straight line, Ya Ure Us Ui ° ar 3 rE 7 Wz —@, hence if a, b, ¢ be in the same straight line, 126 GEOMETRICAL PROBLEMS. NO. XIV. DEC. 1843. (vcosd4+ysindA)sind-0 $(c—«x)cosB+ysinB} sinB-0 (wcosA+ysind)cosA—ax ¢- $(c-a)cosB + ysinB} cosB -a ; vcosA+ysind (c-—«#)cosB+ysnB eee Se : = a ycosA-wsind (c—«#)sinB-—ycosB’ ae aes (sin AcosB+cos Asin B) —y(e—#)(cosAcosB-sin Asin B) —y’(sin A cos B + cos 4 sin B) — wy (cos 4 cos B — sin Asin B) =0; . (ca — #) sin (4 + B) —cy (cos 4+ B) -y’sin(4 + B)=0; or cw — v —cycot(4 + B)-y=0; | hence a + y?— ca —cycot C=0; 9 ie / 2 (w- 5) + (y -£ cot ©) = (S cosee €) Q Vis ae 2 which is the equation to the circle circumscribing the triangle. 16. See No. 14. (P) 17. Let O (fig. 133) be the centre and 7 the radius of the circle which intercepts from AB, AC the two portions ce, bb =+, B respectively ; draw OM parallel to AC; and let 2 AM=znx, MO=y;... ry sin? A= 2, 2 Similarly, 7° — #* sin’ 4 = ae or the locus of O is an equilateral hyperbola whose centre is A, having two conjugate diameters in the directions 4B, AC. Again, let ON be drawn parallel to BC, and let BN = 2’, ON =y'; then if a@ =a, GEOMETRICAL PROBLEMS. NO. XIV. DEC. 1843. 137 and if two hyperbolas be drawn with centres A, B, whose equations are 2 2 2 2 2 eye ie 12 ap ti ae Tg ice a Bi nl af mg 4 sin? B the intersection of the two hyperbolas will give the centre of the circle. There will be four points, viz. one within the triangle, and another between each side and the two remaining sides pro- - duced, 128 GEOMETRICAL PROBLEMS. NO. XV. DEC. 1844. ST JOHN’S COLLEGE. Dec. 1844. (No. XV.) 1. Ir two parallel planes be cut by another plane, their common sections with it are parallel. Perpendiculars are drawn from a point to a plane, and to a straight line in that plane; shew that the line joining the feet of the perpendiculars is perpendicular to the former line. 2. The sides containing a given angle are in a given ratio, and the vertex is fixed; shew that if the extremity of one of the sides moves in a ays line, so does the extremity of the other. 3. b? a 144 GEOMETRICAL PROBLEMS. NO. XVI. DEC. 1845. a>m eo a ~—— Os ee tt a! = y’ =_x#+t-—ecH < Vb _ a*m’*, Of a= Va me — 6’, (2) 1h b? according as Q is in the hyperbola or its conjugate. Eliminate (m) between equations (1) and (2) and using the positive sign in (2) we have b* 2am » aim "9 2 jp f 12 Ariat) 2 2 ae a (« vor vy care y*) a2 +am b? 2a°m a’ m 12 LP AA , ae 2 a 3 (*”- ny + Ey) = atm; therefore by addition, 2b? a’ m? a*m 1g tar age 2 f2 wy (2 a 7 y ) = 20% or a” + x y by subtraction, ‘ © 4ma’y’ = 2a°m*, or am =2a’'y ; 4y"4 ote ( = } =, (3) Again, using the negative sign in equation (2) ee ee GR atm? 2 hoe 2) 79 Melee ale nomena bn DRC y*)=b + am ; Dish) ae Oat a*m” at 28 ae. ie 2 2m? a = (e gins + 1 i‘) am — b° ; therefore by addition, 2b° a oa a*m atm te , 9° (2 Fara oY ) =2a°m, or 7? + by subtraction, 4ma'y’ = 20°; at a‘ b? f2 ke Yv + : ot Seaarsaaar 9 Ae? Aha? y’® ; 4 ap!4 or y*(1 4 |= Os (4) a GEOMETRICAL PROBLEMS. NO. XVI. DEC. 1845. 145 equations (3) and (4) are the loci of the intersections of the tangents at P and Q when (PQ meets the hyperbola and its conjugate respectively. uv? ay! If oA (1 +) = 1; when the curve meets the ellipse "9 Ay’ i.) Qa 2 (1 -¥) (144) 1, o£ (1-H )'=0; & - yf =0, ory = b [Seal / 2 and the corresponding values of w’ are , e= = a, and xv = ee J 2 y” A ap’* similarly, the curve ive (1 + — = 1 meets the ellipse when a ve =0, and a = + —=; V2 and since equation (5) is a complete square, two values of y become equal at the points determined by the equation; or the curves touch the ellipse at the six points in which a a = 0, # = +a, and 2’ = +—=. 2 8. Draw AQ, A4’Q perpendicular to AC, AC respectively (fig. 147) ; join AB’, bisect it in D; also join CD; then CD will be parallel to the axis of the parabola which touches CA CB’: but A'C=CP, A'D = DB’; therefore CD is parallel to PB’ and is therefore perpendicular to A'C: hence the tangent AC is perpendicular to the axis of the parabola; or A’ is the vertex and A’Q is the direction of the axis of the parabola. Similarly, 4Q is the direction of the axis of the parabola which touches AC, CB: and since the angles CAQ, CA’@ are right angles, the circle described on the diameter CQ passes 146 GEOMETRICAL PROBLEMS. NO. XVI. DEC. 1845. through the points 4, 4’, and the quadrilateral CAQA’ may be inscribed in a circle. PAlso, if CO PCR = a,°CQ'= AC sec - the least value of which is AC: and if CQ =2AC, sec : = 2, a Scag 4, and a = 120°. 10y — 4 8y? + 2y —1 9. (2 ") a= y +2y 10y — 4 Data 49y° —14y +1 . Was | eae eee Wes a Reet |) a ein foa I ae Jat (SP) oe CH : 5y —2 —] 8 3 . 8w=5y-—2+ (Ty -—1) = 12y- 3, or —2y-1; therefore the equation is reduced to two straight lines whose equations are 4y —v@=1, and 2y+3xH%+41=0, 10. Let the two axes of the cones which are at right angles be taken for the axes of # and y; and a line through the vertex A perpendicular to them for the axis of =: then if a be the semi-vertical angle of the cone whose axis is Aa, ae a will be the semi-vertical angle of the cone whose axis is Ay: and if x, y, x be co-ordinates of any point in the first surface, a, y', % co-ordinates of any point in the second surface : 2 2 2 2 e y+ = x tan*a, or ——— — 2 2 Ss col gee wn y” 12 2 #3 2 2 UV" +s "= y* cot’a, or ————- — => =-—tan’a. (2 #* cota =? (2) GEOMETRICAL PROBLEMS. NO. XVI. DEC. 1845. 14:7 Now by making = x constant, equations (1) and (2) are the equations of the sections to a cone made by a plane parallel to both axes, at a distance (x) from their plane: and if a = Zs 2 they manifestly become > x’ y” : eT TE ee et vst seas ata Sans the equations to two conjugate hyperbolas. 11. If the equation to the curve be referred to polar co-ordinates it will assume the form Pinte) +70 inci (0) +...+°f29 + pfid+fr = 0. When p is the radius vector, and /f,, (O) SP fig ty 0 eerie ae homogeneous functions of sin@ and cos@ of m, (m —1) &c. dimensions respectively, and f, a constant ; ae hi tay oe fr) : iv gece ate + os SOU) 4G 0; i ae eee ti Si (@) eS pie i ¥i is a homogeneous function of sin @ and cos @ of 1 dimension ; 2f,0 Spe ow) a iE fo fo is a homogeneous function of sin @ and cos @ of 2 dimensions ; £85 1 £8 5,1 HO fo to So a homogeneous function of sin @ and cos @ of 3 dimensions. DOM = Similarly, 2p-" = F,,(@) a homogeneous function of sin 0 and cos@ of m dimensions; therefore the equation to the locus of Q is oot =a” F, (0) =a” (cos 0)" F (tan 0), Kae ( 148 GEOMETRICAL PROBLEMS. NO. XVI. DEC. 1845. where #F'(tan@) is a rational and integral function of tan @ of m dimensions; and in order that this curve may be similar to the original curve, the original curve must have for its equation (-)'= cos” (6 — a) F'tan (0 — a). p 12. The equation may be put under the form Ga aK x’ Ne ie es a 2 O32) (GB ES) 2 Bed) ene Tea” Cian: Cc Gen: c a = 0" La ~ 9 y? » ( 2? i 2 y? 2 2y 2 N —=-*>)]) -2\i-~+ns 1 — — — = he & oe es i & b? 1) ea v” y 2 x fe 2 el eet a Si ae eo Gt) fa G1) | vty? ) (= y \(2 y vo ey CG Yy Cede ihe ae ete oh pe ks PR ane AR Sy ke We ay Vs Be ae (= soe b ) therefore the four straight lines asad) GP .y —+—-+4+1=0, —+>-1=0 tao hs Sanaa 4 eo oy ew oy aeeb fae they B are double tangents, and meet the curve in the circle whose equation is #’ +y? = c’. GEOMETRICAL PROBLEMS. NO. XVI. DEC. 1845. 149 13. The line joining the points of contact will always pass through the focus of the parabola; suppose PS'p (fig. 148) a chord passing through the focus; 4 the vertex ; PQ, Qp tangents at P, p meeting in Q; produce PQ to meet the axis AS in T'; draw PM perpendicular to AS, and let LPTA=0, AM=e2, MP=y; then tan @ = z aval v By = - wamcotd, y=2mct0, 2SPQ=zZSTP=6; and the perpendicular from 4A on PQ =a=ATsinO=axsind=m hence % = tan’@ = cot? QpP. a Now a, (3 are the co-ordinates of A referred to the fixed axes Qp, QP; and if 2QpP be constant, the locus of A is a straight line whose equation is 3 = cot? QpP.a. 14. Let w=y—B-—m(#—a) =0 be the equation toa straight line passing through a point a, 6; and P=0 the equation to a curve passing through the same point; then if 3+ m(a#-~-a) be substituted for y in P=0, the resulting equation will be of the form (# — a) P’ = 0, since one value of «is a; P’ will be a function of # and m; and by putting # =a in the equation P’ = 0, an equation will be determined in terms of m; and if the values of m found from this equation make P’ a multiple of (vw —a), P=0 becomes of the form (a — a)? P” = 0, and two values of « will in this case coincide, or u = 0 becomes a tangent to the equation P = 0, 150 GEOMETRICAL PROBLEMS. NO. XVI. DEC. 1845. If the equation P =0 should be of such a form as to make the resulting equation after the substitution of 6 + m(# — a) of the form (# — a)’ P’=0 for every value of m, which will always be the case if P be a rational function of w - a, and y — (3 of which no term is of less than two dimensions, w = 0 will not be a tangent except for those values of m which make P’ a multiple of a—a, since two values of # will only be made to coincide on this supposition, the equation (# — a)’ P’ =0 indicating that two branches of the curve intersect when # =a; hence in this case in order that «=0 may be a tangent to P =0 we must have P = (a — a)? P” = 0. Hence when u = 0 is a tangent to P= 0, if w=0 be sub- stituted in P —-aQR = 0, the resulting equation will be (7 -— a)? P —a(a#-a)Q'.(w@-a) R'=0; or (« —a)’(P’ -aQR’) =0; or «= 0 will in genera] be a tangent to P—aQR = 0, as well as P=0; or the curves will touch one another. If w =0 when substituted in P — aQR = 0 gives the result (vw —- a)’ (P’- aQR’) =0 for every value of m; the value of m which makes w=0 a tangent to P=0, will not make w=0a tangent to P-aQR=0, unless it makes P’- aQ’R’ at the same time a multiple of v& — a, or the two curves will not touch one another unless the equation w=0 to the tangent to P=0 when combined with P—aQR =0 gives a result (@ — a)? R” = 0. If «=0 be substituted in P* -bQR=0, every value of m will give the result (w — a) (P® -bQR’) =0; and the value of m, which makes w=0 a tangent to P =0, will not make w=0 a tangent to P?—bQR =0, unless P? —- bQ'R' be a multiple of «- a; or unless « = 0 when combined with P? —bQR =0 gives the result (vw —a)>R”’ =0. Hence P=0 does not touch the curve P? —-bQR = 0. GEOMETRICAL PROBLEMS. NO. XVI. DEC. 1845. 151 (8) If y-—m’x*-a(y-—#)=0; when y-a#=0, we have (1 —m*)«?=0; or two values of w become =0; and y — v =0 touches the curve y® — ma’ — a(y — #) = 0. If 2 —m’a*® — b(y— x)? =0, this represents the equation to two straight lines passing through the origin ; for therefore y has two constant values; and neither of the lines v will touch or coincide with y — # =0 unless the value of one of the constants is unity. 15. Let a’, 0’ be the semi- conjugate diameters of one of the ellipses drawn to touch the hyperbola in a point P (fig. 149) whose co-ordinates referred to the asymptotes are a’, y’; T'Pt the common tangent, C the centre; then because 7'Pé is a tangent to the ellipse 9 ’9 (eg SEN ae w y also because 7'P¢ is a tangent to the hyperbola, 12 b”? CT =2a, Ct=2y3; «©. >= 22’; x = 2y', or a? =2H"; b? =2y and @?b? =497? y? =4m'*; ..ab = similarly, if @”, 6” be the semi-conjugate diameters of any other ellipse, a”b” = 2m’. Let tangents be drawn from a point h, & in the hyperbola to touch the ellipse whose equation 1s we y OORT ae 152 GEOMETRICAL PROBLEMS. NO. XVI. DEC. 1845. then the equation to the straight line joining the points of contact is : he ky ao, eo signe and if tangents be drawn from a point a, y, to touch an ellipse whose equation is ie e tey giz t+ pms b the equation to the straight line joining the points of contact is UC YY : ew ee 2 ql’? b’’? ( ) and in order that equations (1) and (2) may be the equations to the same straight line, 2 h H, Fees Yi ea, YF; hk ay, mm mires a fy 9 ‘ » 3 22 * a ° ‘ = 3 Dee on epee a *b “a* bh? 2m? 2m? therefore #,y, = m*; which shews that a,, y, are co-ordinates of a point in the hyperbola. GEOMETRICAL PROBLEMS. NO. XVII. DEC. 1846. 153 ST JOHN’S COLLEGE. Dec. 1846. (No. XVII.) 1. MaewnitupEs have the same ratio to one another which their equimultiples have. Give Euclid’s definition of equal ratios. Explain why the properties proved in Book v. by means of lines, are true of any concrete magnitudes. 2. If two straight lines be at right angles to the same plane, they are parallel. 3. With the four lines which contain a + 6, a +, a — b, a —c units respectively, construct a quadrilateral capable of having a circle inscribed in it. Prove that no parallelogram can be inscribed in a circle except a rectangle; and that no parallelogram can be described about a circle except a rhomb. 4, From two fixed points draw lines to the same point of a fixed line, such that the tangents of the angles which they make with the fixed line are as the perpendicular distances of the points from it. Also when the tangents are in any other ratio. 5. Inthe circle, of which 4B is the diameter, take any point P; and draw PC, PD on opposite sides of AP and equally inclined to it, meeting dB in C, D. Prove that AGCt BC: AD: BD: 6. Two similar ellipses have a common vertex and a common direction of major-axes: a common tangent meets them in P, Q; and a perpendicular to their major-axes through the vertex meets PQ in O. Prove that OP = OQ. 7. The circle described from an extremity of the minor- axis of an ellipse, with radius equal to the distance of either directrix from the centre, will touch the ellipse in two points, one point, or not at all, as the eccentricity is greater, equal to, * or less than 4 \/ 2. 154 GEOMETRICAL PROBLEMS. NO. XVII. DEC. 1846. 8. In two hyperbolas concentric and similarly situated, take two points whose abscissas are as the real axes of the hyperbolas. Prove that the locus of the middle points of the line joining them is a hyperbola concentric and similarly situated ; and that the real axes, as also the imaginary axes, of the three hyperbolas, are in arithmetical progression. 9. The lengths (a, 6) of two tangents to a parabola at right angles, are connected by the equation as 1 += =... 4 4 2 bs as (4 latus rectum) 10. If from the focus of a hyperbola as centre, a circle be described with a diameter equal to the imaginary axis, it will touch the asymptotes in the points where the nearer directrix meets them. 11. ‘The quadrilateral PQRS is inscribed in a circle. Join two opposite angles P, R; draw perpendiculars from ,§ on PQ, PR, QR: the feet of these perpendiculars are in the same straight line. 12. ‘l'wo ordinates of a parabola meet the axis at points equidistant from the focus. If the vertex be joined with the point where one of the ordinates meets the parabola ; find the equation to the locus of the point where this line intersects the other ordinate; and trace the curve. 13. ‘T'wo tangents are drawn to a parabola making angles 6, 0 with the axis. Prove that (1) if sin@. sin 6’ be constant the locus of the intersections of the tangents is a circle whose centre is in the focus; (2) if tan@. tan @’ be constant the locus is a straight line perpendicular to the axis; (3) if cot@+cot@’ be constant the locus is a straight line parallel to the axis; (4) if cot 0 — cot 0’ be constant the locus is a parabola equal to the original parabola. 14. Any three tangents to a parabola, the tangents of whose inclinations to the axes are in harmonical progression, will form a triangle of constant area. GEOMETRICAL PROBLEMS. NO. XVII. DEC. 1846. 155 15. In two ellipses (or hyperbolas) concentric and simi- larly situated, take two points P, Q whose abscissas are as their major-axes; and P’, Q’ two other such points. If 00'q' be the angles which the tangents at PP’QQ’ make tan@d tang with the axes; prove that ——, = At tan@ tang If the curves be likewise confocal, prove that PQ’ = PQ. 16. Any number of ellipses (or hyperbolas) concentric, similar, and similarly situated, are intersected by a line parallel to a directrix in PP’P”’...: prove that the extremities of the diameters respectively conjugate to the diameters through PP’P’”... are in a line perpendicular to the directrix. 17. If the above curves be cut by any concentric hyper- bola, whose asymptotes have the same direction as their axes, in QQ’Q”...: prove that the extremities of the diameters re- spectively conjugate to the diameters through QQ’ Q”... are situated in another branch of the hyperbola. 18. An ellipse being traced on a plane ; the vertices of all the right cones of which it might be a section, are situated - in a hyperbola whose imaginary axis is equal to the axis-minor of the ellipse, and real axis equal to the distance between its foci. And conversely, the locus of the vertices of all the right cones, of which this hyperbola might be a section, is the original ellipse. 19. Find the volume of the pyramid of least volume which can be formed by three planes touching a given right cone, and the plane of the cone’s base. 156 GEOMETRICAL PROBLEMS. NO. XVII. DEC. 1846. SOLUTIONS TO (No. XVII) 1. Evucriip, Prop. 15, Book v. Potts’ Euclid, Def. 5, Book v. and note to the definition, Pic. 2. Euclid, Prop. 9, Book x1. 3. (1) Let ABCD (fig. 150) be a quadrilateral figure touching a circle in the points a, b, ec, d; then Aa=Ad, Ba= Bb, Ce=Cb, De= Dd; . Adat+ Ba+Cvo+De=Ad+ Dd + Bb+Cb, or AB+CD= BC+ AD; hence the sum of the two opposite sides is equal to the sum of the remaining two sides. Hence take AC any line less than (a -b+a-—c); describe a triangle ABC having AB = a + b, BC =a+e; and upon AC describe a triangle ADC, having CD =a -—b, DA =a-—c, then a circle may be inscribed in the quadrilateral figure ABCD. (2) Let AB, CD (fig. 151) be two equal and parallel chords of a circle whose centre is EH, then ABCD will be a parallelogram ; from EF draw EF perpendicular to 4B, and produce FE to meet CD in G; then 2zEGD=2z2EFA=a right angle; and hence WG is parallel to BD, and the angles at B and D are right angles; similarly, the angles at A and C are right angles. (3) If a quadrilateral be described about a circle, the sum of the opposite sides is equal to the sum of the remaining sides; and if the quadrilateral be a parallelogram whose sides are a,b; we have 2a = 2b; .. a = 6b, or the parallelogram is a rhombus. GEOMETRICAL PROBLEMS. NO. XVII. DEC. 1846. 157 4, Let A, B (fig. 152) be the given points; draw 4M, BN perpendicular to the given line; produce BN to P, and take NP: BN :: n: 1, where n is the ratio of the tangents ; join 4P, meeting MN in Q; and draw BQ; then tan 2 AQM _ tan eee GN PN: : tanzBQN tanz BQN BN tan Z AQM _ AM tan Z BQN | BN’ then —— = —, or PN=AM;; hence in this case make PN = AM. 5. Since angle CPD (fig. 153) is bisected by PA, CAS RAD: Chee D: and since angle APB is a right angle, PB bisects the exterior angle of the triangle CPD ; AO Pe er Da Gin BD ; Mbence) GAL: AD CBr ehD,» on AO rs BC: 2 AD 2 BD. 6. Let A (fig. 154) be the common vertex, CD, CD’ the semi-diameters of the two ellipses parallel to PQ; CB, CB the semi-diameters parallel to 40; then since the ellipses are similar CDe.C D FOG] CDAD “FQ0 rere Nh OmMOr TCR (40m S So 7. The equations to the circle and ellipse are a” ] 2 aa eee har She aks x v+(y+ by= @? and Kageyama: hence at the points in which the circle meets the ellipse, 2 fin een rae ba. be = as, y+ 2by +b =—, 158 GEOMETRICAL PROBLEMS. NO. XVII. DEC. 18406, zg. 9 2 e : Ay, - b°=0; l1—e eay 6" b° Se et OS BOL 1)" Sa ee b ea awbe and since y has only one value, the ellipse and circle will in general touch one another in the two points in which 5? aw? y\? bt aaaea7e Mee (") -1~(5) or ial / eat — b' oes 2 If ea= = and if 5 = —j-— — 5 ant owe Poh a ( (7) GEOMETRICAL PROBLEMS. NO. XVII. DEC. 1846. 159 which is the equation to a hyperbola whose axes are a +a’, b +06’; which are arithmetic means between 2a, 2a’ and 2b, 2b’ respectively. 9. Let PSp (fig. 148) be any focal chord making an angle 9 with the axis; PQ, Qp two tangents intersecting at right angles; Qp=a, QP=b; latus rectum = 4m; angle PSM = 0; : , and SP = Dee : sin? — Z SPQ = h Sd 4 NE ree pers cay ence @ psing r sin — cos’ — 2 len 0 m = COS P 2g 6 0 2 cos — sin” es b3 1 + {= ——_, = as ms Mm calbo 10. The equation to the circle is (v -ae)’?+y=0'; b and to the asymptotes y= +-—a; hence at the points in a which the circle meets the asymptotes (a — ae)’ + (2 —1)a° =a’ (e— 1); * ea? —-2aevr+a°=0, or (ev -—a)’=0; 160 GEOMETRICAL PROBLEMS. NO. XVII. DEC. 1846. which shews that the circle touches the asymptotes in two b 2a, b points, the abscissa of which is — , and the ordinates — and ay e e respectively; or at the point in which the directrix meets : ee : : a them, since for every point in the directrix w =-. e 11. If from any point in the circumference of a circle circumscribing a triangle perpendiculars be drawn upon the three sides, the feet of the perpendiculars will be in the same right line (Ex. 15. No. xiv); and § is a point in the circle circumscribing the triangle PQR, hence the feet of the three perpendiculars drawn from S' are in the same straight line. 12. Let MP, M’P’ (fig. 155) be the two ordinates ; draw AP’ intersecting MP inQ; AM=a#, MQ=y; AM’=2', M’P’=y'; then w+ =2a; is the equation required. When # = 0, y = 0 or the curve passes through the origin, and limit BAS 44/9: as @# increases y increases, and when v a” = 2a, y 1s infinite, or the ordinate at a distance 2a from the origin is an asymptote. When a> 2a, y is impossible. 4a When @ is negative, — = -—- ; as @& increases y in- & 24+ H@ creases, and when w is very large, the curve approaches to the parabola y’ = 4a (a — 2a) as the asymptotic curve. The curve is that traced in (fig. 156). 13. (1) Let y’ = 4aw be the equation to the parabola; then yy’ = 2a (a + a’) is the equation to the tangent, and if 2a ; ; a —=tand=m; y=m\a + —); y m NO. XVII. DEC. 1846, 161 GEOMETRICAL PROBLEMS. Pp : a : . or i a ee wo OF (1) but when sin @ sin 0’ = a; v v 1) =: ot a 5 ners" a®’ 6 , ] 1 cosec”@ cosec* =— , or = + 1) ( 3 a m m m m 1 1 ’ 1 x ie: ieee ey See ent eg m ™m a mm a _\ i 1 a Pa) apr ‘f 2 i 2 pees hence (a — a)? +y? = - which is the equation to a circle whose centre is the focus, and . Oe, radius = — = a cosec @ cosec 6’. a a a mm = Samick hence 2’ = B (2) Let tan@tan@’ = £6; which is the equation to a straight line perpendicular to the axis at a distance a cot @ cot 6’ from the origin. 8) Let cot? +cot@ =45; .. : o. cs Y mm a which is the equation to a straight line parallel to the axis at a distance ary from the axis. (4) From equation (1) m 2a 2a 1 0 y VJ y? i Aaa’ m 2a 2a P 162 GEOMETRICAL PROBLEMS. NO. XVI. DEC. 18406. 1 2 Aaa! rises May pied Mes 6 m a as? hte y” oo Aa a’ = a®e; hence y? = 4 a (a ~~): 5 which is the equation to a parabola whose latus rectum = 4a, : ao ne) and vertex is at a distance a from the origin. 14. Let P, Q, R (fig. 157) be three points of a parabola whose ordinates are y, y’, y”; 0, 0’, 6” the inclinations of the tangents at P, Q, # to the axis of the parabola; then since the tangents are in harmonic progression, cot 0 + cot 0” =2cot@’; or —+4+ >= a hence y + y” =2y’, which shews that Q is the vertex of the diameter whose ordinate is PR. Let the tangents at P and # intersect in 7’; then the tangent LQM is parallel to PR; TQ= £TV; and the triangles LMT, RPT are similar ; - ALMT =1 ARPT =} APQR =1QV.VRsin QVR ; AG . y — y V?; and = . but —————— an’ QVR QV=e Rh and RV.sin QVR Bra v= (y— Que oe aa! yy" 1 Yai mei and ALMT =—— (y” - y) —— =———.. 16a (y -9) 4 64a Hence ALMT is constant as long as y” — y is constant ; or for any three consecutive points in a series of points P, Q, R, S, &c. which have the tangents of the inclinations of the tangents to the axis of w in harmonic progression. 15. (1) Let a, y, 2’, y’ be the co-ordinates of P, P’, and XY, Y, X’, Y’ the co-ordinates of Q, Q’; then GEOMETRICAL PROBLEMS. NO. XVII. DEC, 1846. 1638 ug 2 tan@ = — a ary t f ¥i mena (4) similarly LAD dg ata, tanO « \y pyri Yi, xX but mS, Gi) + TiN a a b b / a’ y’ y also ——- = 3 ‘Pinar Cooke b AE NT DS @ eed . peer yoo Ye tan@ tang r Sy A Le tan @’ =‘ tan p If the ellipses be concentric and confocal, they coincide ; hence P, Q coincide, and P’, Q’ coincide; .°. PQ’ = QP’. 16. If x, y be the co-ordinate of one of the points as P; b b, then the co-ordinates of D are y’ =— «a, and — is constant for a a all similar ellipses; therefore y’ is constant for all similar ellipses; and the locus of D is a line parallel to the axis, or perpendicular to the directrix. 17. If a, y be the co-ordinates of Q, so that wy =m’; X, Y the co-ordinates of D; then a b Acer = a3 Yossi; b a the positive or negative sign being used according as the series of curves are ellipses or hyperbolas ; fe tak ee ee tf oe = 777 or the locus of D is the hyperbola, or the conjugate hyper- bola, according as the series of curves are ellipses or hyperbolas. L2 164 GEOMETRICAL PROBLEMS. NO. XVII. DEC. 1846. 18. The vertex of the cone will lie in a plane passing through the axis of the elliptic section, and perpendicular to its plane; hence the vertices of all the cones will lie in the same plane. Let S, H (fig. 158) be the extremities of the axis-major of the ellipse; PQ the axis of one of the cones; draw SY, HZ perpendicular to PQ; then if 26 be the axis- minor of the elliptic section, SY. HZ=06°; and S' and H are on different sides of PQ, therefore PQ always touches a hyper- bola whose foci are S' and H, and conjugate axis (6), and P is a point in this hyperbola, therefore the locus of P is a hyper- bola whose foci are S, H. If a’, b be the semi-axes of the hyperbola; S’, H’ the foci of the ellipse, then SHP=e° +020; -. aad = CS’*=CH-. SENT ONE AT Oy a or the extremities of the axis of the hyperbola are H’, S". Again, if S” H” be the axis-major of a hyperbola, whose foci are S, H; then the vertex of the cone will be in a plane perpendicular to the plane of the hyperbola, and therefore in the plane of the ellipse whose axis-major is SH; and if S’Y’, HZ’ be drawn perpendicular to the axis of the cone, S’Y’. H’ Z’ = b°; or the locus of the vertex is an ellipse whose semi-axis-minor is b, and foci S”, H’, or it will be the original ellipse, since in this case S’Y’, H’Z’ are on the same side of the axis of the cone. 19. The volume of the pyramid = 4 (altitude of cone x triangular base) ; and the area of the triangle which touches (a+b+c) the circular base =7 , where a, 6, ¢ are the sides of the triangle, and 7 the radius of the base of the cone; this will be least when a + 6 + € is least ; but a+b+e A B f = 7 cot — + cot — + cot —?; 2 2 4 2 and if A be given GEOMETRICAL PROBLEMS. NO. XVII. DEC. 1846. 165 ae —-C 2 A sin Z COs — oe hes 2) cot — + cot — = — —_ = 2 2 i Be C BC WAM cos —— — GOS cos — sin— 2 2 2 2 is greatest, or when B=C. which will be least when cos ji Hence the triangle will be least possible when 4 = B= C. In this case the area of the triangular base = 3r° (tan 60) = 34/3.7°; hence the volume of the pyramid =1 $3 V/3r} h =/3rrh. where f# is the altitude of the cone. APPENDIX I. 1. To inscribe a triangle in an ellipse, whose sides shall pass through three given points. Let ABC (fig. 159) be the required triangle whose sides AB, AC, BC are to pass through the three points c, 6, a, whose co-ordinates are dz, 63; ds, 6.3 a, 6, respectively; let the centre O be assumed for the origin, and let a, y,3 2 Yo; V3, y; be the co-ordinates of A, B, C, and a, = acos@,; there- fore y, = bsin@,; similarly, let w#, =acos0,, #3 = acosQ; ; therefore y, = bsin @., y3; = b sin @,; and the equation to AB is b (sin 0, — sin 0,) — bsi Se ie gene a (cos, — cos@,) (w — acos@,) ; 0 0 and if tan— = f,; tan age te fo, tan == ta5 2 2 2 b ( - a) b(1 + t,4) y= — ———— ] “ + —————_ Ras diet tee and since AB passes through a point whose co-ordinates are az, bz, we have ab. os (e+ 2) G, (ly 2,4.) oe a or (a4 + a3) t,t, — 7 Osh +t.) +(@-—a3)=0; (1) similarly, a a Maa) tals (t. + t;) +a—-—a,e< 0. (3) b b a liyar = (= a =) = Mz, — (“ *) = Ns a b. a b/ata, b (a — as al bs )= A 5) APPENDIX 1. 167 b (* + “) b (“ =) | = = nr as = ; a b, ae Oh b, A hence Mt, ty a (t, + te) + Is = 0, Mott; — (t, + ts) + Ne = 0, mtyts — (to + ts) +n, =03 by — No to —-n t—-n {ii ae fee ee er an fen Mot; sae 1 m, te —s i Mb, = 1 , ta - Ny (1 ee n,Ms3) t; + Ny, cae Ns ii ee ee ee ie mt, — 1 (m, — mz)t, + 1 = M,N” b; Sap No Pe (1 faeaee! 1,M3) ty + (nm, etre 3) ‘ Myt,— 1 (m, = ms) ty + (A —m ns) ’ hence (m, — mz — mz + 2M, Ms) t,? + 32 oe my, (no -- N3) wz Ny (me a Mz) + Thy Ms + m,n3¢t, +2, —M —N3t+ MN,N3=03 (4) é ) from which equation the two values of ¢, or tan 5 may be de- termined; which will give two positions of the point A, and by drawing AB through c, and AC through 8, the position of the two triangles whose sides pass through the three given points will be determined. Cor. When b=a the ellipse becomes a circle; and a triangle will be inscribed in a circle having its three sides passing through three fixed points. 2. To find the equation to the straight line passing through the two positions of J. Let the coefficients of equation (4) be 4, B, C; then 0 0 Atan’— + Btan> +C=0, 0; . 0, Ge 1 ‘ or A sin’ — + B sin = cos = + C cos’ = = 0; lad -, A(1 -— cos 0,) + B sind, + CC +cos9,) = 0; 168 GEOMETRICAL PROBLEMS. hence (C' — A) cos 6, + Bsin 0, + (4 +C) =0; o (C= A AHB. (=) + (440) <0: or 4 ( — M)) — (M2 — M2) — (M3 — m3) + mM, N_N; — n,m, m; + oe Yi of $2 — m, (2, -+- 3) 34 Ny, (m, + ms) + NzMs3 - mys} + $m +m, —(No+ My) — (Ng +m) + M,NyNs+N,M,mM,=0; (5) and the straight line represented by equation (5) will pass through the two positions of A, which will be determined by the intersection of the straight line with the ellipse. 3. Let a circle be described on the axis-major ED of the ellipse (fig. 160), and suppose a triangle ABC to be in- scribed in this circle whose sides shall pass through the three points a’, b’, c’ whose co-ordinates are 7, On A3» = bs; then using the same notation we shall obtain equations (1), (2), (3) for determining the values of ¢,, ¢,, 4, in the circle; hence the values of 6,, 6,, 0; are the same as in the ellipse ; and if 44’, BB’, CC’ be drawn perpendicular to ED to meet the ellipse in 4’, B’, C’, the three sides of the triangle A’ B’C’ will pass through the three points a, b, ¢ whose co-ordinates are @,, 6,3 G2, b2; as, b; respectively. a = Dike As b D5 4. To inscribe a triangle in a hyperbola whose sides shall pass through three fixed points. Using the same notation as before, let similarly let @ = asecO,, #=asecO;; .. y,=btanO,, y3= 6 tan @; ; and the equation to AB is —btan@ a y fis a (sec 0, — sec 0,) Sa —asec 0,3 ; APPENDIX I. | 169 ry=- 7 ; eee aa tod ty ty +t, a , a Aes asia ts) tatat ag Os (2; +t.) + a;-a=0, or mstt, — (4, +t.) —n, =0; similarly = mgt,ts — (¢; + t) — m2. = 0; m, tot; — (f + ts) —, = 0; hence (m, — m, — m3 — n,m.ms) t,° + $2 +m, (m. + M3) + m (m2 + Ms) — Nom; — m2; ty — (% — 2, — N;3 — MN2N;) = 0; from which the two values of ¢, may be determined. 5. To find the equation to the straight line passing through the two positions of A. If 4¢? + Bt, +C=0; we have (C — A) cos 0, + Bsin 0, + (4 + C) =0, or (4 + C) sec 0, + BtanO, + (C — A)=0; eer CAce C) + Bo 4 (C- A) =0; is the equation to the line joining the two positions of 4. 6. ‘To inscribe a triangle in a parabola whose sides shall pass through three fixed points. | Let ABC be the required triangle, whose sides 4B, AC, BC are to pass through three points c, b, a whose co-ordinates measured from the vertex are a3, b3; a,, b,; a,, b,; let Bis 3 X25 Y23 #3, Y33 be the co-ordinates of 4, B, and C respectively ; then the equation to AB is Yo2- "1 — y, = ——_ (# - a,); Pare) pees \ 1) 5 Yo— Yi Yi V2 = LyY2 or y —=—— # =-—__~;_ but yP =4aa,3 92 = 4a, ; Wo — Vy Vo = UBT 170 GEOMETRICAL PROBLEMS. 4a Yi Yo eee . Yit Ye Yi + Yo and since this passes through the points az, 03 ; eee 40 YiYo ha a ees Y, + Yz Yi + Yo OF N1Y2 — bs (Y, + Y2) + 4aa;=0; (1) similarly yy; — bs (y, + Ys) + 4aa,=0; (2) Y2Ys — 0) (Y2+ Ys) + 4aa,=0; (3) hence y; = Bal op tas 2 seca Y3 = Maes aks 3; Y= ae = eee Yi — 6, y,— 5, y, — bs bY, — 44a, rm (6,6; — 4a) y, + 4a (a, bs — a3b,) 2 bo Yi - 40M, eb (bs =) +O,b,-4aa, — y, -b, ” BEN) 0, Dae De ot Oe 4aa,ty, + [4a 4 (a, + a) bs — (a, + as) b, + (a, + a) byt — 2b,b,b3|y, + 4a 3d, (bb; — 4aa3) — by (a,b5 - azb,)} =0; (4) from which may be determined the two values of y, which will give the two positions of the point J. 7. To find the equation to the straight line passing through the two positions of A. Since y," = 4a; substituting this for y,2 in equation (4) we have 4a, (6, 6, ast bobs 4. b, be air 400) wv, + [4a $(a, + a,)bs — (dy + a3)b, + (a, + a,) bot — 2b, b.b,] y, + 4a } a (b,b, — 403) = b, (a, bs ae a;b,)$ = 0, which is the equation to the straight line passing through the two positions of A; and the intersection of this straight line with the parabola will give the two points required. 8. To construct the triangle geometrically when the three given points are in the same straight line. Let m, n, p (fig. 85) be the three points; draw any line mab through m; through 6 drawn ben, and through e draw ped; join ad and let it meet the line mnp in q; then abed APPENDIX I. T7i is a quadrilateral figure, three of whose sides ab, bc, cd pass through three fixed points in the same straight line; hence the fourth side ad will pass through a fixed point in the same right line. Now when mab changes its position until the points a, d become coincident, the quadrilateral figure abcd becomes a triangle, and gda becomes a tangent from the point q. Hence from the point q draw the tangent gB (Appendix 11. Art. 66) ; join m BA, ACn, BC; and ABC will be the triangle required. Similarly, if qB’ be the second tangent drawn from q, a second triangle A’B’C’ will be determined whose three sides pass through the three points m, 7, p. If the point q falls within the conic section the problem is impossible. 9. When two sides of a triangle inscribed in a: conic section pass through two fixed points, the third side will always touch a curve of the second order. Let the two sides AB, AC pass through the points a,, 6; ; a,, b.; then from equations (1), (2) Art. 1, we have Mz byt a (t, aa t;) + No = O 3 and the equation to BC is : a (a + a) ft, -— (t+ 6) + (a@- 2) = 05 hence puemt 2 ee ts = patois m3, — 1 Mt, —1 (m, + ms) ty — (2 + m,n; + N.Ms) tL, + Nz + Ns 5 (mst, — 1) (mst, — 1) y t? — (mz + Ns) ty + NM; ; (mt, — 1) (mst, - 1) ° and substituting these values in the equation to BC, (a + @) st, — (M2 + M3) ty + n,n; or f, + ts —v lots = aN az red }(m, + M3) t,” — (2 + MyMz + NyMs) by + Ny + nt + (a — #) Sm, ms ty — (m, + m;)t, + If = 0; 172 GEOMETRICAL PROBLEMS. a or $(1 + m,m;) a+ (1— M,M,) & — 5 (m, + ms) y} t? a + [ fm, +m; — (m2 + Ns)$@+ > (2+ mans + My Ms)Yy — (Img + Mz + My + Ms) A] f, a + (1 + mons) a — (1 — Nos) & — © (My + Ms) y = 05 or wt, +v0t,+w=0; where uw, v, w are known linear functions of w and y; and to find the curve to which this line is always a tangent, we must make 2ut,+v=0; .*. v¥ = 4uw; which is the equation to a conic section having two tangents w= 0, w=0; and v =0 for the equation to the straight line joining their points of contact. The same proof will equally apply to the hyperbola and parabola. 10. Ifa polygon of m sides be inscribed in a conic section, and (7-1) sides taken in order pass through » —1 fixed points, the remaining side will always touch a curve of the second order. Let 4B, BC, CD, &c: pass through the points a,, 6, ; Gz O23 .2. Gn-15 b,_,3 then we have m tt, — (4 + to) + Mm, = 0, Motat; — (to + ts) + NM. = 0, Ma yineit, = (2,1 ue t,) +N) =0;5 and the equation to the last side is a (+ a)tt-7y¥(i+t) +(@-«)=0; (1) i—-n f.—-n “4 ee ty: op aeons mt, — i Mt. — 1 ee t cae — M,N») b + (m2 — 2) Ast, + 2s 2 igeliy Soe CEN Ue), ick, 0 ee (mz — m,) t, + 1 — men, rst + os where a3, (335 3, 6; are known quantities. APPENDIX I. 173 = . ts ae nN: t + Similarly, i= ae tea eee a Bema ss mst. a I 4 ty + O1 where a4, 3,, 1» © are known; and ant, + B,, t,, ahs PE A. > rnb a On where ay, Bn ‘ns On depend only upon a, b,; a, by; »++@y-15 b,-,; and are therefore known ; but from equation (1) I(1+2\ 2-4 em tk S02 | a a ob b ; {( +*) i Fh ats +B.) +( --- *t:) (nti + On) = 03 or ut, +vt,4+w=0; where wu, ¥, w are known linear functions of w and y; hence as before v? =4uw is the equation to the curve which is always touched by the last side of the polygon, and is a curve of the second order. 11. If two sides of a triangle inscribed in a conic section pass through two fixed points, find the condition that the third side may pass through a fixed point. mt, to —s (e; + te) + Ne = 0; My tz ts — (t, + t3) +, = 0; 4, — Ns; . t; =————— ; Mt, — 1 , and (mt; —1)¢%, — (4; —7,) = 0; or (m, ts eae? 1) (¢, a Ns) ae (mst, Lane 1) (4; cs ie m1) = Os ore (m, as ms) La fs ate (n,m; ra; 1) t = ¢! Ba n;,M;) t; ae Ns “a ny; = 0 3 but in order that this may pass through a fixed point a, b., we have a or the coefficients of ¢, and ¢; must be equal ; ie N;, Mes —-l=1- NM} 5 or ny M- + N3™M, = rick 174 GEOMETRICAL PROBLEMS. ba+a ba-a, Now m,=- > nhe- : b, a 0, b6 ata; b a-a, m, = — N,=—. Pina wach.) Menai ween baa b\* (@+a, a@-a, aia, a@—4a, hence [(— ———., ; = 2; a} b, bs bs b, b\? (a —a,a Ce) a b, b, _ 443 6, b; Ph ee a b? =a camel 1) If two tangents be drawn from a point whose co-ordinates are a, 6,; a3, bs will be co-ordinates of any point in the chord of contact; and vice versa. 12. ‘To find the position of the point through which the third side passes. Let ab (fig. 161) be a given straight line; and when pairs of tangents are drawn from any point in this line, let the chords of contact meet in the point c; then if b be any given point in ab, and AB, AC be drawn through e, b respectively ; the remaining side BC will pass through a fixed point. To find the position of this fixed point, join be; then if the chord 6AC be made to coincide with 64’B’, the points BB’ will coincide and CB becomes a tangent at B’. Similarly, if A be made to coincide with B’, C will coincide with 4’, and BC becomes the tangent at A’; hence the point a is the point of intersection of the tangents at A’, and B’; and lies in the straight line ab. 13. If three sides of a quadrilateral figure inscribed in a curve of the second order pass through three fixed points, find the condition that the fourth side may pass through a fixed point. Using the notation of Art. 10, ti - 7, 3 5 Mot, — 1 t, — (t, — No = 0; min 3 (Mate ~ 1) te ~ (ts =m) APPENDIX I. 175 (m3t,-—1)t, — (4, — m3) = 0, (m, — m) ft; + (22m, — 1), + (1 — 2,m,) t + 2, — 2, = 0; or ¢;}(m,—m,)t, + (1 — mm,)} + (nem, — 1) t, + (2, —m) = 0; (mst, — 1) {CL — mn.) t + M, -— m4} — (t, — m3) {(m, — m) t, + (1 —2,m,)} = 0, and when the fourth side passes through a fixed point the coefficients of ¢, and ¢, are equal, hence ms; (m, — 2) — (1 — mm) = 23 (m, — m) — (1 — m,n), or M, (N2 — 3) + My (M3 — 21) + mM; (2, — MN.) = 0; 2b" As ae ay Now Mm. — M,N. = — ; eOF ary a \ bb, 3 = Gan ‘ ae ae ees ae hence (a, — a,) bs + (a3 — a2) b, + (a, — as) b, = 0; ". (a3 — dz) b, + (a — a3) b, — $ (a3 - a2) + (a, - az)} bs = 0; ws (dg — G2) (0, — 63) = (a, — a3) (b3 — 62) ; j b, — by dz — Ay or = : b; — b, a3 — ay which shews that the three points must be in the same right line. 14. If m-— 2 sides taken in order of a polygon of m sides inscribed in a conic section pass through m — 2 fixed points ; find the position of the point through which the m —1™ side must pass, in order that the remaining side may pass through a fixed point. Using the notation of Art. 10; ee + Bia t,-1 = ya ae Yn-141 Te n—1 (1) where a,_1, Bn-1> Ya-1 On-1 are constant, since they ofly 176 GEOMETRICAL PROBLEMS. depend upon the co-ordinates of the points through which the first 2 — 2 sides pass; and (mn-1tn — 1) tp_1 = ty, — My-13 (2) for convenience we will represent equations (1), (2), by | x At, + B n—1 ~~ Ci, + D b) “. (ut, — 1) (At, + B) =, - v) (Ct, + D); or (Ap— C)t,t,-(D-Bu)t,-(4- Cr)é, +(Dv—-B)=0; th and (ut, — 1) t,-1 = (& — v); and in order that the m™ side may pass through a fixed point the coefficients of ¢, and ¢, must be equal ; oe D-Bp=A-Cp; and if a,_;, 6,-, be the co-ordinates of the point through which the 2 — 1™ side passes, (=) 7 (Ge ca, \ Tapa emer ey ). aa or = (D ~ (D - 4)b, - (B ~C)a; hence a,, 6, is a point in the straight line in which a,_,, 6,_; lies. Upon the whole we conclude that if a rectilineal figure of nm sides be inscribed in a curve of the second order, and n — 2 sides taken in order pass through — 2 fixed points, there is a certain straight line, through any fixed point of which if the (x — 1) side be made to pass, the remaining side will also pass through a fixed point which will be in the same straight line. In Arts. 11 and 13 the particular cases have been proved when the inscribed figures contain three and four sides re- spectively. If (2 — 2) sides taken in order pass through n — 2 fixed points, and the » — 1" side passes through a fixed point which does not lie in the straight line determined in equation (3) Art. 14, the n side will always touch a curve of the second order. 16. If a hexagon be inscribed in a curve of the second order, and five sides taken in order pass through five fixed points in a straight line, the sixth side will pass through a fixed point in the same straight line. For the hexagon ABCDEF may be divided into two quadrilateral figures; and since AB, BC, CD pass through three fixed points in a straight line, DA will pass through a fixed point in the same straight line. Also since 4D, DE, EF pass through three fixed points in a straight line, the remaining side F'A will pass through a fixed point in the same straight line. M ivf: GEOMETRICAL PROBLEMS. 17. Hence generally, if 22 — 1 sides taken in order of a polygon of 2m sides inscribed in a conic section pass through 2n —1 fixed points ina straight line, the remaining side will pass through a fixed point in the same straight line. 18. If 2n —1 sides of a polygon 4, 4,... Ay, 4, of 22 + 1 sides inscribed in a curve of the second order pass through 2n —1 fixed points in a given straight line 4B, and the gnth side also pass through O the point of intersection of the chords of contact when pairs of tangents are drawn from any point in AB, the 2n + 1™ side will pass through a fixed point in the straight line AB. The chord A, 4,, will pass through some fixed point A in the given straight line AB (Art. 17); and if 4,, 4,,41 pass through any point in the chord of contact of pairs of tangents drawn from A, the side 4,A4,,,, will pass through a fixed point in the same chord of contact (Art. 11); and if it passes through O, the straight line 4,A,,,, will pass through a fixed point in 4B (Art. 12), which will be the point of intersection of the chord of contact of a pair of tangents drawn from 4 with the line AB. 19. By means of Art. 12, a pair of tangents may be drawn at the extremities of a given chord of a conic section. Produce BA (fig. 162) to any point c; and let DDE be the chord of contact of a pair of tangents drawn from e¢ (Ap- pendix 11. Art. 66) intersecting 4B in 6; through ec draw any line cA’B’; join 4’bC’; draw B'C’E meeting DbE in E; then 4E, BE are tangents at the points A and B. If B'bC, meet the conic section in C,; A’C, will meet Db in the same point E. APPENDIX II. ON THE GENERAL EQUATION OF THE SECOND DEGREE. 1. To transform the origin to the focus of the curve. First let the co-ordinates be rectangular, and let ay’ + bey+ca’+dy+ert+ f=o(#, y) =9 be the given equation; transform the origin to the point a, 6 by making ve=a ta, y=y +B; - a(y + BY +b (a +a) (y + B) +(e +a) +d(y' +B) +e(# +a)+ f=, or ay? + bay +ca?+d'y +a + (a, B) =9, where d’=2aB+ba+d; ¢ =bB8+2ca+e. Next, transform the equation to polar co-ordinates by putting a’ =pcos#, y =psing; .. (asin’0+ bsinO cosO +¢co0s")p" + (d'sinO +e’cos®) p + p (a,3) = 0, i Q£\/Q—4PH ‘e ; er rs: or Pp’ + Qe+Pd=0; but Q? —4P® = (d?— 4a) sin’6 + (2d'e' — 4b ®) sinOcos6 + (e” —4c®) cos’0 ; and if a, 8 be assumed so as to satisfy the conditions d?—~4ad=e"—4c0; 2de& —4bH=0; then Q? —4P0 =e? — 4c, and eos ieee + ; (e’ cos 9 + d’ sin ¥ p 20 e* —4c® Pa 40b a OL Ecos (0 - | if [= tand =p rte} M 2 180 GEOMETRICAL PROBLEMS. which is the equation to a conic section from the focus, having its axis-major inclined at an angle 6 to the axis of a, : ve: e’ sec 6 Its eccentricity E = ———__— ; Sales, /e? — 4e® : 20 and semi-latus-rectum L = 43) AY cree 2, eeeeeoreevreseveerescos the two signs corresponding to the equations from the two foci respectively. Now d? —e” =4(a—c)®, and de = 2b0; , , 9) a 1 4 ae eae rg? WS 2) or w— —=—2cot2d= ( etd e d b us b cot20 = ees Set stacsie edie s6 + ianal sag eee ae (4) a2-c#zr/(a-—cy+B = : ) : ; (5) 1 2\/(a—cy +B B+ = pho tree eee eee sees eee see ees (6) ‘i b Be e’ seco A po +l (7 Trae spk tttteeeee ) Peed com F b y) d'e’ / and 7 eee ae Cee (8) / e? —~ 4c b\/e? — 4c ey 2¢ b : a adnate 2. To find a, B and L. d'e’ — 2b® = (2a8+ba+ d) (68 +2ca4 e) — 26 (aB* + baB +ca? + dB +ea +f) = 2abf + (b+ 4ac)aB + 2bea? + (2cd + bea + (2ae+ bd) B + de - 2b (a8 + baB+ca®+ea+dB +f) = — (b°—4ac)aB+(2cd—-be)a+(2ae—bd) B-(2bf—de) =0; APPENDIX II. 181 or ap—ka-hB+g=0 eth ed ateodsobe (9) 2ae — bd pn where h = ———_— = a WANED ho Ade™ b? — 4ae¢ 6? — 4ac Also d? —4a® = (2a + ba + d)?— 4a (af? + baB+ ca’? +dB+ea+f), or d? —4a® = (6? — 4ac)a’ — (4ae-—2bd)a+a@—4af. Similarly, e? —4c® = (0? — 4ac) 0’ —(4cd — 2be) B+ &? —4cf; and putting b? — 4ac =m, d? —4a®=m S(a — h)? + (ak 2)} ; é? — 40 = m {(B— hy + (hk —@)} 2a 2C€ La (a _ h)*? + Taek == ir) Sy (6 —k)’? + ay hk — g); or (a - WY — (B- = "Daeg; (20) and (a-h)(B-—k)=hk-g; from (9) —k ; 5 aM, and w(a—h) =hk—g; -a=h+tzNV/ (hk — g) be oe CLT) B=k+ Vu(hk—g) Hence hé — g and » must have the same sign, and since 2(a —c) 182 GEOMETRICAL PROBLEMS. the two values of » have different signs; and only one of them will make a and £ possible. _ It will afterwards be proved that in the ellipse where m 1 is negative, b (u +=) is negative; and the negative sign Mu alone of equation (6) can be used. Hence in the ellipse . 1 2\/(a —c)? + 6 a./(a+cy+m Me, pe a ae Ih 1 2(a-c) ag rete 1 pea ACRE hee ae b 5 2 and Oh OC et ae fae. { b 1 : Peis af(ateyr+m — 2a +m 1 2c ate+/(a+ePtm a+/a?im where a+c¢=@, or 2 (a? +m) —2a'\/a? +m (a’ —»/a? +m)? = Sa ee E = #E?=(a—h)?+(B—h)° =(u+- me 8). (13) ENA Et iin Jatt haan e ; 2) «5 bee —(a +/a? +m) (= £) (1: \ APPENDIX II. 183 B= AA - B) =~ (Va 4m) (=P ) (15) ome i spay (16) And the co-ordinates a,, 3, of the extremities of the axis-major are found from the equations 1 CE VET eRAGae al +\f/at+m (Bi — WY! = a (B — WY = (a: — hy = = (ah) = J. an Ju (hk —g){. (18) Ifa, a’; B', B” be the two values of a and #3 respectively Te from equations (11), a +a =2h; B+ 3B’ =2k; but a’, 8’; a’, B” are the co-ordinates of the two foci; there- fore h, k are the co-ordinates of the centre. 3. To find the values of d’, e’, and ¢ (a, £). e = (bB + 2ca +e) = Ok +2040) + (Vu + Fe ) Vik =e ~g; ips Re arin ey yy sae 2 : rm hk — pn (ae NBN, sly Me And 2b¢ (a, B) = de’ = we” = (bu + 2c)’ (hk - g). 4, 'To deduce from the above expressions the co-ordinates of the vertex, and latus rectum when 6? — 4ac = 0. In this case the values of A and & become infinite ; let 2cd-be=K, 2ae-bd=H, 2bf—de=G; (a —c) - CE -m then » = ——-——— ; 184 GEOMETRICAL PROBLEMS. 1 (e — ¢) + CET ph and expanding to the first power of m, | 2c m 2¢ (: m ) LS ee em : pea > b 4.a/ 2a ean 2a m Mew iy 0h io 7 ( = Re ) 5 bh b 2ab by 4a a (ae* — bde + cd° + mf) | ) m now hk -g=-—2b and if M=d’?-4af, N =e° — 4cf; mh* — M i (ae — bde + cd’? + mf) ———————— = 48 ue —__—___~" ; ) mk? — N i (ae* — bde + cd? + mf) —__—— ¢ ——___—__.—___“* ; m m* hk-g mh’-'M mk-—N Oy Maen Ve 4cem teh?! hence a= uP mat (14) m m 4Qaa Velo Ef m mM M H ae i m m 8aa 2 HT? 2H s8aa haha, . 1 N K — —- — Kk? —-mN)({1 al eae mm /( if )( ian 9K \ 8ca’ (20) I m (Kk? —mN rie: mN sas ( ) = seeae (1 - ). or L=+ K ar/ea’t Wisi: i and when m vanishes J, = + Kk | a eeseeevrese eee eeesee eee (21) 4/ ea’ APPENDIX If. 185 | m m F’=1 Fip ne? se de = 1 + ora teteeecesseeeee (8) which gives the eccentricity of the ellipse when m becomes very small. 1 H H m mM pie SETHE z ») mm m ( ean 7 ( 3a) Me iia a) yay ell 1 Similarly, 3, = + Be —k); kK «K m m N m N ak oN STDS EARLS Lar m ca 2K? sa?) 2K 8ca? 5. When 6? —4ac = 0, to determine independently the latus rectum, the position of the axis, and the co-ordinates of the vertex of the parabola. From equation (9) (2cd — be)a + (2ae — bd) B — (2bf— de) =0; K G b G or Ka+ HB=G; .. Be GROG or Sr also the two values of », derived from the equation (A) 1. 2(@-Cc) 2a 2¢ nos ines aoe anes: Bs aoe eons & ' Le _d 2aB+ba+d 2a . (2ae-bd) ee” OB +2cate b b(68 + 2ca+e) 2a and cannot therefore = ror 2¢ b hence the only admissible value of 4 is — oe in 2a Again, d?-4a@=—(2Ha-—M); and e”-4ceb=—-(2K B-N); NaH 4 Oe eR Ges Noe ee, 2 CB) 186 GEOMETRICAL PROBLEMS. and the intersection of the two straight lines represented by equations (A) and (B) will be in the focus of the parabola. K H Also oy 2H) ot aK @XB) =G; K H K HOG s’e OH ae ta CARON) SO ee K A = a 5 ‘reed = i A oe ; hence eta) KB -N)-G-—.M-=.N: qe (“*) (2K B-N)=2bf-de+°(d—4af) +9 (e—4ef) ca? +ae?—bde Kk? FT’ | b 4be 4ab’ k? Pree “ 2h 6-N=2Ha-M= — SS eS 4c (@ + ¢) 4a(a+c) N K pa 2 ee (C) 2K 8c(a+e) é M, HH (D) 8H 8ala4e) eoevcevce d’ 2aB+bat+d _ b Again, — e odd Git@ea ae lity Bs leek 2a" (40° +0?) B+ 2b(a+c)a+2ad + be =0; b 2ad+ be or 3 +—a +——__ = 2a 4a(a +c) 0, ose as teat em b c (2ad + be H ees ———— = ——— Sr = - ae eS EK and e b(B+—a)+e e = ( abe ) 2 (a +0) (F) | Ae! fe 2€ lA , but dé = 2b® = ye”? = —Re3 “e?=—4aQ; and L 2@ cosd _ € COse | H cosé a Mn a NS es 4a (a+e)’ oY — — hw, re a APPENDIX II. 187 ve | Kk “, L=—=— ocr *§ cee cee arfa(atecyt 4/e(at+c)i (G) 6. To find the equation to the axis of the parabola. e b ° ° ° Since tan d = — sq? equation (E) is the equation to the a principal diameter. 7. When the general equation of the second degree is re- ferred to oblique axes, to find the polar equation from the focus. Let ay’? + bay+ca®+dy+ex+f= (x,y) = 9, be the equation of the second degree referred to two axes inclined to each other at an angle w; transform the origin to a point a, B by making #@ =a’ +a, y=y +B; . ay? +bay +cu? + dy + ea + > (a, B) =9, where d'=2aB+bat+d; ¢& =bB+2ca+e. Next let the equation be transformed to polar co-ordinates _by making , sin(w — @) , sind = ——_ 0; Yy = cer pte fa 5 a@ = 9 ; sin w sin w -. fasin’@ + bsin Osin (w — 0) + esin® (w — 0)} ° +sinw $d’ sin@ + e’ sin (w — 6)} p + sin’w¢@ (a, 3) = 9, {(a — bcosw + ccos’w) sin’ @ + (bsin w—csin2 w)sin Acos@ + csin®w cos’O} p° + sinw §(d’ — e’ cosw) sinO + ¢’ sinw cosO} p + sin’w@ (a, B) = 0; and representing this equation by Pp’ + Qsinwp + sin’w® = 0, we have 1 Ono ee PH Re i ae : p sinw®P p sin* wD p 2 sinw® 1 ag IQ af GQ 4P® 188 GEOMETRICAL PROBLEMS. Now Q’ - 4P® = {(d' - e' cosw) sin 6 + e’ sin w cos Ot” — 40}(a — bcosw + ¢ cos’w) sin’6 + (b sin w — csin2w) sin@ cos@ + ¢ sin?w cos’@} ; and assuming a, (3 so that the coefficient of sin@ cos may vanish, and the coefficients of sin?@ and cos? be equal in the expression for Q* — 4P@, we have (d' — e' cosw)? — 4@ (a — bcosw +e cos’ w) = 6) sin’ o — AED sin’ gs). moet. aes ee) and 2 (d' — e cosw) e’ sinw — 4 (b sin w — ¢ sin2w) = 0, ... (2) or (d — e' cosw) e = 20 (b — 2c cosw), ......0.. (3) and d” — 4a® — 2d'e’ cosw + e% cos2w = 4 (¢ cos2w — b cosw) = — 4c@ — 2(d' — e' cosw) &' cosw; id? = 4a® =-e = 4e@, 62.6. 8. (4) and d’e’ — 2b®@ = cosw (e? — 4c®). O16 COs ew Hence, uttin SS pees Ft ae ea = tan o eeeceeesceceeeeee 5 P 8 e’ sin w [A ’ ( ) A é PoE ae 1 —e'sinw seco cos (6 — 0) + sinw»/e” — 4c® p ‘ 2 sinw PD al , Ve? — 400 = &' sec 8 cos (0 — 8) (6) or - Se hg a ORL Maa Mere rete Me 30509502: which is the polar equation to a conic section from the focus, 4D whose latus-rectum =——_______ ; wes trainee froin ein ole a ieee ean Has \/e? — 400 ah e’ seco CCCENCTICILY (fy te Sta 5 yes vel ee eee een BA / e” — 4c and inclination of the axis-major to the prime radius = 0. Also from (3) e' (d’ - e’ cos w) = 20 (b — 2¢ cos w) ; . é@*sinwa = 20 (6 — 2c cos), APPENDIX II. 189 ———oor—= poor and EF = vy) ener cers Ae Pees Laren (9) b — 2ecosw” 8. To find p, we have e* sinwp = 2(b — 2c cosw) ® ; d’ : a =sinw + cosw, and d® —e?=4(a-—vc)®; e” 3(u? — 1) sin’w +psin2w}=4(a—c)O; (u? — 1) sin?w + sin2w 2(a—c) OWS Qa as See ee ea uw sInw b — 2c cosa 1 2(a—bcosw + ccos 2 or ae ann EOE ONO COS 28) 5 — 2 cot 2d, ... (10) Mw bsinw — € sin 2w 1 1G and uw+—=+ 4+ (n--] Mh Mu 2»/(a—bcosw +c)? + + (0? — 4ac) sin’w 1 Or w+ -= + 1 i mm bsin w — c sin 2w a) It will afterwards be proved that in the ellipse and parabola the negative sign only can be used in the expression 1 for uw +—. kL 9. To finda, B, and L. z From (Art. 2) d? — 4a® = m {( = het = (hk - 8)| > e? — 4c = m\(B- by + (hk - 2}. de —2bh= —m }(a—h)(B —k) — (hk - g)}; but d? — 4a =e” — 4c, and d’e’ — 2b@ = (e” — 4c) cosw (a - Wy - (B- We 29 Ay yy, eee (12 190 GEOMETRICAL PROBLEMS. b — 2e cosw (Ak —g). (13) cosw (3 —k)’ + (a-h) (8 —k) = —~h si Also Are = Bonus for dividing (12) by (13) and w= 1 2(a—c) {i — 1) sinw + 2pu cosw| = =o SS oe -__--—q—_—<———— << um 9 F J cosy +2 b — 2c cosw sin w sin (w — 0 from which \ = — COSw = alae. sind ; core 6-2 @ . (8 — k) {cose oe = Se Ol age DESEO OS nd(hk—g)3 se.ee (14) h ay es ence (3 ) ; ae re , acd similarly, (a — h)* = ; aes ee an (w — 0) (hk —g), (15) sin’w 2 (bsinw — c sin 2w) d (AE) = k)? —— hk - poe) = ase) sin?d b sin 20 ( 8) bsinw — csin 2 1 or (AE”) = cere seme) (w+ +) (hk — g). ...... (16) Let a-—bcosw+c=a, b — 2ccosw = b’, and (6° — 4ac) sin?» = m’; therefore in the ellipse 2\/a? +m’ 1 2(a@ — 2esin*w) ES Ch a aS ea pie ta aac ’ i SID MM 6b sinw 1 (a — 2¢ sin?w + \/a? + = or — = — Mh b’ sing : a 2¢ sin®w @ af J a? oe ia) nh OD sinw b’ sinw ; APPENDIX IL. 191 arfa? +m 2 (a? +m’) ~2a'/a? +m hence 2? = ———_ >= = a’ + \/a? +m’ _ fae 40D ernie aeg elem)! vy a GLT) m Vitae v( | : = SsIN®. "y * FR? w sad i b E? Seah RS Ata — (a + V/ a? + mi) ( b =) cos ecccececs (18) 2= A(1 Bory oe a) a + fa? +m - 7 hk —- Bos — (a —/a? am a +m) (“8 “= —£). vig) La #(1- E)'= B (1 - BE’); a 24m) (hk — rerg se (a vial + my ) ae" é) Pe eee (20) hk - AE? = —2\/a? +m’ (~=—*). bs natal sf (21) b — 2acosw hk-g (a = hy? = a tan (w - 8) (4), ...@2) 2 hie — (3 - py? = EO tard ( —£); ee (23) Nw or (BG -—k) = b' sin w b (eee pa) . sin w hk-g ; ae ‘ (8 -— k)’ = (a — b cosw + eos 20 ~ V/a" + m)( and (a—h)’ = (0 — b cose +4 00520 ~ /a+ mt) (8) 5 sin? w And if a, Br be the co-ordinates of the extremities of the axis-major ; 192 GEOMETRICAL PROBLEMS. (B, - k)’ = = B-W = (jevee) (8 - hy’ , ‘ m+2c(a'+/a2+m)| (hk-g oF (By — ay — | EEE I (E). ea Similarly, ry mt hele + vies ml hk -—g , Ce iat wammeray reer crt) hk - g\’ And (area of the ellipse)? = 7?_4?B? = — 7m’ ( P 2) ; aie Sa? WB thereforearea of the ellipse = — 4/4ac —B°.sinw ( 5 2) . (26) It may be observed that as in the former case ieee “—*) (~—*) ; ry Pee ee WT he le Sia 2em 10. ‘To deduce from the above the expressions for the position of the axis, the co-ordinates of the vertex, and the latus rectum, when b? — 4ae vanishes. : a@ — 2c sin®w—\/a? + m' In ‘this case we'have), =. se se ee b’ sin w / * 9 m — 2csin*w — — : F a 2c sInw (: ) ihe b’ sin » i b’ 4a’e sin? w 2esnw (3 m ) (1 or haps 3 a 7 Pan ve . 0 b Aa’‘e ) Bu (mk? — Hence (6 —k)? = — bib CP) : sIn w 2em APPENDIX Il. 193 oO ~ K Wi my IG? m N gy age) at oe) m Kk? ie # f, = =a) ea, Wem sa’. 0 8K) | : N K or 6) = Spa Al Celaie. ©. 6 cleie eiesl al alatevets (2) M H imilarl ea oh noe d vce ecb veg eon Similarly, a fe aoe (3) 1 m!' Also Bi — kh = 5 (B- #), and EP =1+, 443 aa 1 m’ m sin? w i ae aE ee Ve, 8a? 8a?’ re Kk K fe ( 1 =a) ' ™m sin? w . 5 ger eee m 5 eee — ‘mm 8ac 2K* ( 8a”? } N K ‘ ee = — — —_(a - sin’ w 2K Faita § ) N = — — —, (a —bcosw +c cos’w); ....... 4. aK sare | ) (4) imilar] sue b + *w) (5) similar a= —— — ——_ _ (€ — 8 COS &) + GCOS’ w). ea-cee Ja SITS Sa? a. : 2c sin w And from equation (1) «= — ee ae ce sin? w a—bcosw+ccos? w sin? 6 = ; , 0s? 6 = ———— a ; a a B N K cos? 6 Get eumiey ie sae ° M Hoos" (w — 6) (6) a, =. oT ao > FF eePMeeeFe ee Boe 194 GEOMETRICAL PROBLEMS. = hae ? m 2em (a — fa? +m’) (" k? — * 2cem 16ca® m? (K? —-mN Pa K? —mN = (Ea) ues ES) | or L when m™ is small; sin? w K (1- m N haviont 2k sin® wk ar/eat Sy and when m vanishes, LZ = When m is very small, B SAM tek m (4ac — b’) sin?'w A> 4a"? 4a” or as 6 —4ac diminishes, the ratio of the axes of the ellipse approaches to /4ac — Bb. SID w 2(a—bcosw+c) OS eiG) OO. 910,078 68S Cees S - (8) 11. When b’ — 4ac = 0, to determine independently the position of the diameter, the co-ordinates of the vertex, and the latus rectum of the parabola. From Art. 5, d?-—-4a0 =—-(2Ha-M), e? -4cb=-2KB-N); and de’ - 2b0=Ka+HB-G; ‘ eee and Ka+HP-G=-(2KB8-N) cose; or 7 (@Ha) += (2K) + (2K B- N) cosw =G; ie) JE Renate. lee? ap + 008) (2K B - N) =G-=M- =n; APPENDIX Il. —b hence — ees ; nen) (2kKB-N) = 2bf—de+ AG = 4af) + > (@ - 40f) 3 c@+ae—bde K* HT’ te Come ahe 408° N K N jel = = = —— 5 é ° 1 S 29K 8ca ORE Aba’ (1) pa emt Mper HG) Me i K i simuarly, Ar et CeCe aie rele shale ole) ere (2) , d : Also — = (cot w +) sinw; and since b? —- 4ac=0; equations 10 and 11, Art. 8, give b — 2¢ cos w — 2csinw p= 5 cr 2esin w b — 2c cosw b bcos w — 2€ ays COU et a eee OlG - uy, en x 2c sin w sin w (b — 2c cos w) Geb bcos w — 2c_ and — = —- or ———— ; e 2¢ b —2ccosw dd gab+ba+d 6 (2cd — be) UG ee eee Pee a ok e bB+t2cate 2c 2c(bB+2ea+e) 2e 3 2ce!' b ee ‘ which cannot = nat hence the only admissible value of « is Cc 2c sin w / (b — 2 cos w)” + 4¢ sin’'w s/c sinw 1 din le sy bcosw — 2C¢_ also — = = 2c 2¢ce b — 2c cosw — 2ecsinw ; and sind = b — 2c cosa 196 GEOMETRICAL PROBLEMS. K bcosw-2c 6b Abecosw —4ac — 40 2a) ' 2ce b-2cecosw 2c 2c(b-2cecosw) 6” i OK hence e¢ = — ——; 4ca % oe, : : L 2@cosd e”’sinwtand.cosd e'sinw sind and Le ee ee E e e(b-—2ecosw) 6b -—2ccosw c sin sin’ w K but sin COE “ = + ——=— ..,...... (3) Ja L sins N K K sin? GF ay 6} a eens + Fa 5 dg aceeee. (4) o sin wt ooh oe Soa Sa L sin (w — 0) rd H Hsin’ Oy RG et Ne pe oe a ate 2 sinw OM 8 Eee iF: 8a? 2 b Kk Again e =b + 2Ca+e= —— 3 B 4ca’’ q sin 0 (=") and ————_— = ~ {— ]; sin (w — 0) b / bK a 2 —— 23.0% Oise Wein cece pel On B+ HO pac © (6) is the equation to the principal diameter ; and the intersection of the two straight lines 2Ha-M=2KB-N; K’ cosw and Ka+ HB- G=—-cosw(2KB- N)=- wean) Aca’ will be in the focus of the parabola. Let A be the vertex, S the focus, 4.Sa (fig. 163) the axis of a parabola, and let the equation be transformed so that ZPSa=0-6; then when the axes are inclined to each other at an angle w, de’ — 2b® = (e” — 4c®) cosa; “. (d — e' cosw) e = 2 (b — 2c cosw) O; APPENDIX I, 197 fe 7 , 2e sin w or pe*sinw = 2b; but w= —-- at . ye esi @.e” p= - Bane. . and is negative since a@ and © are positive. Hence inv order that the latus rectum of the parabola may be positive we must take the equation j _ Ve = 400 meat e’ seco ans + cos (0 - 3} p 20 a/ e” — 4c) therefore e’ sec 0 is negative ; K sinw K sinw but e’tand = ———_—3 .*. e’ sec 0 = ———-> 2a 2a' sind which is negative; hence & and sin 0 have different signs. From equation (6) if b be positive, CD (fig. 164) is the direction of the axis, and the parabola will assume the position A, P, or A,P, according as sind is positive or negative; 1.e. according as K is negative or positive. If b be negative, C’D’ is the direction of the axis; and the parabola will assume the position 4,P, or A,;P; according as sind is positive or negative; i.e. according as K is negative or positive. This equally applies to Art. 6, where the axes are rect- angular. 12. If any relation among the coefficients of the given equation should make d’ and e’= 0; since d?—e? = 4(a —c) ®, we have ®=0; and the equation when transformed to polar co-ordinates becomes Pp’ =0, or Jasin’ @ + bsin@ sin (w — 9) + ce sin’ (w - 0)} p? = 0, and since p is not always =0; this equation determines two values of (8), or the equation represents two straight lines passing through the point a, £6. 198 GEOMETRICAL PROBLEMS. 13. When d and é are each = 0, the origin is the centre ; and d’e’ — 2b = — maf3 — 2bf, d® — 4a® = ma’ — 4af, e? —4c0 = mB? — 4cf; . m(a— f') =4(a-e)f, and cosw (mp? — 4ef) = — maB—20f; 9 tt a see: . A= 2S cq! +Va?+m), m 2 pe OR Nae 2 B= °F (a! = J/ al of m’), m 656-2 2b ies 2 WAS (w wi 9) 228 b sin w m 2(b —2 De iad Wid Sala 1 ity A SIn @w m 2 (b — 2c cosw) sInw B = tan ue m 14. ‘To find whether « = 0 or « , when b — 2c cosw = 0. a’ — 2csin®w — \/a? + m’ Se ’ b’ sinw a =a+c-—beose, b'=b— 2c cosw, or b=b' + 2ecosw; . a =a+cec—0 cosw — 2c cos’w =a — €c0s2w — b cosa, and a’ — 2c sin’?w=a —c— OU cosw: also a? +4 m’ = a? — 2accos2w +e’ cos’ 2w — 2b cosw (a — ecos2w) + b” cos*w + (b? + 4b'e cosw + 4¢° cos’w) sin? w — 4ac sin?w a ~2ac +¢° — 2b' cosw (a — ccos 2w — 2esin’w) + b” ll (a —c)’ — 2b’ cosw (a —c) + b? = (a — € — B'cosw)? +b” sin? w. Now if a>e; (a —e —0B' cosw) - / (a —c— b'cosw)*? + 6” sin? w i b' sinw b’ sinw — ——______,-~ = 0, when Bb’ = 0; 2(a—-c—b'cosw) ~ ‘ APPENDIX II. 199 and when a or c Le ae Ma — (a 4 fa? 4m) hk - SOR — fa —ccos2w+(a—c)} (2 =-2(a ~ ecos’w) Fe BSL ate 7 Pain Nie Ja? +m’) al ) B li: hk-g : hk - g = - fa — ccos2w ~ (a0) = ~ 2esin*w ( 5 ). When a therefore when m is negative, mM — H? must be negative, otherwise every value of # would make y impossible; hence (b—2ccosw) tano is negative, since a is supposed positive; and (b — 2¢ cosw) (.+-) = (b — 2c cosw)p. (* si bh we is negative; but 1 (b — 2c cosw) (u + -) = pr 2\/a2+m' be ce ee sin w .. when m is negative, the negative sign only of /a™ +m’ is admissible. When m is positive, and a positive, the positive or negative slen must be used according as mM—H” is positive or negative ; but this is the condition that the curve may or may not be continuous for all values of #; for if mM> HT”, every value of # will make the quantity under the radical in the value of y positive, or y is possible for every value of w, and the curve is unlimited from # = + © tow =—o. If mM eee (3) 2H =8a(at+e) AG pa 7088 Hcosé 2a 4a (a +c) and cos 6 and tand may be considered to have the same sign ; S40 + 4ac_ . seco = : b b (2ae—1 b — Beer, oo eA ah) aie 20d Tee eee (4) 8ar,/ce(a+ecy? 4,/e(a+c)3 4n/c¢ (a +c)8 16. (1) When the axes are oblique, let the origin be transferred to a point a,, 2, as before, and the equation then reduced to polar co-ordinates ; Sa sin’ @ + bsin sin (w — 0) + sin? (w — 0)} p + sin w Sd’ sin@ + e’sin (@ — 0)} =0; or § (a —bcosw + ccos*w) sin’9+sin w (b —2ecosw) sin Ocos@ + csin’wcos’6t p + sinw §(d' — e’ cosw) sin @ + e sin w cos 0% = 0; hence {2c sin w cos@ + (b — 2¢ cos w) sin 0}? p + 4c sinew }(d' — e' cosw) sin @ + e’ sinw cos Ot = 0 ? Later 2.2% “ees 5 APPENDIX II. d’ — e cosa 2¢esinw let Tyre Le ee es einen oes ae tAN Ch eeetsen (1) e sinw b — 2c cosw Be Jae Sple sin® (0 — 8) p + sed 5 cos (8 - 6) = 0, sin? 0 28 and sin? (0 - 8) p + ~ 5 cos (8 — 0) = 0; which is the equation to a parabola whose axis makes an angle 6 with the axis of w, and latus rectum e’ sin? 6 9aL=- ¢ cos 0 (2) To find Ais (3; and 1 , We have — = cosw — e 2c sin’ w _ bcosw urd Reo ncos ye b —2ccosw 2a3,+ba+d bcosw— 2c bB, + 2ca, +e b—2ccosw’ or 2a 2ae — bd b cos w — 2€ hence — — —— ~~ = ————__ b 6(b8,+2ca,+e) b-2ecosw H 2a bcosw—2e her. b — 2c cosw 2ab—2b? cosw+2be 2(a—bcosw+e) 2a’ b(b — 2c cosw) 6 —2e cosa ee Abs Ho’ and e’ = ieee ner ia or b +2ca,+e= ee2eee 2 2ba’ Bi : 2ba' @) is the equation to the axis ; , ° e ; Hsinw , HAH sinw /c sin also L = —— tandsinéd = —~ sin 6 = ——,— ve Vase cae 2¢ 2ba ba’ Ja since sind may always be considered positive ; HT sin? w. /c pred bc a Ca 3 204 GEOMETRICAL PROBLEMS. Hesin’w —(2ae—bd)csin*w (be — 2ed) sin’ ab/eat 2br/ca’t Bs As in the last case e” = —- 2K 3,4 N, reo! Nae o @K 2K 2K 8Kb%a?’ H b b? d —=+—$ v2 Le aoe a ae K 26 4c” N Kb” N K : or Des ac eaepats a 5 T Salie M2 — bcos we + ccas w) N oe N K Ksin’w ee (4) 29k sae 8a? M HH Hsin’? w ph aie eee 9H saa 8a similarly, a, = 17. Let b*<4ac, and the co-ordinates rectangular, to find the polar equation from the centre, Change the origin to the centre by putting w=a' +h, yay th; . ay* + ba'y' + cx? + b (hk) = 0. Again let this equation be transformed to polar co-ordinates by putting v= pcos 6, y= p sin @; “. (asin® 0 + b sin @ cos 6 + ¢ cos’) p*? + h (hk) = 05... (1) - or [a — {(a — c) cos’ @ — b sin 0 cos Ot |p’? + P (h, k) = 0; ; b and if tan20 = — ——-, we have a-cC fa — (a — c) sec 25 cos @.cos (0 — 20)} p? + hb (h, k) = 0; ... (2) but the polar equation to the ellipse, referred to the centre, whose axis-major makes an Z 0 with the axis of w is $1 — EK’ sin’d — E” cos 6 cos (0 —20)} p= B’; APPENDIX ILI. 205 hence equation (1) may be made to coincide with this equation by putting VEPs) 4.8 re _ (@—¢) sec 26 (3) Fo ee an = 3 eee BOK BP (hy k) 2 —2F’ sin? 6 2a 2— by Bertier ie Seger sweeney gon Whom 5 OF = 3 E? -?2F’sin?é a—e E*®cos20 a-e , qe a+e a+c a an cag eae, ESE = = SS iE (a —c) sec 20 / (a +c)? +m /ar+m the negative sion being inadmissible since a : wy ie —1s negative and cannot = — / a? +m KE , hence — = 1 + eo i /a? +m 2\/a?+m (a —4/ a? +m) or EF? = | ; (4) al +fa® +m m also ie eer) eee ee ree 2 a ee ee (a —c) sec26 . (a’ aes +m) hence B? = ES ae aera. {ho} os aoe RUS ID ae ae aoe gee (6) | A, ki rg see P= B1-E)= aati? eG ae AME ERYG ¢ When 6’?<4ac, asin’ @ + b sin@ cos @ + ce cos’ @ 1 = \(2a sin 8 + b cos 6)? + (4ac — b*) cos* 6} a is positive ; or from equation (1), @ (h, k) is negative. 206 GEOMETRICAL PROBLEMS. From Art. 9, if e' =bB+2ca+e; and d(a,B)=0 whatever be a, 8 we have Is ‘ 2 d? — 400 = m {(B — k)? + = (hk = @)}, and whena=h, B =k, e becomes=0; and = ¢ (h, hk); . = 40¢ (h, k) = 207 (hk - g), or ge ie - (=-*). —=—.. hk Hence 4°? = — (a +\/a? +m) atR) B= - (a' —\/a* +m) (“> =i Pep neeee (9) Tee Se (== S| eee T+ (10) KF? In the ellipse @ (h, k) is negative, and since P is positive, equation (3) shews that in the ellipse (a — c) sec 20 is positive ; . (a —c) sec2d = + fa? +m. In the hyperbola equation (3) becomes Ee (a —c) sec 20 E? (a —c) sec20 = -— —_, or = = __; - B @ (h; k) B? p (h, k) and since in this case m is positive, hk-g M-mh* Mm-— H’ pan--2 ta ote ae which when a is positive, will be negative or positive according as the curve extends indefinitely on both sides of the axis of y or not. 4a 4am 1 — E” sin’ pee eee 2 B p (A; k) 2 a+e 1 ine ha) 7a" eyes APPENDIX IT. 207 and if @(h, k) be positive, (a —c) sec20 is positive =./a’?+m; , Ap Stat Meee 2 a . 2 fa? +m see eet meg OF ==3... (11) EE fa? +m a +/a? +m pm. Eh, ¥) 9 Va? +m ~ @) (a —-c) (a—c)sec20_ P(r, k); (12) aE , Cae é m AP =~ (ofa? +m em +a) (—E S) eases (15) If @ (h, k) be negative, (a—c) sec 20 is negative = — Var +m; , ro) oe A ae i g , and EF? = Vat me 3 1D Jae +m Va?em—a BR Ep (hy kh) _ 2 (h, k) (a — c) sec 20 ti ot 5 SAC EON? k); (16) R? o) 12 4 = —— = _ ANS k); (17) Kk? - 1 Mate yg A or BE = (al + a" +m) ( —£), Ne aE (18) : ee hk — Jie (Va +m - a) ( ; ae amare: (565 18. Let 6? <4ac, and the co-ordinates inclined at an angle w, to find the polar equation from the centre. 208 GEOMETRICAL PROBLEMS. Change the origin as before to the centre, and transform the equation to polar co-ordinates, and the resulting equation becomes {(a — bcosw + ¢ cos’w) sin’ + 'sin wsin@cosO + csin’w cos} p° + sin? wd (h, k) =0; or § (a—bcosw + ccos’w) +b’sinwsin@cos0 — (a—bcosw + ccos2w)cos’O tp + sin’w.o (h, k) = 0; hence §(a’— ¢ sin’w) + b’sinw. sin cos@ — (a’—2csin?w) cos? p” +sin’w. p(h, k) = 0; “. {(a'—csin® w) — (a — 2c sin?w) sec 26 cos @ cos (8 — 20)} p” + sinta@ (heh) 20s «s.r eee - b' sinw by putting tan 2. => ee Lee eee 2 YP 8 a — 2c sin? w (2) and the polar equation to an ellipse from the centre, whose axis-major makes an angle (0) with the axis of a, is (1 — E* sin’ d) — E® cos @ cos (9 - 20)} p? = B’; and equation (1) will coincide with this equation if 1— E’sin?d (a — c sin’ w) Be” sin’wo (hy ht)’ E? ‘ — 2c sin® w) se and — = — & La ee wah eos eee (3) B* sin’ w p (h, k) 29 —2K sin? 6 2a — 2esin* w 0) WOM econ rune: mreronprarenne cde ge E? — 2 EF? sin? 3 a’ — 2csin?@® ’ 9 — fi a’ hence = : 5) eae ? E*® cos26— a’ — 2e sin’ w Q j a or. = = -}- > P ie (a’ — 2c sin? w) sec 26 APPENDIX II. 209 5. (a’ —2esin?w) sec26 is positive from equation (3); hence equation (2) gives (a’~ 2c sin’w) sec 20 = J (a — 2csin?w)? +b sin?w =\/a? 4m’; a’ 2 or -._=1+ = 5 yD "a m’ , aa m A WD} : abel FEI eervrereoerrssees200 (4) — E® sin? w. o (hs By a2 sin® or (a’ — 2csin* w) sec20. om joe —/a? +m’) (h, k) 2(a’ —»/a? +m’) Ne. (5) a 7 h, k A? = 2(a +a" +m) AE) 5 (6) 1— E m —2(a'—/al’ +m’)? p (h; k) P= B (1 - £’*) Cae To ee ee (7) : m’ 2o(h, k hk — and as before, eR) aes AE 8), m h hk A= —-(a+Va +m’) (- =) Pa Mine Fob: (8) aca a ean) (== *); BATT 2 80} pra VET (=o) m b .. (10) In the hyperbola, equation (3) becomes 1 —E” sin? 6 a —csin’w oe i? bs (a’ — 2c sin*w) sec 20 DR sint?wd (hk) Be sin’wop (h, k) O 210 GEOMETRICAL PROBLEMS. Hence, (a) if d (h, &) be positive (a’ — 2c sin’w) sec 20 is positive ; 2 a he —_ = eee eS ee a at ee BE DE a? erg BO * (a — 2e sin’) sec 26 tee m’” EP sin’w. db (A, &) - i P(hsk) 2 ee ee x _ i ihe (a’ — 2c sin’w) sec2d (Jame mi —a) m oe 4 h,k A’ = = = 2 (/a® +m’ +m +a ee Be iy scidas eemeee (12) Cia Reg AY Ape. or A? = - (a + \/a? + m’) (“== y teeter eee (13) eiieratseet +5) B= (a - /a* +m’) ——*. Toda cscss ss4 beta eee eee (8) If @(h, &) be negative, (a’ — 2csin’w) sec 26 = ~ J a” SS and oe 2n/a® +m’ Trg git eee mt 7 1 B= -2(a' + V/ a" + mi een = (a 4+ VSa? +m’ my (“F 2 k ate _ Pe = ~2(/ 4m a) P29 WF +m'—a') a 2) . (16) : b’ sinw Also since tan20 = — ae a —2csin*» we have, (IV LE Bi=3G, tan 20'= 0; .. d= \0,.0n 90. | b’ sinw 2) If ac tan2o = — ———____— 2) : a — 2csin*w (6 — 2c cosw) sinw 2ccos’w — b cosa APPENDIX Ii. ot} (3) If a’ — 2csin*’w = 0, or a — b cosw + € cos 2w = 0, tan20 = ©, and 6 = 45. * (4) If a’ — csin’w = 0, or a — b cosw + € cos’w = 0, 1 — E’sin?d RP 0: .. E =cosec 0. (5) To deduce the latus rectum, and co-ordinates of the vertex of the parabola by making m vanish. The co-ordinates of the extremities of the axis-major, are A’ sin’ (w — 0) eNiay? cyte sin*d a a t= os es =-=h— ii sin’ a sin? w hk-g H* —-mM Kk? -mN also £2 See ee Se ee ee eee See b . 2am 2cem? nd when m’ becomes very small a y ‘ (a —\/a®+m'y m’* 7 ae 4 a at nearly ; iy cee m’* aN. (k? —mN) sin* w 8a” 2em l6ca® i K sin’ w m N hence ed een osanen CAR) 4r/ea? 2K? : Ksin’w and when m vanishes L = 4 ————. ...........200- (B) 44/ea’3 a —bcosw + ¢cos2w —1\/a? +m Also tand = —_ Mi : 6b sinw m’ a’ — 2esin’w — (7+ =) P 2a (1 m A a a ae -+- ——— ; . b’ sin w b! } ha’e 212 GEOMETRICAL PROBLEMS. pa os Wi a / a? +m’ ( 2a’ cosec 20 = — Ss Se ee es b sinw b sinw 2 ° , . i tano ce sinw m — b sinw S107) ee ee (A ees nh eee 2 cosec 26 b ; m a+ 2a e ) , ee c sin*w m m or sin?) = j ae trap hved i. 73 a 4ac 2a* ce sin’ w m = —— 11 + —,— (a — b cosw + cos 2w) 75 a 4a°C eae \ (K?—mN | hence (3,= J (20+) ( E. YS {i +25 (a-beosw+ ccos2e)| a in © 2g 2cm*> Ja 4a’*e K pie ; mN mm m (eee 20) hy ee — _ — = eRe eS TPT @ — OCOSW + € COS ZW m m K? 4a” 4a7e 1 ere mN wm m eee ee ca +— +, (a — bcosw + ccos2w)?, nm m Oke 8a 8a N K ee oN i= (a — 6. cO8w +0 COS a): eer B, = =~ ae ( C08"). sere (C) Similar! za (c~b * eo): t(D) imilar = —. — — (ec — bcos @ COS" @). wee When the two axes are tangents to the curve, M=0, hk - g (~—-) (“): b 2am i 2em N =03-. and hk - either of which expressions may be put for Soe ah the values of A, B, L. 19. To determine the position of the axis, the co-ordi- APPENDIX II, A, bes nates of the vertex and the latus rectum of the parabola whose equation is ay —bxy +cx* —dy—ex+f=hp(a#,y) =09; where 6° = 4ac; d’=4af, and e =4cf; the co-ordinate axes being inclined to each other at a given angle w. Change the origin to a point a, 6 in the curve; and trans- form the equation to polar co-ordinates ; Sa sin’ @ — bsin @ sin (w — 0) + ec sin’ (w - 0)! p + sinw jd'sin@ + e' sin (w — 0) = 0; where d’=2a8-ba-~d; e=—bB+2ca-e; and @ (a, 3) = 0; or }(@+ bcos + ¢ cos’w) sin’@ — sinw (b + 2c cosw) sin @ cos +c sin’ w cos’6} p +sinw }(d — e’ cosw) sin 8 + e’sin w cos0? =0; and since 6? = 4ac, $(b + 2¢cosw) sin @ — 2c sin w cos 0%” p + 4esinw {(d —e’ cosw) sin 9 + e’ sinw cos} = 0. d’ : 1 —é€ COSw 2esnw Let we = ——___ = tano; é sin w b + 2¢€ cosw 4¢ sin? w ‘pi 4ce’ sin? w Snare (0 —d)p oT REE So come ape which is the equation to a parabola whose axis makes an angle 6 with the axis of #, and latus rectum e’ sin’ 2e’sinw 2h = — ———_ = ~ '+—______ gjn 6. ¢ COs 0 b + 2c cosa / CRs d 2afg—ba-d 2¢ sin* w bcos w + 2c Also = = ———— = COS . e 2ca —b3—- b+2ccosw b+ 2CCOSw therefore by reduction 2(a+bcosw+c)(bB - 2ca +e) =db + 2ae + (2cd + be) COS w 5 214 GEOMETRICAL PROBLEMS. but db=2ae, and 2cd=be; - —2(a+bcosw +c)e' = 2d(b + 2c cosw) ; 2d sin w sino hence 22 = a+bcosw+ec’ Ve. sing and sind = ——~—=====- =~} /a+bcosw +e dr/c sin? w (1) —_— (a + bcosw +c)3- weseveseveeeev ee eee The axes of # and y are both tangents to the parabola ; and if a,, 6, be the distances from the origin at which the curve meets the axes, a d Ci teo b 2b OJ — = 2b,, WE ere - = mule : a 1 a a, heey fa _ — sin’ w 2b, -— sin’ w a a ay >. Lb = wm = om —_ b c\? 2b, bv 71s (1+ cos w + © 1 + —cosw + (=) y 2a,"b,? sin’ w (2) (a,? + 2a,b, cos wm + b,°)3 Hence if 4B, AC (fig. 165) be two tangents to a parabola meeting in A, and CB be bisected in D, since (2AD) = a, + 24,5, cosw + b,’, 7 24B' AC. sin?w _ (4 ABCY % Tiere oe The equation to the curve may be put under the form aha) ae eos ys bos: APPENDIX If. CY a Aay a Baten pS poo 1 = ee : @ “ b, a,b, @ y h aaa J) = 1, ence _ es (8) 4a," b,? And if w = 90, the latus rectum = ——-——. . Geren Se 20. To find a, 6 the co-ordinates of the vertex. d” — sad (a, B) = (2bd + 4ae)a; or d? = 4bda; e” — 4c@ (a, B) = (2be +4ced)B; .. e? = 4beB; and —(a+bcosw+c)ée =(b+2ccosw) d; —(a+bcosw +c) d' = (bcosw + 2c) d; ae b ) —- | — + — COS Ww a d (b cos w + 2c)’ a\a OS ———— FO 46 (a +b cos ey? b b Ge ( i eon c) 4-(1 +7 cosw + *] a a a a, 6,” (b, + a, cos w)* ore 1 1 (0, 1 @) (5) oY (a)? + 2a,b, cos w + b,*)’ b, a,’ (a, + 6, cos w)? (a,” + 24,6, cos w + b,”)? Similarly, 6 = (e) (¢) If = =m, then ee 1 Qa 1 +m COs ® 215 2 ) ; hence if m be constant, w is constant; and if parabolas be drawn touching a two given lines 4B, AC so that the chord of contact BC is always parallel to a given line, the locus of the vertices of the parabolas is a straight line passing through 4. 21. To find a,, 6, the co-ordinates of the focus. L sin (w — 6) i d\/ ae sin? w =a 7 ae 2: SIMes 2(a +b cosw +c)’ 216 GEOMETRICAL PROBLEMS. 4ced(a+bcosw +c) _ cd OF 10) = ee ee ee ee es 4b(a+bcosw+c)? b(a+bcosw +c) b,’ a, ee “+ 20 b, COS Ww Ee) ba aD; Fig (ER A. tae Gee SB EER oe atelate e a,” + 2a,b, cos w + b,” hence a, = Similarly, 3, = Misael Migs MPa riya 9), al By a (3) ay b, by a hence if a series of parabolas be drawn to touch two given straight lines so that the chord of contact may be always parallel to a given line, the locus of the foci is a straight line passing through 4. 22. If § (fig. 165) be the focus, and AS = py; bisect BCs in D, then p,’ = a,’ + 2a, B, cosw + B,” : (a,6,)” = (677 4+ 22a,0,.cos *) Ora OL ucoea erat) (a,*° + 24,6, cos w + 6,*)” a,b, t= >= = Os tee eae a,” + 2a,b, cosw + 6; em: (4) AB. ¥ hence eee or 24S8.A4D=AB.AC; ... (5) and if SE, SF be drawn parallel to AC, AB respectively ; AS* = AEF .AB= AF.AC and a circle may be described roungsihe pointes Bow, I, Co 0... . Seema eee (6) 23, When a parabola slides between two straight lines inclined at a given angle w, to find the locus of the focus. 2a," 6b,” sin? w 2p; sin?w 2a,/(3, sin? w ms (a,° + 24,6, cosw + 6,”)# ‘iy a,b, pi > APPENDIX LL. ays pr 4 sin‘ w 1 2 COS w 1 4 sin @ ee aie + ——— + — = iy op L? 2 ay” a, 3, Br cL? is the equation required. 24. When a parabola slides between two straight lines inclined at a given angle w, to find the locus of the vertex. Let x, y be the co-ordinates of the vertex in any position ; then w = a,b, (b, + a, Cosw) b, a,” (a, + b, cos w)” 4 D) Up ee 4 5) Mm Yr; 2a," b,” sin’ w ip calc Se aa ual J & . 1 where r,” = a,” + 2a,b,cosw + 6,7; and if — =m, we have ay x (m + cos w)* L 2sin?w (1 + 2m cos w + m’)3” y (1 + m cosw)* L Qsin?wmr/1 +2m cosw +m Qu (m + cos w)* On == a /1 + 2m Cos w + mM 1 2 (- + cos o] 2y m an TY 7 edie a (- + cos 0 + sin? w Mm ‘ L where a= - : sin’? w 2 2 Ags th wv a) * 9 hence m + COSw = Q{—+- = + sin’ w) ; (cao a 1 : pe % La 5 ve (o+2 7+ sintw) m a” a a” 918 GEOMETRICAL PROBLEMS. 24 (x / xv* % ; a al es ke = + sin‘w} ~ 608 @ a \a@: a 4 era. Ves sine ) = cose} = 1 25. ‘To find the equation to a conic section passing through five given points 4, B, C, D, E. Produce BA, CD (fig. 166) to meet in O; and let OAB, ODC be the co-ordinate axes; O4=a, OD=b, OB=da', OC = 0b’; then the equation to the conic section is tt 4 \ fare y é ~ —— 1 — pa 1 | = 05 ray + (43 ) (+4 ) and in order that this may pass through the point EL whose co-ordinates are a’, 6’, we have “t th i} rab" + + (— oo We ,-1) =0, al’ b” a’ b” wv Yy . v y by ae oa ey al Ae 571] (= v1) =0, el Ga ese Jey ; oe a’ is the equation required. Cor. Hence may be found the equation to the parabola which passes through four fixed points. The general equation to the conic section is ee en eae as GH faces ee Ae =0;: es Vier a Er or Ay’ + Bay+Ca’?+Dy+Ex+1=0; and if dX be assumed so that B’ = 4AC, we have ae, Quy nae (- i, “| E *) ies say eae et gay ay eee aa err TT bb’ Opal J ‘ which is the equation required. APPENDIX II. 219 Also the equation to a line through the origin parallel to , 2C Pe ie the axis of the parabola is y + Fs «= 0, (Art. 11, Equation 6). Therefore the line y= + \/ —, & is the equation to a aa line through the origin parallel to the diameters of the two parabolas. 26. To find the locus of the centres of all the curves of the second order which pass through four given points.’ The equation to the conic section is Cae a v Yy SS Rp Oy ON WERE eS ig KO and if h, k be the co-ordinates of the centre, the equation from the centre becomes ne) (fs EVR ee eee 1) “44 (2 J vibe ODODE HE + (S451) Natet ate) jo hh : fig. 1 oe +5(5 b! 2k? = Bh? 1 1 eg aie Te een pane ee cn BAAN ck (; Hi >) x1 which is the equation to a conic section, the co-ordinates of i + me 0 whose -centre are viet 5 220 | GEOMETRICAL PROBLEMS. D Aig eo ie eile atte oh 2k es — O% site * i aa im B SS a bb’ a (9) 27. If w=0, v=0, w=O0 be-the equations to three straight lines, then Su + / av “e V/ Bw = 0 will be the equa- tion to a conic section which touches the three straight lines, where a and (3 are arbitrary constants. Since \/u+/av + / Bw = 0, Uu= av + Bw —2\/abV/vo; “. (u-—av— Bw) = 4aBve, or u+a'v’ + Bw — 2auv — 2Buw — 2aBvw =0; and when wu = 0, a’v? + Bw —2aBvw=0; or av— Bw=0; therefore the straight line w= 0 will meet the conic section in the straight line whose equation is av — Bw =0; or in one point only ; hence w= 0 touches the conic section. Similarly the straight lines v = 0, w= 0 will touch the conic section. 28. If w=0, v=0, w=0 be the equations to the three lines BC, AC, AB (fig. 196), which touch a conic section in the points a, b, ¢ respectively ; since av — Bw =0 is the equa- tion to a straight line passing through 4 the intersection of v = 0, w = 0, and also through the point of contact with w = 0, it is the equation to da. Similarly the equations to Bb, Ce are u— Bw =0, u—av=0; and at the point of intersection of Bb, Ce, (wu — Bw) — (w-av) =0, or av — Bw =0, which is a point in Aa; therefore da, Bb, Ce meet in the same point. APPENDIX II. via! 29. To find the condition that the straight line whose equation is w + Av + Bw=0, may touch the conic section whose equation is /u AL ae / Bo = 0. We have wu =av + Bw + 2\/ a/v; hence at the points of intersection with the line w+ Av + Bo = 0, (A4+a)v+ (B+ B)w + 2V/aBV/ vw =0; (2) this will in general, when combined with «+ Av + Bw =0, give two values of #, and intersect the curve in two points ; but when equation (2) is a complete square, / (A +a)v+/(B + B)w=0, or (A+ a)v=(B + B)o; therefore w is determined by a simple equation and has only one value; in which case equation (1) becomes that of a tangent. Hence (4+a)(B+ B)=af; or oF imo, which is the condition required*. 30. To find the condition that a 3 —1=0 may be the a equation to a tangent to the curve ay +bveyt+cue?+dy+ex+f=0 referred to two tangents as axes. a (3 (5) + baf eal + ca® (=) +48 a + ea (=) +f=0; =. a (1 -*) +ba8(=) (1 -*) +ca° (*) +28 (1 -*) +ea - +f =0; a or (a B’—baB+ea’) (“) "(2a B'—bap+dB—ea)” +aB"+dB+f=0; (1) a " > and in order that — + F —1=0 may be a tangent, - must a a only have one value ; and equation (1) must be a square ; * This condition is given by Mr Hearn in his “‘ Researches on Curves of the second order.” Waray GEOMETRICAL PROBLEMS. +. 008? —baB4+dB —ea)*= 4(aB*—baB + ca’) (aB'+ dB +f); or (b° — 4ac)a’ 3 — 2(2ae — bd)af’ —2(2ed — be)a’ B +2(2bf — dejaB + (@- 4af) 3’ + (2 - 4cf)a° =0; and when the axes of w and y are tangents d’ — 4af=0, e’ —4cf=0; . aB —-2hB -—2ka + 2g=0, .....--20-- (2) which is the condition required. | When b? —4ac = 0, we have H3 + Ka-G=o0. ... (3) But HB + Ka — G=0 is the equation to the chord of contact BC (Art. 61); hence if from any point D in the chord of contact of a parabola, (fig. 167) DE, DF be drawn parallel to the tangents to meet them in FE, F; EF will be a tangent to the parabola. 31. If a conic section touches four given straight lines, ‘ : v viz. the axes of #, y, and the two straight lines - + oe 1.0; a vo ¥ —+=-1=0; to find the locus of the centre. a B From equation (2) we have afs —2hB -2ka+ 2g = 0; a! 3' — 2hB' -2ka’ + 2g =0; . (a B’ —aB) — 2(8' —B)h- 2(a -a)k=0; (4) which gives a relation between h and k, and is evidently that ’ ! of a straight line passing through the points = z 2 3s or the straight line joining the middle points of the diagonals of the quadrilateral figure formed by the intersection of the four straight lines. (a) Hence if a conic section touches five lines a, 8, y; d, €> let any four of them, as a, 2, ry, 6 form a quadrilateral figure q,; the centre will be in the line 0, joining the middle points of the diagonals of q ; similarly, let ¢ and any three of the others, as a, 3, y form a quadrilateral figure q.; the centre will be in APPENDIX II. 223 the line 6, joining the middle points of the diagonals of q,; therefore the intersection of 0,, 0, will be the centre of the conic section. 32. From Art. 11. If #, y be the co-ordinates of the focus of a parabola, Ke+ Hy -G=-2Kycosw; and Ha= Ky; and Ka+HB—-—G=0, (Equation 3, Art. 30); K(ew-a)+ Aly —- B) = —2Ky cose, aH or —(t-a) +y-B = — 28 cosa; Y .@+2coswwy +y —av—Py=0, ...... (5) which is the equation to a circle. Hence the locus of the foci of all parabolas which touch three given straight lines is a circle passing through their points of intersection. 33. (a) From equation (2) it appears that if any tangent EF (fig. 167) be drawn between two given tangents 4B, AC to a curve of the second order which has a centre, and the parallelogram AEDF be completed, the locus of D will be a hyperbola whose asymptotes are parallel to 4B, AC. From equation (3) it appears that if the curve be a para- bola, the locus of D will be a straight line, viz. the line joining the points of contact of the two tangents AB, AC. (3) Since =, : are the co-ordinates of Q the. middle ~ point of EF’, when the curve has a centre, the locus of Q is a hyperbola; but when the curve is a parabola, the locus of Q is a straight line. 34. To find the equation to a conic section which shall touch five given straight lines. | Let two of the straight lines be taken for the co-ordinate axes, and let GEOMETRICAL PROBLEMS. 224 or u =0, u' = 0, vu’ =0 be the equations to the three remain- ing straight lines; then the equation to the conic section is /u+ /au + / Bu" = 0; where a and B are constants to be determined. afte “og ord. Be let Sra ee eee 9? 57 ce ] ] 1 ui | Sa BaP Ss eg -u+Au+Bu’ =da, and A = eps 3 1 1 1 1 b” 7 Mon 1 1 ] ] ss oe re ° e « / a a a Similarly, if 4 = na - eae 7 7 tie Ge Cae att ad. aa u+Au + Bu" =n y; and when the axes of w and y are tangents, (Art. 29) a B a B —+.+4+1=0, ~+=+1=0 ee: egy idii Be i a ] r ar B — as (- =| 1 1 ( J =) a ae Wand \ ak tee a 3B 1 ——__— + ——____ + pee ( LE | (; 7) b’ b Bie lerab oe and ifa=d (“-<) (7-55), 1 1 i a] 1 1 i aol (ce med A Car eee tH pes By. hi ( arte & | t (9 = bearb. nes-4 He hey 1 ee ae Es Sas) * bee ae a Pp | ae ee 5 (7 =) ae 6 ) i (= a - -p)4@-m (pa) -° APPENDIX II. papas, (2 ue *) A i Oe '. q=——____———__,, B= eS We ) ( ia (= cae =) 5 Be a’ =) NL ay a se a ae hence (= ~~ =) (i — 7 ( ae oc 1) a a RIA Teeter gy So ee a — 1 rH & 7) ; 7) ies b /(} 5 1 ) oY : ‘ aS eae ee sw i LAB eS — # a’ all 5) (et is the equation required. 35. Find the equation to the conic section which passes through a given point, and touches four given straight lines. Let two of the given straight lines be taken for the co- i —1=0, be the equations to the two remaining straight lines; and a”, b” the co-ordinates of the given point ; then the equation to the conic section is ordinate axes; and let — + eens th) aie ard a b’ 1 1 1 ] SAP TEN 7 Cg Pit re ae ee 1 1 | or A=(—-=), pet he: a a b’ h te Bp. be i ence os Boe O> (Art. 20): 926 GEOMETRICAL PROBLEMS. and \/aa” + 4/ 3b" + Jo Earn; a from which equations a and ( are determined, and these values when substituted in equation (1) will give the equation required. There will in general be two conic sections corresponding ” " a to the two values of a and 3; but when — + are 0, or a the point a”, b” is in one of the given straight lines, a, 6 have only one value respectively. 36. Find the equation to the conic section which passes through two points, and touches three given straight lines. Let n/a ae ay aie (1) be the equation to the conic section; and a’, 0’; a’, b” the co-ordinates of the given points; then Jo as aad + VB #03 ek ae ok dae n/ fee poi+Vaa" +9/ Bb" = 05 from which a, (3 may be determined; and equation (1) is the equation required. 36.* Find the equation to a conic section which passes through three points and touches two given straight lines. Let the two given straight lines be taken for the co- wh Tee. : ordinate axes, then if — + Ae 1=0, be the equation to the # ) chord of contact of the two tangents, the equation to the conic 2 een es section 1S e ao : — 1) + Aay = 0, where a, 3, A are to be we determined. Ifa, 5b; a,b’; a’, b’; be the given points, APPENDIX II. JO R227 Vad — VAL tea bo Vab sic BATES ———— oan CORN oh / aby o cle faa” / bb” wd dee s/ ab Similarly, Sain «Amara: g imilarly : Shee aie, (2) k : 1 1 A from which equations —, and B may be determined, and the a equation to the conic section is ab (= +4 1) (= +2 1) = 0 a B a’ B een From equation (1) it appears that when a conic section passes through two given points, and touches two given lines 4B, AC; then if BC be a chord of contact, and the parallelogram ABPC, (fig. 167) be completed, the locus of P is a hyperbola; unless bb’ ab =a'b’; in which case B= peti. a and the locus is a aa straight line passing through the origin. The locus of the middle point of the chord of contact BC is a hyperbola. 87. Find the equation to a conic section which shall pass through four given points, and touch a given straight line. Let the given straight line be taken for the axis of 7; and let the equations to the four lines joining the given points be v os poles oe <4 7 -12u =0, &e; ich re then the equation to the conic section is uu’ +ruUuU" =0= Ay’ + Bay+ Ca? + Dy+Ea+rFk; and since the axis of # is a tangent, H” =4CF, r 1 1 1 1 but Se = + J ans ~E={- +5) +r(Gtcn).FPar4as a a P2 228 GEOMETRICAL PROBLEMS. lyeu Lape sii) ® 1 r {Cee a) +s Ge a)f sleet ae] oN G a a a aa a a To Eaten Leh ehh) Cape d 1 I 1 mle =~ =| At 2 5 oer Saat aes rts, | ee rte 7 A+ (-—<, = 08 a. @ a a a a aa aa a a Bed Wee 1 1 I 1 2 2 1S “(5 7) r $2( tt izes it As A (--= =0; a @4 aa aa aa aa aa aad a@ a hd Be ets, eek hao Ly el\efeteee ey . pag N'+2 Gara Gore, toa tape (aa A+ (-—— =05 a a |\a’ a a a a a/\a@ a a a 66s Ne Nhe uke bey a Bale 1 ea \e/ os © (5-2) +(G-a)C-2) t8e- e-em 1 1 1 1 {1 1 1 1 ai , ” ”) , ” “Wt a a a a a a a a which determines two values of i. , 37. (a) To find the points of contact, when a conic section passes through four given points, and touches a given line. Let the four lines joining the given points meet the tangent jin: P,,-P.3\ Ps; P, respectively (fig. 188); and let 4 be the required point of contact; then (Art. 54.) if A be sup- posed for convenience to the left of P,, ~ APPENDIX II. 229 Pea dP, AP.” ] ] ] 1 On ee ee es apes Paes) AP: PB yr hence —————— = OPOUPEE AP, APS or if AP, = 2, P,P. = Qe; hag a=: Ce de dag Qa a3; — ay w(u+a,) (@+ a) (@ 4%)” — AnA, i V/ Ao Qy (@_ — &3) (@, — G3) Az + A, — G3 (a. + @, — G3) j from which the two positions of A may easily be determined by a geometrical construction ; and the problem is reduced to that of the determination of two conic sections each of which passes through five given points. If w be negative, A will be to the right of P,. 37. (@) Having given two tangents to a parabola and the direction of the axis, find the focus and vertex. Let PT, QT (fig. 210) be the two tangents, PM the direction of the axis; draw QW parallel to PM, and make ZSPT = 2TPM, and 2TQS = 2TQN; then PS, QS will intersect in S' the focus of the parabola. Draw SY, SZ perpendicular to TP, TQ respectively ; then Y and 7’ are points in the tangent at the vertex; join YZ and draw SV perpendicular to YZ; V will be the vertex required. 37. (vy) To describe the two parabolas which pass through four given points. Let A, B, C, D (fig. 211) be the four given points, pro- duce AD, BC to meet in G} and BA, CD to meet in F'; join BD, AC; then a pair of tangents drawn from A and JD in- b id tersect in FE; (Art. 62); and Art.25. Cor. l. y= + ~* v 230 GEOMETRICAL PROBLEMS. is the equation to the two lines parallel to the diameters of the two parabolas; hence if FK = F'K’ be a mean proportional between F'4 and FB, and FL a mean proportional between FD and FC, the axes of the two parabolas will be parallel to KE, and K'L; bisect AD in G; draw GH parallel to K’L meeting EF in H; then HD, HA are tangents, and if GH be bisected in P, P will be a point in the parabola; and since the tangents HD, HA at the points D, 4A, and K’L the direc- tion of the axis of the parabola is known, the focus and vertex may be determined 37 ((3). Similarly the parabola may be described whose axis is parallel to LK. 87. (6) To describe a parabola touching four given straight lines. Let AB, BC, CD, DA (fig. 212) be the four given straight lines; produce BA, CD to meet in H; and AD, BC to meet in £; about EAD, FCD describe two circles; these will intersect each other in S' the focus of the parabola (Art. 32) ; from S draw two perpendiculars SY, SZ on EA, ED re- spectively ; then YZ will be a tangent at the vertex; and if S'V be drawn perpendicular to YZ, V will be the vertex required. | 38. Let y+mart+e,=0, y+ mer+c,=0, Y+Ms0 +C3=0, Y+mv+eo=0 be the equations to four sides of a quadrilateral figure in- scribed in a conic section whose equation is ay’ +bey+cxv’?+dy+exrt+f=0, then the equation to the conic section may be put under the form (y+m,v+e,)(y + m3 + C3) + d (y-+m0 +C,)(y-+m, 0+ Cy) =0; (A) or (1 +A) y? + {(m + ms) +A (m, + m,)? wy + (mm; + AmyM,) B + ie, +0; +A(Q+# Oe) LY + {C3 + m3, + A (Mey + M4Co)t @ + Ces + NCoey = 03 APPENDIX LIL. O31 and in order that this may coincide with the given equation to the curve, we must have the five following relations : a {(m, + ms) + A(m, + m,)i =F +A); (1) a(m,mz +AM,M,) =c (1 +A); (2) a fe, +63; +X (G4, + €,)} =a +A); (3) a {c,mz + C3m, + A (Cem, + Cym,)t = € (1 +A); (4) @ (C,C3 + AC.¢,) = f (1 +A). (5) Equation (4) equally represents the equation to the conic section when YM +C2= 0, YtMeL+ Cc, = 0; are the equations to the diagonals of the quadrilateral. 39. If any three of the quantities m,, m,, m3, m, be given, the fourth may be determined by equations (1), (2); hence if three sides of a quadrilateral figure inscribed in a conic section be parallel to three fixed lines, the fourth side will also be parallel to a fixed line. 40. In a conic section inscribe a quadrilateral figure, such that one of its sides may pass through a fixed point, and its three remaining sides be parallel to three fixed lines. Let P (fig. 168) be the fixed point; draw any line 4B parallel to the first side, and BC, CD respectively parallel to the second and third sides; then the remaining side DA of the quadrilateral figure will always be parallel to a fixed line. Hence, through P draw PD'J/’ parallel to DA, and 4’B’, BC’ parallel to 4B, BC; the remaining CD’ will be parallel to CD, and A’B'C’D' will be the figure required. 41. In a conic section inscribe a triangle similar and similarly situated to a given triangle. Describe as before a quadrilateral figure whose sides 4B, BC, CD (fig. 169) shall be parallel to the three sides of the triangle; join DA; then DA is parallel to a fixed line: draw RN GEOMETRICAL PROBLEMS. any other line A’D’ parallel to AD, and bisect AD, A’D’ by the straight line aa’; draw ab, be parallel to 4B, BC; then ac will be parallel to CD; for abe may be considered as a quadrilateral figure whose evanescent side at @ coincides with the tangent at a; and since the tangent at a, ab, be are parallel to DA, AB, BC, the fourth side ac will be parallel to CD. 42. If ABCDEF (fig. 170) be a hexagon inscribed in a conic section, draw the diagonal 4D, and let DE, EF be parallel to AB, BC respectively; then ABCD, DEFA are two quadrilateral figures, having three of their sides DA, AB, BC, and AD, DE, EF parallel to the same three lines re- spectively ; hence the remaining sides AF’, CD will be parallel to a fixed line; and therefore to one another. 43. If three of the quantities c,, c., ¢,, c, be given, the fourth may be determined by equations (3), (5); hence if three sides of a quadrilateral figure inscribed in a conic section pass through three fixed points in the axis of y, the fourth side will pass through a fixed point in the axis of y; and the axis of y may be taken in any direction; hence if three sides of a quadrilateral inscribed in a conic section pass through three fixed points in a straight line, the fourth side will pass through a fixed point in the same straight line. b : 44, If m,+m;=—; equation (1) gives m, +m, =—; a a and when 6b = 0, the axis of & is parallel to one of the axes of the conic section; hence if m,= —m,, m,=— m,; or if two opposite sides of a quadrilateral inscribed in a conic section be equally inclined to the axis, the two remaining sides will be equally inclined to the axis. In the same manner it appears that the diagonals will be equally inclined to the axis. 45. Let two conjugate diameters CP, CD (fig. 171) be taken for the axes of w and y respectively, then b = 0, and if APPENDIX II. 933 pq be any chord parallel to CD, and y= mv, y = m;a be the equations to Cp, Cq respectively ; since pq is bisected by CP, m,=—m,; hence m= —m,; and if Cp’, Cq be drawn through C parallel to the second and fourth sides, p’q’ will be bisected by CP, and therefore parallel to CD. Hence if three sides taken in order of a quadrilateral in- scribed in a conic section be parallel to Cp, Cp’, Cq drawn from the centre C, join pq, and draw p’q’ parallel to pq; then the fourth side will be parallel to Cq’. This will be equally true if the diagonals of the quadri- lateral be taken instead of the second and fourth sides. 46. When the first and third sides coincide with AB, which is parallel to CP, then the points p, q coincide in P, and CP bisects the evanescent chord joining them; and the remaining sides of the quadrilateral will be in the directions of the tangents at A and B. Draw Cp’ parallel to the tangent at 4, and p’q’ an ordinate to CP; then Cq will be parallel to the remaining side of the quadrilateral or to the tangent at B. Hence it appears that if Cp’, Cq’ be conjugate to CA, CB respectively ; AB, p’q' will be parallel to a pair of conjugate diameters. 47. If the first and third sides coincide with 4B, (fig. 172) the second and fourth sides will be in the directions of the b tangents at A and B, and if m, =m, = porte +m, will be a constant for all chords parallel to 4B; but if PP’ be drawn through C parallel to 4B, since the tangents at P and P’ are b parallel, m, = m,, and m,+m, cannot = unless the tangent at P is perpendicular to the axis of w; and C'T' is parallel to the tangent at P; hence m, +m, = cot AT'C — cot BTC is constant for all chords parallel to 4B. 48. If the origin be changed to any point in the conic section, a, b, c remain unchanged, and f= 0; hence if m,, m,, 234 GEOMETRICAL PROBLEMS. m, be given; m, and A will be known from equations (1), (2) ; C1 C3 and = — from the equation (5); but if p,, po, ps3, ps be C204 the perpendiculars from the origin upon the four sides i 4 3 a C2 ae EY Ata ties) erences Ps VG + m,*) (1 + me) | Pops / (1 + m,°)(1 + m;*) and is therefore constant. Hence if perpendiculars be drawn from any point of a conic section upon the four sides of a quadrilateral inscribed in the conic section, p,p; will bear an invariable ratio to pop. (Senate-House Problems, Thursday, Jan. 7, 1847, 1... 4. Quest. 14.) The same will also be manifestly true of any quadrilateral, three of whose sides are parallel to three fixed lines. d d 49. If c¢,+ce,=-—; then Cait Cameras and when d = 0, the a origin is the middle point of the chord which is the axis of y; and if c; = — C,, C, = — Co. Hence, (1) if A (fig. 173) be the middle point of a chord of a conic section, and PQ, RS meet the chord in two points B, b equidistant from 4; the two remaining sides RQ, PS will meet the chord in two points C, ¢ equidistant from A. In the same manner it may be shewn that PR, QS' meet the chord in two points equidistant from J. (2) If PQ, SR be any two chords passing through 4 the middle point of BD; (fig. 174) RQ, SP will meet the chord BD in two points C’, ¢ equidistant from J. Similarly QS, PR will meet BD in two points C’, c’ equi- distant from 4. APPENDIX II. 235 (3) If PQ, SR pass through A and become coincident, SP, QR become the tangents at P and Q; hence if A be the middle point of a chord, the tangents at the extremities of any chord PQ passing through A will meet the first chord in two points equidistant from A. (4) If the points B, D (fig. 175) be made to coincide, the line CBAc will become a tangent at 4; and if PQ, Sk meet the tangent in two points E, e equidistant from 4; PS, RQ will meet the tangent in two points C, ¢ equidistant from A; and also PR, QS will meet the tangent in two points C’, c’ equidistant from A. (5) If the points Q and R (fig. 176) coincide, then QR becomes a tangent at @; and if PQ, QS meet the chord or tangent CAc in two points E, e equidistant from 4, SP and the tangent at @ will meet the chord or tangent in two points C, c equidistant from 4. SOniAGL yore awl, then epee and if A be the Q origin, and ABC (fig. 177) the axis of y, AB. AC = f, hence a if the first and third sides PQ, RS of a quadrilateral figure inscribed in a conic section cut a straight line ABC in two points E, F such that AE. AF = AB. AC; then the second and fourth sides SP, QR will cut ABC in two pen G, H such that AG.AH = AB. AC. Similarly if the diagonals SQ, PR meet the line ABC in G, H'; AG’ .AH’ = AB. AC. : (2) Ifthe points E, F coincide in O (fig. 178) so that AB.AC = AO’, and PQ, SR be drawn through O in any direction, then PS, QR will meet ABC in two points G, H such that 4G. AH = AO’. (3) If RS be made to coincide with PQ (fig. 179), then PS, QR become tangents at P and Q, and will intersect ABC in two points G, H such that AG. AH = AO?’ is constant for all chords drawn through O. 236 GEOMETRICAL PROBLEMS. (4) Ifc,=0, then c,= ; the first side QP (fig. 180) passes through the origin 4, and the third side RS is parallel to ABC; hence if PS, QR meet ABC in G and A, AG.AH= AB. AC for all positions of PQ, QR. (5) If P and Q coincide, then AP (fig. 181) becomes a tangent meeting a chord ABC in A: PR is any line; RS parallel to ABC; then AG. AZ is constant for all positions of PR. 51. If d =0, the origin is the middle point of a chord; and if ¢,c. = ae. then ¢, + ¢; +A(c, + c) =0; €,C3 + ACC, = —C,C,(1 +A) or C,(C2 +3) + ACL (eC, +¢,) = 0; C; tC ey C) (c, + €3) — C2 (C, + €,) (< A =) = 0; ‘9 mt « hence €,(€y + 3) (Co + C4) = C2(e, +3) (€; + C), Or €,¢,(¢, — €)) = €3¢,(€2 —C))3 .*. Cg, = CCn = — ue a > hence if A be the middle of the chord BD (fig. 182), and PQ, QR meet AB in C, C’ such that AC. AC’ = AB’; then RS, PS will meet AB in ec, ce’ such that 4c. Ac’ = AB’. (2) If c,=0, or the first chord PQ passes through 4, (fig. 183) c,= , and QR is parallel to 4B; hence if RS, SP meet AB in c,c; Ac.Ac = AB’ for all positions of S§P, PQ. (3) If PQ be a diameter passing through A (fig. 184) ; the points Q, R coincide; and if PS, QS be lines drawn to any point S' of the curve, from the extremities of a diameter PQ, meeting an ordinate to PQ in c’,c; then de. Ac’ = AB? and is invariable for every position of §, APPENDIX IL. 237 52. From equations (1) and (2). m m b Ms. +m ale If peg £3\-thenss— ee re m,mMs Cc MoM, ot or the first ay third lines are parallel to the axis of x, then ; butif m, = 0, m;=0, —+ — =-; hence if PQ, RS' be two parallel chords; (fig. 185) the sum aloe the cotangents of the angles which PS, QR, or PR, QS make with PQ in the same direction is constant. (2) If PQ, RS coincide, PS, QR become tangents, or the sum of the cotangents of the angles which a pair of tangents at the extremities of any chord parallel to a fixed line makes with that line in the same direction is constant. his | Rateale Spas etic eA a C, C3 C2C4 4% ; and since 1 | usted , then —+-— is constant; and will be the same whether we oe, _ take the second and fourth sides or the diagonals of the quad- _ rilateral. (2) If c,=0, then c,= 0, and the first and third chords PQ, HS (fig. 186) pass through the origin 4; ies PSs t ABC t D h- th ted QR mee in two points E, such tha AD + — i constant. (3) If PQ, RS coincide, (fig. 187) PS, QR become tangents, and the tangents at the extremities of any chord PQ passing through a fed point 4 will eo a given line ABC in : 1 two points 7’, 7”, such that —— + —— is invariable. AT i? 54. Ifd=0, and f=0; the axis of y Se a tangent 1 1 Te to the curve, and from equations (3) and (5) - = aa —-=—+4+— C2 C2 om 238 GEOMETRICAL . PROBLEMS. hence if the four sides of a quadrilateral inscribed in a conic section meet a tangent at A (fig. 188) in the points P,, P,, P,, P, respectively, and the direction AP, be considered positive, 1 is 1 ] Hit APP a AP AP, gent in Q,, Q,; ; and if the diagonals meet the tan- —— ——__—_ ——S|= Es ———— = — oe 55. If c and f=0, the equation to the conic section becomes ay’? + bay + dy + ex = 0; which is that of a hyper- bola having one of its asymptotes parallel to the axis of #; and ee ei or if A be the origin in the curve; and the m, mM; Mm, M, four sides of a quadrilateral figure meet a line passing through A and parallel to one of the asymptotes in P,, P,, P3, P,; then AP,. AP; = AP,.AP,; and if the diagonals of the quadrilateral meet the same line in Q., Q,; AQ AQ = APeeaPe 56. Ifa=0, then X= — 1, and the curve is a hyperbola having the axis of y parallel to one of the asymptotes; and if a, and c=0, m,m, —m,m, = 0; in this case the axes are parallel to the two asymptotes. Hence if a quadrilateral figure be inscribed in a rect- angular hyperbola, the product of the tangents of the angles which the first and third sides make with either of its asymp- totes, is equal to the product of the tangents of the angles made by the remaining sides with the same asymptote. 57. If a and d=0, the axis of y becomes one of the asymptotes, and c,+¢;=¢c,+c,; hence if four sides of a quadrilateral figure inscribed in a hyperbola meet one of its asymptotes in the points P,;, P., P;, P,; AP, + AP; = AP, + AP,; as PsP 2 tates also P,P,= P,P;; or the two adjacent sides intercept the same portion of the asymptotes as the remaining two. APPENDIX II. 239 If the diagonals of the quadrilateral, meet the asymptote in Q,, Q,; then AP, + AP; = AQ, + AQ): or P,Q, = P3Qi5 similarly P,Q, = P,Q: 58. If aand f=0, ¢,¢;=¢,c, In this case the axis of y is in the curve and parallel to one of the asymptotes ; and if the four sides and the two diagonals meet the axis of y in P,, Pp, Ps, Ps, and Q,, Q, respectively ; then £0 Fs AP, = ARS APs aQ, AQ: and when P,, P; are fixed points in the axis of y, AP,.AP,, and AQ,. AQ, are invariable. 59. If b and c =0, the curve becomes a parabola with the axis of # parallel to the axis of the parabola; and from i if 1 equations (1), (2) we have — +— =—-+-——. Hence the m, mM, M, M, sums of the cotangents of the angles which each pair of oppo- site sides, and the diagonals of a quadrilateral figure inscribed in a parabola make with its axis in the same direction are re- spectively equal. 60. Since the form of equations (1), (2) which connect M15 Mz, Mz, m, is similar to the form of equations (3), (5) which connect ¢,, Co, €3, €,; and the quantities are similarly involved in equation (4); if any relation be found between My, Mz, Mz, M,; a corresponding relation may be determined between ¢,, €2, C3, C43 and vice versa. The following example will be sufficient to illustrate the preceding remark. If two tangents be drawn to a conic section the tangents of whose inclinations to the axis m,, m, are such that m,m3, or 1 1 m, + ms, or — +-— is constant, the locus of the intersection m Ms of the tangents will in each case be a conic section. From this we may infer that if a pair of tangents meet any diameter at distances ¢,, c; from the centre such that c¢,e¢;, 240 GEOMETRICAL PROBLEMS. 1 Lae : : Or C, + C3, or — + — Is constant, the locus of the intersection of C; C3 the tangents will in each case be a conic section. 61. If AMN, APQ (fig. 189) be any two chords drawn through a fixed point 4 to a conic section, the straight lines joining the points of intersection of MP, NQ, and of MQ, INP will be the chord joining the points of contact of two tangents drawn from A. Let A be taken for the origin; and let the equations to AMN, MP, APQ, QN respectively be y + maw =0; YtML+C.=05; Yt+mMvr=0;3; Y+mMC0+o,=0;3 hence the equation to the conic section is (y + 7,2) (y + m3”) +A(Y + Mx + C.)(Y + Me + C,) =0; (1) and in order that this may be identical with the given equation to the conic section ay’ + bay + ca’? + dy +ex+f=03; we must have ad(c, + c,) = d(14+A); Ad (C.M, + Cym.) =e(1 +A); arene, = f(1 +X); 1 1 d M, mM e Sob oes Cayce, ip C2 C4 © f j but for the intersection of y + m7 -+¢,=0; y + m,x + ¢,=0, we have 2) +(e conan Spe py a a a ee or —Y+-7+2=0; Loree C; C2 Cy 4 es tf hence the second and fourth sides intersect in the straight line dy +ex+42f=0, which is independent of m,, m,; and since equation (1) will equally represent the equation to the conic section when y + m,# + ¢, = 0, y + m,v + c, = 0 are the equa- tions to NP, MQ; the locus of § the intersection of MQ, NP is the straight line dy+ev+2f=0; and the locus R of the intersection of MP, NQ is the same straight line; therefore APPENDIX It. 241 dy +ex+2f=0 is the equation to the straight line joining Rand S. Since J2,S' is independent of the position of AMN, and APQ, let P and Q move up to 7, and M and N to T’, so that AT’, AT” are tangents drawn from A; then MP, NQ become coincident with the chord of contact 7'7"”; and since they always meet in the same straight line, RS’ must be the chord of contact of two tangents drawn from A. The equation to the chord of contact of two tangents drawn from A is dy +ex+2f=0. 62. Since RST” will be the same straight line in whatever direction the lines AMN, APQ be drawn; let AP’Q’ be drawn very near APQ; then PP’, QQ’ are ultimately tangents at P and Q; but PP’, QQ’ will intersect in the chord of contact RS; hence a pair of tangents at the extremities of PQ will meet in the chord of contact of two tangents drawn from A. 63. If a pair of tangents be drawn at the extremities of two chords PQ, MN passing through A, the straight line joining the points of intersection of each pair will be the chord _of contact of two tangents drawn from A. 64. When tangents are drawn from A, the chord of con- tact is RS passing through §'; similarly when tangents are drawn from £# the chord of contact is 4.8’ passing through S$; but when pairs of tangents are drawn from any point in AR, the chords of contact will all pass through the same point ; hence the chords of contact will all pass through S. Conversely if pairs of tangents be drawn at the extremities of any chord passing through S', they will intersect in the line AR, 65. The line joining the points of intersection of pairs of tangents at the extremities of the chords MQ, NP is AR; hence MP, NQ, and also MN, PQ meet in the line joining the points of intersection of pairs of tangents at the extremities of the chords MQ, NP. Q 942 GEOMETRICAL PROBLEMS. 66. To draw a pair of tangents to a conic section from a given point dA without it; and also to find the chord of contact. Through A draw any two chords AMN, APQ; join MP, NQ intersecting in &; join MQ, NP intersecting in S; then RS is the chord of contact of a pair of tangents drawn from A; and if RS meet the conic section in the points 7', 7”; AT, AT’ will be the tangents required. 67. To draw a tangent at a point A in a conic section. Through A draw any line CABD to cut the curve (fig. 190); find the chords of contact PQ, RS of pairs of tangents drawn from two points C, D without the conic section; let QP, SR intersect in J'; join 7'A, this will be the tangent required. Since the chords of contact of pairs of tangents drawn from any point in CD pass through the same point, the point T’ is the intersection of all chords of contact; and when tan- gents are drawn from points indefinitely near A and B, the chords of contact approach to the tangents at A and B; and since the chords of contact always pass through 7’, the tangents at A and B pass through 7'; hence 7'4 is a tangent at the point 4. See also Art. 89. It may be observed that the tangents have been drawn by simply joining points; so that the compasses are not required. 67. (a) Let 4, B, C, D, E (fig. 213) be five points in a conic section; produce BA, CD to meet in H; AD, BC to meet in A; and let AC, BD meet in LZ; join HK, AL, KL; then the tangents at 4 and ( meet in HK. (Art. 64.) Similarly if HC, 4D meet in K’, and CD, HA meet in H’; the tangents at 4 and C meet in A’R’. Hence if HK, H’K’ intersect in 7, T'A, T'C will be tan- gents at the points 4 and C. Again, the tangents at A and D meet in HL; and the tangents at A and B meet in KL; hence if 7'A meet in AL, APPENDIX II. 243 KLin T’, 7”; T’A, T’'D will be tangents at A and D; and T’ A, T’B tangents at A and B: and four tangents T'A, T’B, TC, T’'D have been drawn at the points 4, B, C, D of the conic section which passes through the five points 4, SC; Dy Fe: 68. If RST" be the chord of contact of a pair of tan- gents drawn from a point 4 without a conic section (fig. 189) ; AMN any chord passing through 4, S' any point in RT’; join NSP; then AP, MS will intersect in a point Q in the conic section, and the locus of Q will be the same conic section. 69. If M (fig. 191) be the vertex of a parabola whose axis is MB; and 4M = MB; draw BST'R perpendicular to MB; then BR is the chord of contact of two tangents drawn from A; and since 4M does not meet the curve again, SP must be drawn parallel to AM; and the locus of the inter- section of AP, MS will be the parabola MP. 70. If PAC, PDB (fig. 192) be drawn to the extremities of the axis AB of a conic section from any point P in a straight _ line PE perpendicular to the axis, DC will meet AB in a fixed _ point F. Let DA, BC meet in G; then GF is the chord of contact of a pair of tangents drawn from P; and the chords of contact of pairs of tangents drawn from any point in PE will pass through a fixed point in the axis 4B; hence the point F is fixed. (2) Ifa pair of tangents be drawn from G, PF will be the chord of contact, and since it passes through F, G is a point in PE; hence DA, BC intersect in PE. (3) If BC be produced to meet PE in G, and DG be joined it will pass through 4. 71. If the curve AD is a parabola, we must suppose B removed to an infinite distance; in which case PD, GC become parallel to the axis; hence if from any point P in a line PE Q2 244 GEOMETRICAL PROBLEMS. perpendicular to the axis of a parabola, PAC be drawn through the vertex, and PD parallel to the axis; DC will meet the axis in a fixed point #. In this case dF = AE. (2) If CG be drawn parallel to the axis, DG will pass through A. | 72. If A be removed to an infinite distance PQ, MN (fig. 189) become two parallel chords; and MP, NQ, as well as MQ, NP will always intersect in a straight line, which will be a diameter to the chords. 73. If AMN, APQ (fig. 189) be any two lines drawn through A to meet a surface of the second order in M, N, and-P, Q respectively ; the intersection R of MP, NQ will be in the plane of contact of the enveloping cone whose vertex is 4; and the locus of # will be the plane of contact. For the section of the surface made by the plane passing through APQ, AMN will be a conic section, and the locus of R will be the line of contact of two tangents drawn to the conic section from 4; and this will be a straight line on the plane of contact of the enveloping cone whose vertex is 4; and the same will be true when the lines AMN, APQ are drawn through 4 in any other direction; hence the locus of R is the plane of contact. In like manner it may be proved that the locus of § the point of intersection of MQ, VP is the plane of contact of the enveloping cone whose vertex is 4. 74. If a hexagon be inscribed in a conic section, the points of intersection of the opposite sides will all lie in the same straight line. (Pascal’s hexagram). Let the sides of the hexagon ABCDEF (fig. 194) taken in order be represented by a, 3, y, 0, ¢ ¢; and let their equations be respectively U,=0, Ug= 90, U=0, Us= 0, UL= 0, Ue= 0; APPENDIX II. 245 also let w, = 0 be the equation to the diagonal BE; and U=0 the equation to the conic section; hence UU, + A{U,U, =m,U; Uj Uy, + AgUgus = M,U; vs (mu, — mu.) U =X, Ua, 6, — AgUgUs Uy. (1) If uw, and u;=0, the values of # and y will be the co- ordinates of the point of intersection of the sides a and 0; and from equation (1), (m,u, —_m,u,)U =0; but U cannot =0, since the intersection of q@ and 9 is not a point in the conic section; .*. 1%, —m,U,=0; or a and 0 intersect in the straight line whose equation is MU, — MzUz = 0. Similarly, if in equation (1) we put w, and w,=0, the lines 8 and ¢ intersect in the straight line m,w, — mu, = 0. Now m,u, — m,u;= 0 is a straight line passing through yy, ¢; hence the intersections of a, 3; B,e; ¥, ¢ lie in the same straight line whose equation is m,w, — m,u, = 0. (a) If the six chords of the conic section be taken in the order AB, BE, ED, DC, CF, FA, it will appear that the ‘intersections of AB, CD; BE, CF; ED, FA are in the same straight line. Hence BE, CF intersect in cf. Similarly 4D, BE intersect in eb; and AD, CF in ad. In like manner by varying the order of the points 4, B, C, D, E, F, we may find several series consisting of three different points which lie in the same straight line. 75. The three diagonals of a hexagon circumscribing a conic section meet in the same point. (Brianchon’s theorem). Let ABCDEF (fig. 170) be the hexagon; a, b, ¢, d, e, f the points of contact; then (Art. 63) AD is the chord of contact of tangents drawn from a’ the intersection of de, af; similarly BE, CF are the chords of contact of tangents drawn from a’, a” the intersections of de, ab; and ef, cb respectively ; 246 GEOMETRICAL PROBLEMS. and a’, a’, a” lie in the same straight line (Art. 74); there- fore 4D, BE, CF meet in the same point. 76. Let ABCD (fig. 195) be a quadrilateral figure in- scribed in a conic section, this may be considered as a hexagon whose sides at the points A, C vanish, and are in the directions of the tangents at 4 and (’; then since the opposite sides of the hexagon intersect in a straight line, if AB, DC be pro- duced to meet in E, and AD, BC to meet in F, the two remaining sides of the hexagon, viz. the tangents at A and C will intersect in the line EF. Similarly the tangents at B, D intersect in the line EF. If the lines inscribed in the conic section be taken in the order ABDC4A, and the tangents at A and D be considered as two evanescent chords passing through A and D; then the intersections of AB, DC; BD, AC; and the tangents at d and D will meet in a point; or the tangents at 4 and D will meet in HG. Similarly, the tangents at B, C meet in HG. These properties have been already proved (Art. 63). 77. Let ABCD (fig. 195) be a quadrilateral figure cir- cumscribing a conic section, and touching it in the points a, B, c,d; (1) The quadrilateral figure may be considered as a hexagon whose angular points are A, a, B, C, ec, D; in which case the three diagonals AC, BD, ac meet in a point. (2) Let A, B, b, C, D, d be considered the angular points; then the three diagonals AC, BD, bd meet in a point. Hence ae, bd pass through the intersection of AC, BD. (3) Let A, a, B, 6, C, D be considered the angular points; then 4b, Ca, BD meet in a point. Similarly, 4c, BD, Cd meet in the same point. And in the same manner it may be proved that Be, Db; and also Bd, Da intersect in AC. APPENDIX II. 247 78. If a triangle ABC (fig. 196) be described about a conic section, and touch it in the points a, b, ¢, since ABC may be considered as a hexagon whose angular points are A, c, B, a, C, 6; the diagonals Aa, Bb, Ce intersect in a point. Hence if a conic section inscribed in a triangle touches two sides BC, AC in the points a, b; join Aa, Bb to meet in D; and produce CD to meet AB ine; ¢ will be the point of contact of the side AB. 79. Let a, b, c,d (fig. 195) be the points of contact of a quadrilateral figure ABCD circumscribing a conic section ; produce 4B, DC to meet in £; AD, BC to meet in fF’; also let AC, BD meet EF in H and K respectively ; and let AC, BD intersect in G. (1) ab, de, (Art. 63) intersect in the line joining the points C, A of intersection of the pairs of tangents at the extremities of the chords be, ad; hence ab, de intersect in AC. (2) ab, de intersect in the line HF joining the points - E, F of the intersection of the pairs of tangents at the ex- ~ tremities of the chords ac, db (Art. 65); hence ab, de intersect in ff. Similarly ad, bc intersect in K. (3) ac, bd (Art. 64) will meet in the line AC joining the points of intersection A, C of the pairs of tangents at the extremities of the chords ad, be. Similarly ac, bd meet in BD; hence ae, bd will meet in G, the point of intersection of AC, BD. 80. A conic section is to be inscribed in a quadrilater al ede ABCD (fig. 195) so as to touch the side AB in a; to find the points b, c, d of contact with the three remaining Hes Join AC, BD meeting in G; let AB, DC meet in L; and AD, BC in F'; produce AC to meet EF in H and aG to meet DC inc; join Ha, He meeting BC, AD respectively in 6, d; and the points b, c, d are determined. 248 GEOMETRICAL PROBLEMS. 81. When a pentagon is described about a conic section to find the five points of contact with the sides, Let ABCDE (fig. 197) be the pentagon, and a the point of contact with the side AB which it is proposed to determine ; then da, aB, BC, CD, DE, EA may be considered as the six sides of a circumscribing hexagon; and the three diagonals Da, AC, BE meet in a point. Hence join BE, AC, meeting in G; draw DG meeting AB in a; then a is the point of contact with the side 4B. In like manner the points of contact with the remaining sides may be determined. 82. Having given five points 4, B, C, D, E of a conic section, to determine a sixth point in any given direction DF passing through one of the given points. Let F (fig. 198) be the sixth point which is to be deter- mined; then ABCDFE is a hexagon inscribed in a conic section; and the opposite sides BA, DF'; AE, CD; and BC, FE meet in three points K, K’, K” which are in the same straight line. Hence produce BA, DF to meet in K; CD, EA to meet in K’, and let CB meet K’K in K”; draw K”E meeting DF in F'; F is a point in the conic section. In like manner a point of the conic section may be determined in any other direction passing through D; and the conic section can be traced by a consecutive series of points. 83. ‘To determine geometrically the centre of a conic section which passes through five given points. Through any given point as D, draw (Art. 82) a chord DF ina direction parallel to BC; bisect DF’, BC: the line joining the middle points will pass through the centre. Si- milarly, find a chord DG parallel to HA; the straight line joining the middle points of DG, EA will pass through the centre, and the intersection of the two diameters will be the centre required. Cor. If the two diameters are parallel, the centre is at an infinite distance and the curve is a parabola. 84. To find geometrically a consecutive series of points in a conic section which shall touch five given straight lines. APPENDIX II. 249 Find (Art. 81) the five points of contact 4d, B, C, D, EB - with the five given straight lines; and determine (Art. 82) a consecutive series of points in the conic section which shall pass through the five points 4, B, C, D, HE. The centre may also be determined by Art. 83, or Art. 31. 85. Find a consecutive series of points in a conic section which shall pass through three given points, and touch a straight line in a given point, (fig. 199.) Let A be the given point in the given straight line AK’; B, C, D the three given points; B&# the direction of a chord passing through B; £ a point in the conic section, which it is proposed to determine; then ABECD may be considered a hexagon inscribed in the conic section, whose sides are the evanescent chord at 4 in the direction of the tangent AK’, AB, BE, EC, CD, DA; and since the opposite sides meet in a straight line, if 4B, DC meet in K, and BE, DA in K”, then the tangent at A and CE will meet in K’ a point in KK"; hence produce 4B, DC to meet in K, DA, BE to meet in K"", and let the tangent AK’ meet KK” in K’; join K’C, it will cut BE in the required point £. And in like manner we “may find a point in the conic section in any other direction ~ drawn from B. 86. Find a consecutive series of points in a conic section which shall touch four straight lines, and one of them in a given point. Find (Art. 80) the three points of contact B, C, D of the remaining sides; and determine a series of points in the conic section which shall pass through B, C, D; and touch the first side in the point A. (Art. 85). 87. Find a consecutive series of points which shall touch three given straight lines 4B, AC, BC; and two of them BC, CA in given points a, b, (fig. 200.) Find (Art. 78) the point of contact c with the side AB; and let ed, be the direction of a chord of the conic section 250 GEOMETRICAL PROBLEMS. which passes through ¢; produce ba, ed to meet in #; join EB and let be meet EB in £'; then if d be a point in the conic section, cdab may be considered as a hexagon inscribed in the conic section, whose sides are the evanescent chord at ¢ in the direction eB; cd, da, the evanescent chord at a in the direction Ba, ab, be; and the points of intersection of the opposite sides are in the same straight line; hence be, ad meet in EB; therefore produce be to meet HB in F; join Fa, which will intersect ed in the point required. 88. Find a series of points in a conic section which shall pass through a given point C’; and touch two given straight lines 0.4, OB in given points 4, B, (fig. 201). Let AD be the direction of a chord of the conic section through 4A; D the extremity of the chord which is to be determined ; then ADCB may be considered a hexagon in- scribed in the conic section whose sides are the tangent at 4, AD, DC, CB, the tangent at B, and BA; and the opposite sides meet in the same straight line. Let CB meet O4 in K, and AD meet OB in K’; then DC will meet AB in the line KK'; hence produce AB to meet KK’ in K”, join K”C meeting AD in D; then D will be a point in the conic section. 89. Having given five points 4, B, C, D, E, to draw a tangent at A to the conic section which passes through them, (fig. 202). Since ABCDE may be considered as a hexagon inscribed in a conic section whose sides are AB, BC, CD, DE, EA and the tangent at 4; K, K’, K” (the points of intersection of AB, DE; BC, EA, and of DC and the tangent at 4) lie in the same straight line; hence join KK’, and let DC meet KK’ in K’, K” A will be the tangent required. The construction here given is more simple than that in ALGOO 90. When a conic section passes through five given points APPENDIX II. 251 A, B, C, D, E; find the extremities of the diameter to which a chord AF drawn from J in a given direction is an ordinate. Find F (fig. 203) the extremity of the chord AF (Art. 82) ; and O the centre of the conic section (Art. 83); bisect AF’ in G; draw a tangent 47’ at the point A (Art. 89); and draw OG meeting AT in 7; then OGT is the direction of the diameter to AF’; and if OD’ = Od’ be taken a mean propor- tional between OG and OT, D’, d’ will be the extremities of the diameter. 91. Find the extremities of the diameter which is parallel to AF, or conjugate to D'd’. Through the centre O draw OK parallel to 4F'; and from E one of the given points, draw D'E, WE meeting OX in H, K; take OL = Ol a mean proportional between OH and OK ; then (Art. 51, 3) Z and 7 will be the extremities of the diameter. 92. Find the extremities of a chord GH drawn in any given direction. Through the given point 4, (fig. 204) draw the chord AF parallel to GH (Art. 82); draw the diameter D’d’ to which ’ AF is an ordinate (Art. 90), and let it meet GH in M; take E one of the given points, and join DE, dE meeting GH in K, L; take MH = MG a mean proportional between MK and ML; then (Art. 51, 3) G, H are the extremities required of the chord GH. 93. Having given two semi-conjugate diameters CP, CD of a conic section, to determine the magnitude and position of the axes. (fig. 205). Let aA be the axis-major; CP, CD the given semi- conjugate diameters; draw PT parallel to CD which is there- fore a tangent at P; let CP=a, CD=0, 2ACP=0, 24CPT = 2PCD =a; then 6’ cos (a — 0) = ; a sin @; 252 GEOMETRICAL PROBLEMS. b b’ sin (a — 0) = — a cos; . 6? sin2(a— 6) =a" sin26; b? draw PE making zEPT’= z2CPT; and make PE=~;; jon CE; then 2CPE=7-2a; and if 4PCE=9¢; L£PEC=2a-$9; Jed eA i¥: sin @ sin 26 ' PC a* sin@a-—@) sin(@a—20)’ or CA bisects angle PCE. Cr’ ae: Hence make z T”"PE = 2zCPT; and PE “ap? Join CE, and bisect 2 PCE, this will give the direction of the axis-major. The axis-major CA always falls within the acute angle a he PCD’; for tan. tan (a - 0) =—; a P=205 .. tan (a -8) = — tan (= - 6) < tan (= -0) s TT 4 TT hence a< a since 9 may be taken a5 : Draw CB, PM perpendicular to CA, and PN perpen- dicular to CB; then if the tangent at P meets CA, CB in T’, IT’; CA, CB are mean proportionals between CM, CT’ ; and CN, CT. To find the foci, take BS, BH = CA; and §, H will be the foci required. 94. Find the magnitude and position of the axes of the conic section which passes through five given points. Find a pair of conjugate diameters D’d’, Li (fig. 203) (Arts 90, 91); and thence Art. 93, the axes and the foci of the conic section may be determined. Cor. When the curve is a parabola, the vertex and focus — may be found by Art. 37. (y) APPENDIX II. DIO 95. Having given five points, draw a tangent to the conic section which passes through them, from a given point without Te Let H (fig. 206) be the given point without the conic section; A, B two of the given points; join HA, HB; and find F, G the extremities of the chords (Art. 82); join BA, GF meeting in K; and AG, BF meeting in L; then KL is the chord of contact of a pair of tangents drawn from H (Art. 61); find P, Q the extremities of this chord (Art. 92) ; and HP, HQ will be the tangents required. 96. In like manner we may find the extremities of any chord, and draw a tangent from a given point either in or without any of the conic sections described in Arts. 84—88 ; and also find the magnitude and position of the axes, since in each case we are enabled to determine five points of the conic section. 97. To inscribe geometrically in a given conic section a triangle whose three sides shall pass through three given points 383. C: If from any point P in the conic section Pa,, Pb, (fig. 207) be drawn through A, B; then the straight line a,6, will always touch a conic section. (Appendix 1. Art. 5). Let a,b, a.b., a3b3, a,b4, 4,6,, be any five positions of this line; and from the given point C draw two tangents CED, CD’E’ to the conic section which touches them (Arts. 95, 96); let these tangents meet the given conic section in the chords DE, D’E’ respectively, join DAF, EBF; DAF’, E' BF’; then DEF, D'E’F’ will be the two triangles inscribed in the conic section whose sides pass through the three given points A, B, C. 98. In a conic section which passes through five given points, it is required to inscribe a triangle whose sides shall pass through three given points 4, B, C. Let P be one of the given points; through 4A, B, draw 254 GEOMETRICAL PROBLEMS. the chords Pa,, Pb, (Art. 82); join a,b, Similarly, we can determine a, b,, a3b3, a4by, a,b; by taking the four remaining points instead of P; from C as before draw the two tangents CED, C’D'E’ to the conic section which touches the five lines Ab), Arbo, a3b3, A404, A5bs 5 find D, EF, D', E’ the extremities of the chords CED, C’D'E’ (Art. 92) ; join DA, EB meeting in F'; and D’A, E’B meeting in /”; DEF, D'E'F’ will be the two triangles required. 99. In like manner we may inscribe geometrically a triangle whose sides shall pass through three given points, in any of the conic sections enumerated in Arts. 8488, although the conic section itself is not traced. 100. To inscribe in a given conic section a polygon whose m sides taken in onne shall pass through 7m fixed points, 27 Papo bere BS If (m — 1) sides of a polygon taken in order pass through (n — 1) fixed points, the 2 side will always touch a conic section. (Appendix 1. Art. 6). From any point P in the conic section let a polygon be formed whose sides successively pass through 4,, A,...4,_,3; and let a, 6, be the position of the nm side; in like manner let a,b,, a3b3, a,b,, a;6; be four more positions of the m™ side, and from A, draw as in Art. 97, two tangents B,B,_,, B’,B’,_,; to the conic section which ‘touches the five lines a,b), a.b., a3b3, a,6,, a;6;; these will be the 2" sides of the two polygons which can be inscribed in the conic section so that its sides may pass through the m given points; draw B,_,B,_. through 4,_,, B,.B,.; through A,» &c., B;B, through A;, B, B, through A,; then BA will pass through A,, Ten B, B....B,3 ByBz..:B,' will be the two polygons agin 101. Since the construction can be completed as in Art. 98, when only five points of the conic section are given, or in any of the conic sections described in Arts 84—88, we are thus enabled to inscribe in any of the above cases a polygon whose m sides taken in order shall pass through 7 fixed points, although the conic section itself is not actually traced. APPENDIX II, 255 102. To find under what limitations the problem is possible. Since the n™ side which passes through A, is a tangent to the conic section which touches a,b,, a,b,...a;,6,; if A, be without this conic section, we can draw two tangents, which will give two polygons. If 4, is a point in the conic section, there will only be one tangent, and only one polygon can be described. If 4, is within the conic section, it is impossible to draw a tangent from 4,, and the problem is impossible. Hence if 4 be the point of contact with a, b,, and 4A, be drawn, find B the extremity of the chord 44,; and there will be two polygons, one polygon or none according as A, falls beyond B, upon B, or within B. 103. (a) If two chords AB, AC (fig. 208) be drawn from a point A in a curve of the second order equally inclined -to a given line AD, the line joining the extremities B and C - will always pass through a fixed point. (Senate House Pro- blems, No. 15, Thursday, Jan. 7, 1847, 1...4.) Let AD be the axis of w; then if y— m,v =0; y+me=z0; y-mwew=0, and y—me —c,=0, be the equations to 4B, AC, the tangent at A and BC re- spectively, the equation to the conic section is (y — m,a@) (y + m,xv) +A(yY — mx) (y — me — C,) =0; or (1+A) y’—A (My +m) LY + (AM.M,—M,) & —ACLY+AAM,C,L=03 and if the equation to the conic section be ay’ + bey+cuv’+dy+ex=0; we have AC, d 4 fi q A Ms Cy ee OS SS) CO CS 1h, ) 4 aD X(m, + m4) 6 0 RCE Nae ica Oe d° 256 GEOMETRICAL PROBLEMS. d e ; hence c, = 5m a zs s Ms = - 3 b Qs ; d therefore the equation to CB becomes y + 5 = mM, (« 4 -) : which is the equation to a straight line passing through a fixed point whose co-ordinates are ——; — . Let the angles BAD, CAD be indefinitely diminished ; then CB ultimately becomes a tangent at D; hence the chord CB always passes through a fixed point in the tangent at D. (8) If AD be a chord of a surface of the second order; let any plane pass through AD, and let two chords 4B, AC be drawn in the plane section (which will be a curve of the second order) equally inclined to 4D; the lines joining B and C' will all pass through a point in the tangent, at D to the plane section, which will be a line on the tangent plane to the surface at D. Hence the lines similarly drawn in different plane sections will intersect in a series of points on the tangent plane to the curve surface at D. If four lines 4B, AC; AB’, AC’ be drawn in two dif- ferent plane sections, BC, B’C’ are two lines in different planes; and if they meet at all they will meet in some point in the intersection AD of the two planes. Hence all the possible intersections of the chords passing through the points B, C so drawn that 4B, AC may be equally inclined to a fixed chord AB in a surface of the second order, will be a plane curve on the tangent plane to the surface at the point D, and the line 4D. 104. Let AD be a fixed line, 4C, AB two lines equally inclined to 4D; BC the chord joining the points B, C; B’C’ any other position of BC; produce BC, B’C’ to meet in E; DE will be a tangent at the point D; which may be drawn at a given point D in a conic section. APPENDIX II. 257 105. If AB, AC ‘be two chords of a conic section, to find ; Let y+mev=0; y+mv=0; Y + Mew + C= 0; and Ly + m,2 = 0; be the equations to AB, AC, BC and the _ tangent at y respectively ; then the equation to the conic section is (y + mx) (y+ myx) +A (Y + Me + C,) (Y + Ma) = 0; + z ‘3 ‘ a 4 and making this coincide with the equation to the conic section ne ay? + bay + ca +dy+ex=0, we have a cane ms) + A (Mz + m,) = b (1 ae rd); a (m,m; + Am m,) = c (1 + r); arncgm = e{1+Ad); r 4 ; a ( eMs Cc ~ | CCy j or m,m _ — = - S ; 13 \ a d = Bit Prt a a (m, + m _ 1) ri, 2S 2 3) {—- M. +-— = —. pay bie ta ie (2) | Hence if m,m;, or m, +m, be constant; we have an equation of the form 8+ m,a+c¢,=0, where (, a are con- | stant ; and the line y + m,a” +c, = 0 passes through a fixed | i point a, PB. ani Let B + ma + c, = 0, be rhe general equation connecting iF N i m, and C3 q | . mM; (— - | sale (“ xi 4 er a ac, e +B be, | (m + ms) -1) mre ise | | : o 258 GEOMETRICAL PROBLEMS, and eliminating c,, an equation of the form m,m, — A(m, + m3) + B= 0, is determined where 4, B are constant. If this relation holds, the line BC will pass through a fixed point ; *, (m, — A)(m; — A) is constant, or {— — —]{— — =] is constant. iby aes) ANE, CD ’ | ed deh If B=0, — + — 1s constant. Mm, Ms 106. If the angle BAC be a right angle, m,m; = — 1; and BC will pass through a fixed point. To find the position of the fixed point; let C approach to A, then AC becomes a tangent, and 4B a normal; hence CB becomes a normal, and the fixed point lies in the normal. 107. To draw a normal to a conic section from a given point in the curve. Let A be the given point (fig. 209); draw any two chords AB, AB'; and AC, AC’ at right angles to AB, AB’ re- spectively ; join BC, B’'C’ meeting each other in O; then 4O will be a normal at the point 4. (Art. 106.) ¢ APPENDIX III. 1. Ir CL’ (fig. 214) be a given chord in a circle, it is required to draw through a given point G a chord CGL equal to C’L’. Let O be the centre of the circle; draw OH perpendicular to C’L’; with’centre O and radius OA describe a circle; from G draw CGML touching this circle; then since CZ and C’L’ are equally distant from the centre, CL = C'L’, There will be two positions of CGML corresponding to the two tangents through G. 2. If FQ, FR (fig..214) be two fixed lines; and from any point K”’ in a circle C’K’L’, K’C’, K'L’ be drawn parallel to FR, FQ respectively; then C’L’ will be constant for all positions of K’. For z2CK’L'’= 2QFR and is constant; hence C’L’ is constant. | 8. In a given circle, it is required to inscribe a triangle so that its three sides may pass through three fixed points P, Q and R. Let ABC (fig. 215) be the triangle required, whose three sides 4B, BC, CA pass through the three fixed points P, Q, R respectively ; join RP; produce it to F' so that FP.PR=AP.PB; join FQ, and make QG. QF' = QC. QB; produce BF to K; join CK, KL; then since a circle may be described about ast, A and kK, LARP=24ABF =24ABK = 2ACK; hence KC is parallel to FR. 260 GEOMETRICAL PROBLEMS. Again, since a circle may be described about the points C, G, F, B, £BCG+2BFG = 2 right angles =2KFQ+4BFG; | ZEFQ=2BCG=2zBCL=2BKL; hence KL is parallel to FQ; but when K’ is any point in the circle and K’C’, K’L’ (fig. 214) are drawn parallel to FR, FQ, C’L’ will be constant (Art. 2); hence if through G the line CGL be drawn equal to C’L’ (Art. 1), the point C will be one of the angular points of the triangle; through Q draw QCB, and through P draw BPA; join AC, then ABC will be the triangle required. The two positions of CGL will give two triangles which satisfy the conditions of the problem. If OG be less than OH, the point G falls within the circle whose radius is OH; and the problem is impossible. 4. To inscribe geometrically a triangle in an ellipse whose three sides shall pass through three fixed points. This problem is reduced to that of inscribing a triangle in the circle described upon the axis-major. (App. 1. Art. 3.) 5. To inscribe geometrically a triangle, in a hyperbola, whose sides shall pass through three fixed points. From Appendix 1, Art. 4 we have, A - (a + a3) tt. — posh + i) + Gz - t= 0. a b Let = ns = se 2 et a3 a Bs ; as then (a+ a) ht - By(i +t) +(a—a)=0. (1) 2 2 e ° a b, a” b, Similarly, if ag=—; B.=— ap; aay Beton oy 3 2 9 5 Qo5 a = 9 Tao Qa, Dy b a, B Baas we have (a + az) t,t; — (32 (4, + ts) + (@ — a2.) = 03; (2) (a + a) tots; — By (¢, + ts) + (a — a) = 0; (3) APPENDIX III. 261 which are the equations we should obtain for determining the angular points of a triangle inscribed in a circle whose radius is (a) so that the three sides may pass through three points whose co-ordinates measured from the centre are ane 0, Go a 0.05 OF). OB) a 3 3 9 9 oy ne 2 Go 10 a, U0 as hence if CA (fig. 146) be the semi-transverse axis; P an angular point of the triangle inscribed in the circle 4PD so that its sides may pass through the three points above deter- mined; then zACP=80,; and CR =a sec QO, =the abscissa of the corresponding angle of the triangle inscribed in the hyperbola so that its three sides may pass through the three points @3, bs; de, b.3 ay, b,. 6. To inscribe geometrically in a parabola, a triangle whose three sides shall pass through three given points. Substitute for y1, Yo Ys in equations (1), (2), (3), Art. 6, Append. 1. the values 2a¢,, 2a¢,, 2a¢, respectively ; and let a (a — a3) abs SESW Mem, Toes 3; =———, a+ Gs + Os a (a — as) ab, a, =————, B,= 4 a + Ay a+ Ay a (a — a,) ab, OA ei age eee MT 3, = 3 a+ a, a+ a, Sie (a - ats) t, ts — GBs (¢, a t) -F (a = Gz) => 0, &ec. hence ¢,, ¢, f;. are the same as for the angular points of a triangle inscribed in a circle whose radius is a, so that its three sides may pass through three fixed points a3, G3; a, Bo; a, 1, respectively. Hence if (fig. 15) be the vertex of a parabola; B the focus; with centre &, and radius FB describe the circle BDA, and let C be one of the angular points of the triangle 262 GEOMETRICAL PROBLEMS. inscribed in the circle BDA, so that its three sides may pass through three points whose co-ordinates are as, [355% Gemaicrne a,, 3,; draw BF perpendicular to AB meeting AC produced in F; then LCEB=0, £CAB= zs and BF = 2at;.= Ys; or BF is equal to the ordinate of the corresponding angular point of the triangle which is inscribed in the parabola so that its sides may pass through the three given points a, b;; Go, b,3 yy Oy. Hence in each of the three conic sections, the problem is reduced to that of inscribing in a given circle a triangle whose three sides shall pass through three fixed points. 7. From the equations in App. 1. Art. 10, it will easily be seen (1) that when a polygon is to be inscribed in an ellipse so that its 2 sides may pass through m given points a@,, 6, ; Ay bos 22. Ans b,; the problem will be equivalent to that of inscribing in the circle described upon the axis-major, a polygon whose m sides taken in order shall pass through the n fixed points a a , 5 03 A25 y O23 eee Ons Beil bee Qy > (2) When the polygon is to be inscribed in a hyperbola ; let a circle be described upon the transverse axis; and in this circle let a polygon be inscribed, whose n sides shall pass through the 7 fixed points 9 3 pty Ae 2 3 aD ai eer ae Ch rh Bie J then if P be any angular point of this polygon, as in Art. 5, CR (fig. 146) will be equal to the abscissa of the correspond- ing angular point of the polygon which is to be inscribed in the hyperbola. (3) When the polygon is to be inscribed in a parabola ; let E (fig. 15) be the vertex; B the focus; with centre E APPENDIX III. 263 and radius BE describe a circle; and in ihwe circle let a polygon be inscribed whose v sides shall pass through the 2 fixed points a(a—a) ab, a(a—-a,) ab a(a-a,) ab, 3 a ) ESR 3 | 3 a+ a, a+ a, a+ ay a+ a, a+ a, a+ a, let C be one of the angular points of this polygon; then, as in Art. 6, BF will be the value of the ordinate of the correspond- ing angular point of the polygon which is to be inscribed in the parabola. 8. When only five points of the conic section are given, the most simple method will be to determine the axes of the conic section, which will give geometrically in each case the circle and points through which the nm sides of the subsidiary polygon are to pass; from whence the abscissz of the cor- responding angular points of the polygon inscribed in the conic section; and the angular points themselves may be de- termined. For the Geometrical construction when the polygon is to be inscribed in a circle, the reader is referred to the Liverpool Apollonius by J. H. Swale. Bie A ve a aa Gf) Pa oe ny Pay Ale 2 Sy Se Rs Ja 6 Metcalz & Falmer, Irtho Metcalfe & Palmer, L tho 3D 4 D See a, X NA \ : Q N\ : : cS / rH | ~*~ * VE | S one ‘ pes Pees $ 8, oc 38 yy NS ve ii