THE SOLUTIONS OF GEOMETRICAL PROBLEMS, CONSISTING CHIEFLY OF EXAMPLES IN PLANE CO-ORDINATE GEOMETRY, PROPOSED AT ST JOHN'S COLLEGE, CAMBRIDGE, FROM DEC 1830 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.XLVII. Cambribge.: Ptrinted at te (titeraistti) tress. TO WILLIAM CRACKANTHORPE, ESQ., 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. PREFACE. THE Examination Papers which the Author has selected 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 residence, and according to the plan which he adopted in the solution of the Trigonometrical Problems, he has endeavoured to place them before the reader in a proper form for the inspection of the examiner. 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. 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 PRJEFACE. 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 conic 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 n sides taken in order shall pass through n fixed points. A simple and concise geometrical solution of M. Cramer's problem has been extracted from " The Liverpool 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 comparatively 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. 1ii 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 connexion with the corresponding results when the co-ordinates 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. viii PREFACE. In a subject which for ages has exercised the skill and ingenuity of the most profound mathematicians, little can 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 deserving the attention of every one who wishes to study Geometry. T. G. (CA 13 RIDG E, ANov. 1847. GEOMETRICAL PROBLEMS. ST JOHN'S COLLEGE. DEC. 1830. (No. I.) 1. PARALLELOGRAMS 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 circumnference. 4. The line joining the centres of the inscribed and circumscribed 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, t3, 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 a2 a + f3b + y7c = abc. 9. Describe a circle through a given point and touching a given straight line, so that the chord joining the given point A 2 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 a (2 + y2) + b2 (t + y) = 0 draw the lines represented by the equations y' - 2xy seca+ 2 = 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? 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)2 from the angle A of the required point = (b2 + c2- 17. If the centre of the inscribed circle of a triangle be fixed, and a, 3, y represent the distances of its angles from any fixed point in space; then whatever position the triangle assume, the expression a2a + t32b + 72c is invariable. GEOMETRICAL PROBLEMS. NO. I. DEC. 1830. SOLUTIONS TO (No. I.) 1. EUCLID, Prop. 35, Book i. 2. Euclid, 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 L ADB is a right angle = zADC; also L ABD = z ACD, and AD is common to the two triangles 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 perpendicular to AB, and join AD, Ad: then z ADE- = ADB = z C; /},7 - + f+C A A+ B - C or D= AC+B+C A+ B-C. or L DAE= ---- - C= ---2 2 2 A also dAe = -: 2.. L DAd = L DAE - LdAe --- In like manner it may be proved by joining DB, dB, DC, dC that z DBd=, and z DCd =a. 2 2 If L B be less than z C, Z 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,.. DjA = 4DDE' = 4DA. DB or DA = 4DB; 4B hence AB= 3BD, and BD - which determines the position of the point D. A 2 4 GEOMETRICAL PROBLEMS. NO. I. DEC. 1830. 6. Let C (fig. 4) be the centre of the larger circle; A 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 P(Q 2PA ZPOQ =2 PCQ, or -; PO PC' 2PO.. arc PQ = -- arc PA = arc PA; PC. 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 A of a circle; C the middle point of the arc AB; join AC, CB and draw CM, CN perpendicular to AB, AT respectively; then CN= CA sin z CAT= CA. sin z CBA = CB. sin z CBA = 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 r r a A- ' - B' = C; sin- sin sin2 2 2 hence a2a + 42b + y2 -- 2 + b si2 a+ i(C)s sin) sin -) ( a A b B c C\ = 2 2 cot -+ - cot - + - cot - sinA4 2 sin B 2 sin C 2 2ar2 { A B C a b c s iXA cot - + cot- + cot since- sinA 2 2 2 sinA sin B sin C GEOMETRICAL PROBLEMS. NO. I. DEC. 1830. 5 A B C but rcot- S - a, rcot- - -b, r cot = - c, 2 2 2 a + b + c anb+c where S = ( A B C\ S r cot - - cot cot = S; 2a 2a or a2a + 32b + yc = A r - - x area of A ABC sin X1 sin A = — (-/ sinA =ab,. sin A 2 9. Let P (fig. 6) be the given point, and AB 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 AB, 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 AB; and Z PDB = the angle in the alternate segment cut off by DP = L CPD = the given angle. 10. Let b, c (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 z Cab = z Bac. (POTTS' EUCLID, p. 293.) Now if a be the angular point properly determined, and b the angular point in the side AC, then ac + cb is least when Z Bca = z Acb: similarly, if a, c be two angular points of the triangle, ab + be will be least when z Abe = z Cba: hence the perimeter of the triangle abe will be the least possible when Z Cab= Bac; Z Cba = Z Abe, and Acb = Bca. Now if Aa, Bb, Cc be drawn from the angles A, B, C perpendicular 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 L Aac = L ACc = - A and Bac - Aac = A. 2 2 6 GEOMETRICAL PROBLEMS. N O. I. DEC. 1830. Similarly, the circle described on the diameter AB will pass 7r~~~~~~~~~7 through the points 6, a; and z Aab = L ABb - - A;. Cab =-Aab = A,orLlac= Cab; similarly, z Cba = Z Abe, and L Aeb z, Bea; therefore the triangle abC has the least possible perimeter. 11. See HTYMERrs' CoNIC SECTIoNs. Art. 13. ___ X b'y 12. (1). + - xy$ = 0; a a,Q a 2 al a / 2 t 2 a 2b the equation to a circle the co-ordinates of whose centre are b2 b2 b2 and whose radius -. Hence if Aze, Ay (fig. 8) be the co-ordinate axes, take AlB, BC each = - and with centre C and radius describe a circle AED; it will pass through A and meet the axis of x in the point D such that DB = BA; and L zACB D LDCBJ L. 4 2 hence AED' is a quadrant of the circle having the chord produced for the axis of v. (2) Since y - 2Qvy sec a + v2 = 0, we have (Y)I - 2 sec a + (sec a)' = (tan a)' sec a t tan a = tan (5 L xl GEOMETRICAL PROBLEMS. NO0. I. DEC. 1830. "/ which are the equations to two straight lines passing through a the origin and inclined at angles 45 + - and 45 - - respectively 2 2 to the axis of a?; hence the angle between themn a a~a 4(5 s 4+ - ( ) ca. 13. Let 0 (fig. 9) be the origin; AB the straight line meeting the axes of x and y in the points A, B respectively; draw OP perpendicular to AB and let AB = a, OP b, QA = m, OB n; then equation to the line AB is where m'+-n2=a a'da n n= 2A O0B =AB.OP = a'; m + n NS12 r 2 -I a6, m - n V a= - 2 a. b'; or 2m=L (/a2+ 2ab bL-Va 2- 2ab), and 2n W= ( a+ 2ab 9=,& - 2ab); therefore if Va&a + 2 ab + Va2 -2 ab =!2 acand\N/a2+2a b -,\la2 2ab=2/3, we have m = A= a, or a /3, and the corresponding values of n are zL/, La; hence - =6L= o, or 6- J= -1; each of af)3 f3 a which equations will give the equations to four straight lines by the four combinations of the double sign I. If OA, OA', QA", QA"' be taken each equal to a, and OB, OR', OB", OB"' each equal to 3; then Al, BA", A"B", B"A; A'B', B'A"', A"'B"', PB"A' will be the straight lines required. 14. See HIYmE1s' Co-\Tic SECTIONS. Art. 104. The equation to the ellipse is y ( -e2) a? - a?2) = 2p (I + e)> - (I e2) IV2 and when e = -, yII 4pxa, which is the equation to the parabola. 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, c respectively: then A APB = - AB. Pc. sin PcB =- A ACB = 1(1B. Cc.sin PcB), Cc Bb Aa.. Pc = -; similarly, Pb = and Pa Join ac, then PC = 2 Pc, PA = 2 Pa, and Pa: Pc:: PA: PC, or ac is parallel to AC, and c: BA:: ca: CA:: Pc: PC::: 2;. BA = 2Bc, and the side BA is bisected in C; and in like manner it may be proved that a, b 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 xv, y be the coordinates of P; PM: AC:: PC: Cc, or PMI = AC; b c b y=-, and AM: MP:: A: ca::-:-; 3 2 2 c b cy c..*. t -: - and =v = - 2 2 b 3 16. Let PA = a, then Aa =: and since BC is bisected in a, 2Aa' + 2Ba2 = AB2 + AC', or + — = b + c2; 29 2 17. First let P (fig. iI) be a point in the plane ABC, 0 the centre of the inscribed circle whose radius is r; join PA, PB, PC, PO, AO, BO, CO; then if PA = a, PB = 3, PC= 7, z AOP =, we have GEOMETRICAL PROBLEMS. NO. I. DEC, 1830. 9 a2a + (3b + 7y2 = a(P02 + A02 - 2PO. AO. cos POA) + b (PO2 + BO2 - 2PO. BO.cos POB) + c (PO2 + CO2 - 2PO. CO. cos POC) = (a + b + c) PO2 + a. O42 + b. BO + c. CO' -2PO(a.AO.cosPOA + b.BO.cosPOB+c. CO. cosPOC) = (a + b + c) PO + abe r A+B3 A ++C p a.O.- cosO +b Bcs (0+7r- ) +c cos(7- -0) sin- sin - sin2 2 2 (from Quest. 8) = (a +b +c)PO' + abc sinA 2 2 \22 = (a + b + c) P02 +- abc ar f A C LB\' - jPO. --- <!cos - cos Co - (sin - cos -+ - cos - sin -} cos O sinA\ 2 2 2 2 2 = (a + b + c) PO' + abc -4PO si cos - - sin ) cos 0 a A sin+A 2 2 = (a + b + c) PO2+- abc, B +C A since sin — = cos -. 2 2 Next let P be without the plane ABC; draw PQ perpendicular to the plane, and join QA, QB, QC, QO, then PA' = QP2 + QA2, PO' = QPe + QO2; and aa+f'b+7y c=a(QP+ QA2)+b(QP2+QB2)+c (QP +-QC2) = (a + b + c) QP2 + a.QA' + b.QB-+c. QC = (a + b + c) QP: + (a + b + c) QO0 + abc (by the first case) = (a + b + c) PO + abc, and is therefore invariable, since P and 0 are two fixed points. 10 GEOMETRICAL PROBLEMS. NO. II. DEC. 1831. ST JOHN'S COLLEGE. DEC. 1831. (No. II.) 1. IF four magnitudes of the same kind are proportionals, 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 AB of a right-angled triangle ABC, and from tile centre E the radius ED be drawn at right angles to AB, shew that the difference of the segments on the two sides equal twice the sector CED. 7. 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. GEOMETRICAL PROBLEMS. NO. II. DEC. 1831. II 10. Find the length of the perpendicular from the origin of co-ordinates upon the line whose equation is a (m - a) + b (y - b) = 0, and the part of the line intercepted between the co-ordinate axes. 11. The equation to a circle isS y2 + 2 = 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 x, the equations to the equations to the other sides are y = ax + b, and y = a'x; 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 r, i' be the radii of these circles, shew that rr' =R2. 12 GEOMETRICAL PROBLEMS. NO. II. DEC. 1831. SOLUTIONS TO (No. II.) 1. EUCLID, Prop. 25. Book v. 2. Euclid, Prop. 10. Book xi. 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. sin 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; join AF, BF; then BF = BA and LBFA= LBAF= L AFD, or Z DFB is bisected by the straight line AF. Similarly if AE, BE be joined, z DEB is bisected by AE. Also if with centre A 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 circumference ACB. Bisect the arc AB in D, and join CD; then since arcAD = arc DB, Z 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, ADCB the semicircle described upon the hypothenuse AB; then since AE = EB, AAEC= A BEC; 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 AED = quadrant BED. GEOMETRICAL PROBLEMS. NO. II. DEC. 1831. 13 7. Let O (fig. 16) be the centre of the circle inscribed in the AACB; join OA, OB; then z AOB = Sr - (OAB + OBA) A +B 7r -C Vr C ~ -7 ---= -r - -r -; 2 2 2 2 and is constant. Hence the locus of the point O is a segment of a circle containing an angle 7r C 2 2 and the z ADB in the alternate segment /7r C\ '7 C r'- + - - -- _ \2 2J 2 2; therefore the angle subtended by AOB at the centre E = T - C. When the A A CB is right-angled, q7r 7i zC=-; and LAEB= - C =-; 2 2 therefore AOB is the quadrant described upon the hypothenuse AB. 8. Let P (fig. 17) be any point in the line AP which meets the plane in the point A; draw PM perpendicular to the plane; join AM, and through A draw any other line AQ in the plane; draw PQ perpendicular to AQ, and join QM; then if 0, 0', be the inclinations of AP to AM, AQ respectively, PM PQ sin 0 and sin 0' =4AP AP' but PQ2 = PM2 + MQ2 since Z PMQ is a right angle, or PQ is greater than PM; therefore sin 0 is less than sin 0' and 0 is less than 0'. Now 0 measures the inclination of the line AP to the plane, which is therefore less than the inclination of the line AP to any other line AQ drawn to meet it in that plane. 9. Let y = ax + b, y = a' + b' be the equations to the two straight lines; then if 0, O' be the angles which they respectively make with the axis of, tan 0 = a, tan ' = a', 14 GEOMETRICAL PROBLEMS. NO. I. DEC. 1831. a -a and sin (0'- 0) = cos cos0'(tan0' - tan 0) \/(1 + a) (1 +a'2) which is the sine of the inclination of the two straight lines to each other. 10. Let 0 (fig. 9) be the origin, and A, B the points in which the given straight line meets the axes of x and y respectively; then the equation to the line OP drawn through 0 b x perpendicular to AB is y= -; and if xar y' be the co-ordinates of the point of intersection P, we have a (x' - a) + b ( —b ), \a ) b x' or w =a, and y' = =b; a therefore the length of the perpendicular OP = / = 4a' + b +. Again in the equation to the given straight line, make a2 + b2 y = 0, the corresponding value of x is OA = --; and by a as + b2 making x = 0, the value of y is BO=;. B. AB: v/AO' + BO' = -- ab 11. x2 - a? + y - ay - 0, a\ a\ 2 a2.'. - - 4- -- - -- the equation to a circle, the co-ordinates of whose centre are a a 2' 2 Let y = mcm be the equation to the diameter which passes through the origin, then since this straight line passes through a a a a a point - we have -=m.- or m= 1; and y =x is the 2equati2 2requ equation required. GEOMETRICAL PROBLEMS. NO. II. DEC. 1831. 15 12. Let ABC (fig. 18) be the triangle, A the origin, and BA produced the axis of x, the equations to AC, BC are y = a'x, and y = ax + b;.. tan (CAx) =tan (wr - A)a' or tan A= -a', and tan CBx =tan B = a; a -a hence tan C = tan (CAx - CBx) = - i [ aa Draw CM perpendicular to AB, and in the equation to BC make y = 0, then the corresponding value of x = - AB b b = - -or AB = -: let x', y' be the co-ordinates of M, then a a a x = ax + b or x' =-= AM t a -a a'b b /12 + a'2 = CM = a <= a- AC' = a/' y -a a-a a -a b b ba' also BNl = JB- AB i - + 4 -; a a - a a (a' - a) ' ab v/1 + a a (a - a) 13. Let the bounding radii AB, AC (fig. 19) of the quadrant be taken for the axes of x and y respectively; bisect the Z 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 AB, AC: and if AE = ED = p, the radius will also be = p, and the equation to the circle is (X - p) + (y _ p)2 =p2 or o 2 + y _ p (0 + y) + p2 = o. Let this circle cut the quadrant in the point P whose co-ordinates are h, k;.h. h2 +k - _2p (h+- ) + 0, or R?- 2(h + ) p + p2 = 0; from this equation the two values r, r' of p may be determined, and rr'= R?2, r + r' = 2 (h + k). 1 6 GEOMETRICAL PROBLEMS. NO. III. JAN. 1833. ST JOHN'S COLLEGE. JAN. 1833. (No. III.) 1. MAGNITUDES 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 E 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 E, H, and the straight line FACB in F, G; shew that a circle may be made to pass through the four points F, E, G, H. GEOMETRICAL PROBLEMS. NO. III. JAN. 1833. 17 9. If through any point O within a triangle, three straight lines be drawn from the angles A, B, C to meet the opposite sides in a, b, c respectively; prove that Ac.Ba. Cb = c B. a C.bA. 10. Find the equation to a straight line which cuts a given circle, when the lines drawn from the points of intersection to the centre contain a right angle, and one of them is inclined to the axis of x 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 1, ~2 the diameters of the circles inscribed in the triangles formed by joining the vertex and the middle point of the hypothenuse; shew that 1 1 I 1 St 2 a, 13. Find the equation to the straight line drawn from a given point to bisect a given equilateral triangle. B 18 GEOMETRICAL PROBLEMS. NO. II. JAN. 1833. SOLUTIONS TO (No. III). 1. EUCI)D, Prop. 9. Book v. 2. Euclid, Prop. 20, Book xI. 3. Let AB (fig. 20) be divided into two parts in the point C; E the required point such that LAEC = z BEC; then if AC = c, BC = c', CE = p; z ECB =, z AEC = L BEC =, -P-= sin (Q+ ) p _ sin (-0) ) c sin 0 c sin 0 2cc' /1 1\ ) sin ((p + 0) - sin (P -_ 0) - c)P = i- =2 cos (p, or p -, cos which is the polar equation to a circle c - Cf 2ccwhose diameter =- is in the direction CB. C - cc Produce CB to D, take CD =, and upon CD describe a circle; from any point E in its circumference draw AE, EC, EB, then will z AEC = z BEC. 4. Let AB (fig. 21) be the given base; draw AC perpendicular 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 AD, and make z DAE = Z ADB, then AEB will be the triangle required; for z ADE = z DAE,.-. AE = DE, and AE + EB = DB the sum of the sides. 5. DECF (fig. 22) is evidently a parallelogram, and z CDE D= z DCF = z DCE;.-. DE= E = 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', z ABQ = O, AP = a, AQ = 2a, AB = c; '. p2 + c ~ 2cpcos0=a2; p +c2- 2cp'cos 0= 4a2; GEOMETRICAL PROBLEMS. NO. II. JAN. 1833. 19 and if p' = 2p, we have 4Pp' + c - 4cp cos = 4 a2, but 4p2 + 4c2 - 8cp cos0 = 4a2; 3c '. 4p cos = sc or pcos 0 -. 4 Hence in AB take BM = P AB or AM-= - draw MP 4 perpendicular to AB, meeting the inner circle in P; join BPQ, and BP will be equal to PQ. 7. Let 0 (fig. 24) be the centre, CAO = 0, AO = a; then CA = 2a cos0, and CP = n (C); A.. AP = (n + 1) CA = 2 (n + 1) a cos 0, or the polar equation to the locus of P is p =2 (n + 1) a cos 0, which is the equation to a circle whose centre is in the line AO, and radius = (n + 1) a. 8. Draw DK (fig. 25) touching the circle at D; then z DFG = z KDE = z EHG: hence a circle may be described through the four points F, E, G, H since EFG and EHG will be angles in the same segment and upon the same base EG, and are proved to be equal to one another. Ac sin AOc Bc sin BOc 9. (Fig. 26) 9. (Fig26) AO- sinAcO' BO sinBcO; Ac AO sin AOc Bc BO sin BOc Ba BO sin BOa Similarly a C 'sinCOa Cb CO sin COb Ab AO'sinAOb' hence by multiplication Ac. Ba. Cb ------ = C 1 or Ac. Ba. Cb= Be. Ca.Ab. Bc. Ca.Ab 10. Let PQ (fig. 27) be the straight line whose equation is required, C the centre, r the radius of the given circle; CA B 2 20 GEOMETRICAL PROBLEMS. NO. III. JAN. 1833. the axis of v; L PCA = a,.% L QCA = 90 + a; and the coordinates of the points P, Q are a cos a, a sin a; and - a sin a, a cos a respectively; therefore the equation to PQ is a cosa - a sina y - a sn a= -i (- - a cosa) sin a -a - a cosa sin a + cos ae (,, - a cos a); sin a + COSco a sin a - cos a a sin a +cosa a+ sin a + cos a' hence y=tan (a - 45)?+ - sec (a - 45) is the equation required. 11. Let h, k be the co-ordinates of the given point without the circle; a?, y', a", y" the co-ordinates of the two points of contact, then the equations to the two tangents are 'v+ y'y = a 2, and a?+ yy=a; 2 and since they both pass through a point whose co-ordinates are h, k, we have ha' + ky' = a2, hav" -+ ky' a2; and the equation to the line joining the points of contact is y" - y " ' h Y-Y= Y (a? -?'); but (V~ X X -..y-y'= - (a?-a') orhAi + ky= hx' '+Jy'a=a the equation required. 12. Let ACB (fig. 28) be the right-angled triangle, D the middle point of the hypothenuse AB; AC = a, BC= a,, AB = c; a, A A B 2 - = tan - = a t~an - 82 a tan - 2 2 2 2 A B A B cot - cot - jcot - cot - 1 1 2 2 ii 2 21 or a, a, c sin B sin / GEOMETRICAL PROBLEMS. NO. III. JAN. 1833. 21 2 cos — - 2 cos' - 11 2 2 1 ( cos- cosB) c sin 4 sin B sin A sin B c (cos A - cos B) a,- a, (csin A) c sin B a al 1 1 a, - a 1 1 2 a, a a) a, 13. Let a,,3 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; X y then if AD = m, AE = n, the equation to DPE is - + - =1, m n and since this passes through the point P whose co-ordinates /a 3 are a, p, we have- - - =1, m n and A EAD = n sin A = 1- A CAB = 2 -sin A); 2 2 2 as a'.mn = -, hence an +/3m=mn -, and an - 3m = 4 J/(an + 3m)2 - 4a/3mn a4 a2 8a/ =_ - -02aa --- 4- - 1 a4 2 a.. 2an =- 1 _- - I /8a/\ a2 a 8 a a' 2/3m 1 1 - 2) and mn t;.'. -- a( Vi -- ) a Vi - a), and the equation to DPE becomes 1 ( / 8a B - "-H) 't 21 T 8=1) a(I v1 6\I- )X(~ I V)/. 22 GEOMETRICAL PROBLEMS. NO. IV. DEC. 1833. ST. JOHN'S COLLEGE. DEC. 1833. (No. IV.) 1. FIND 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. If a 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 quadrilateral 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 respectively, prove that QR2 + Qr2 = AB2. 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, n, p respectively. If t,,,, t,, t1p be the lengths of the lines drawn from m, n, p touching the circle, shew that tmnttp = An. Bmn. Cp = Amn. 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. 23 10. Investigate the line or lines represented by the equation y' + (x - a) y2 + (,2 - a2) y + (v - a)2(x + 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', t", y the coordinates of the other two; prove that the area of the triangle = (W -Y'X 24 GEOMETRICAL PROBLEMS. NO. IV. DEC. 1833. SOLUTIONS TO (No. IV.) 1. EUCLID, Prop. 1, Book IIn. 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 determined, and the circle is therefore determined in magnitude and position. 2. Euclid, Prop. 18, Book xi. 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, l ABC = - AC. BE. sin a: Also A ADC -- AC. DE. sin a; therefore by addition, the area of the quadrilateral ABCD = 2 (AC. BD.sin a): this will be the greatest when AC, BD and sin a are respectively greatest. Now if the quadrilateral be inscribed in a circle, AC, BD are the greatest possible when they are two diameters; and sin a is greatest, when AC, BD are at right angles. Hence if two diameters AC, BD be drawn at right angles, and AB, 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 the quadrilateral figure cannot be inscribed in a circle. (y) If a circle can be inscribed in A BCD (fig. 31) touching the sides AB, BC, CD, DA in the points a, b, c, d respectively, then Aa - Ad Ba = Bb, Dce Dd, Cc = Cb,.A. a - Ba + DB + Cc = Ad + Dd+ Bb + Cb, or AB + CD = BC + DA; 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 AB to E making BE= AB; join AC, CE; then AACE = 2AACB = o ABCD equal the given rectilineal figure. 5. Draw CDN (fig. 33) perpendicular to the base, meeting AB, Rr in D and N respectively; then since Rr is bisected in N, RQ" + Qr2 = 2 (RN2 + QN2) =2 (CTN2 + NQ2) (since z CRN = 2 a right angle = L RCN) = 2 CA2 = CA2 + CB2 = AB2. 6. See Appendix i. Cor. Art. 1. 7. Let Amp(fig. 34) =0, Anp = p, z Bpm- =, tn Am sin q Bp sin 0 Cn sin/ A n sin0 Bm -sin b' Cp sin ]' Am. Bp. Cn therefore by multiplication, A- B. C 1, An. Bm. Cp also t =Am. Bn, t = Cn. An, t =Bp. Cp;.'. tm t. t = (Am.Bp. Cn) (An. Bm.Cp) = (Am.Bp. Cn)2, or t.t. tp = Am. Bp. Cn = An. Bm. Cp. 8. Let the fixed point A 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 CA and equal to it; join CB and with centre C and radius CB describe a circle, this will be the circle required. For CB2 = CA2 + AB2 = 2CA2, and circles are to one another as the squares of their radii; therefore the circle whose radius is CB: circle whose radius is CA:: CB2: CA:: 2: 1. 26 GEOMETRICAL PROBLEMS. NO. IV. DEC. 1833. 10. The equation may be reduced to the form (y + x - a) (y2 + v2 - a2) = 0, which is satisfied by making separately y + x - a = 0, and y2 + x2 - a- = 0. If CA, CB (fig. 36) be taken for the co-ordinate axes, the former equation represents the straight line AB, 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 AB, AD (fig. 37) of the parallelogram ABCD be taken for the co-ordinate axes; join AC, BD, intersecting each other in E, draw EF parallel to AD, and let AB = a, AD = b; then the equations to the diagonals AC, BD are y =-, and - + - 1, and if x', y' be the co-ordinates of a a b their point of intersection E,, b' xa y y = -- + - 1; a a b 2 ', a 2 y' b a 2 b 2 AE AF AC hence = B- or AE - 2' BE EF BD and BD - or BEand the diagonals of the parallelogram bisects each other. 12. Let the equation to the circle referred to the centre C (fig. 38) be x2 + y2 = a2, and the equation to the chord PQ, y = m + c; then the equation to CR drawn through C per1 pendicular to PQ is y = --, and if X, Y be the co-ordinates m of the point R, C cm/ Y= -mY+ c, or Y - andX= - - + m 1 + m2 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 IV2 + (Mx + C) 2= a2,or (I + M2) x2 +2 2m cx + (c2 - a2) = 0; therefore if v', x"be the two values of xv which are the abscissoe of the points P, Q or IR is the middle point of PQ. 13. Let ABC (fig. 39) be the triangle, Ax,, Ay the coordinate axes; z BA x = 0', CAxc 0", AB = p', A C = / then the area of the triangle ABC ~AB. AC. sin z BAC= p'p91fsin (0"f-0') (p" sin 0" I. p'cos 0' - p" cos 0". p' sin 0') = —(y~ xy') 28 GEOMETRICAL PROBLEMS. NO. V. DEC. 1834. ST JOHN'S COLLEGE. DEC. 1834. (No. V.) 1. WHAT 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 magnitudes 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 incommensurable 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 invariable. 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: /5. 7. ACB is a triangle whose base AB is divided in E and produced to F, so that AE: EB and also AF: FB as AC: CB. Join CE, CF and shew that z ECF is a right angle. 8. The point C is the centre of a given circle, and E 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 pI P2t, p, be perpendiculars from any point within a triangle on the sides; PI, P2, P3 perpendiculars from the angular points on the same sides respectively, prove that + P = 1. P1 P2 P. 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. If a 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 xi. and Definition 6. Book xI. 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 AE, EF perpendicular to the two planes BAC, BAD respectively; then A.E is perpendicular to the plane BAC and therefore to AB. Hence AB is perpendicular to AE, AC, AD which are consequently in the same plane. Similarly AF may be proved to be in the same plane with AC and AD; and right z EAC = right z FAD;.. EAF = zDAC. 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 = PD2- pD2= (PC2 - CD) - (pC2 CD2) = CP2 - Cp2, and is therefore invariable. 5. See Potts' Euclid, p. 73. Taking Euclid's figure, AH = F = a;..EF=.=a = a + - =V/AB2 + AE2= BE, which give d c.4 which gives Euclid's construction. GEOMETRICAL PROBLEMS. NO. V. DEC. 1834. 31 6. Let AB (fig. 43) be the given straight line; 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 = /AB2 + BC2 = 5. B, and AD: AB:: AB: AC:: 1: 0/5. 7. Produce AC (fig. 43) to G; then CE, CF bisect the angles 4CB, BCG respectively;.L. z ECF= zBCF +J z BCE = ( z BCG + z 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 L PEC is a right angle. Draw any line CQ'Q meeting the circles in Q', Q; join EQ, EQ'; then z CPE = L CQ'E is greater than Z CQE, or z CPE 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, E two points in the diameter AB at equal distances from C; PDQ any chord through D; join CP, CQ, EP, EQ; then PD' + PE2 = 2 CP2 + 2 CD' = 2AC2 + 2 CD2; DQ'2+ QE2 =2 CQ2 + 2 CD` - 2 AC2 + 2 CD"; 2DQ. DP = 2AD.DB =2 AC - 2 CD2; therefore by addition PQ2 + PE2 + QE2 = 6 AC' + 2 CD2, and is invariable for every position of the chord PQ passing through D. 10. Let P (fig. 46) be the point within the triangle ABC; join AP, BP, CP; then 32 GEOMETRICAL PROBLEMS. NO. V. DEC. 1834. BC Pi BC Pi AA -_ BPC = pi Pi.Pi. PI= B z~PCp,2 P1, 2 P1 C similarly znAPC = z32L ABC; and AAPB = AABC; P3 therefore by addition ABPC d- APC + AAPB =,ABC= Pi z+3'AABC; Pi P2 P i 2P or -+ —2+- 1 P1P2 P3 11. Let AM, AN (fig. 47) be taken for the axes of iv and y; and v, y the co-ordinates of F; then iV + y = I the 2~~~~~~~~~~~~1 perimeter of the parallelogram - P 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 perpendicular to AB, and let AM = x, PM = y; then AP2 =iv2 + y2, BP2 =(a -V)2 + y2, DP2 = '-& - (a - y)2, CP2 = (a _ V)2+ (a _ y)2; AP, +2 fBP2 + DP2 + CP2 X2 ~2+ (a _,V)2 + y2+ (a - y)e 3 =C2; C2 4a2 X2 - 2ax + 2y2 -2ay= and x - + _ = the equation to a circle the co-ordinates of whose centre are a a_ and radius e ic_- 2a2. GEOMETRICAL PROBLEMS. NO. V. DEC. 1834. 33 Hence in order that the problem may be possible c2 must not be less than a2. 13. Let the straight line joining the two points A, B (fig. 49) be taken for the axis of x; C, the middle point of AB, the origin; AC = c; then the equations to the two lines AP, BP are y = m (x + c), y = m'(x - c) respectively. And if (a) be the angle between them; Y Y '- m t - c x + c 2cy tan a -- m - C 1 + y1 + y. 2. x + y2 _ c = 2 c cota y, and x2 + (y - c cota)2 = (ccoseca)2; 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 = c coseca. 14. Let AB, BC, CD, DA (fig. 50) be the four sides of the base of the parallelopiped; ab, be, cd, da the corresponding sides of the top; then ACca will form a parallelogram whose sides are two diagonals AC, ca of the base and top, and the two sides Aa, Cc 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 AB = a, AD = b, BAD = a; BD d, AC = d', we have d2= a2+ b2 -ab cosa, and d'2 = a2 + b2 + 2ab cos a;... d2 + d2 = 2 (a2+ b2), or in any parallelogram, the sum of the squares of the diagonals is double the sum of the squares of the sides. Hence Ac2 + Ca2 = 2 (AC2 + Cc2), Bd2 + Db_ = 2 (BD' + Bb') = 2 (BD2 + CC2);... A2 + Ca2 + Bd2 + Db2 = 2 (AC2 + BD2) + 4 Cc2 = 2 2(AB2 + BC2)} + 4 Ccs = 4(AB2 + BC2 + Cc') = the sum of the squares of the twelve edges. C 34 GEOMETRICAL PROBLEMS. NO. V. DEC. 1834o 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, c respectively: then AB = V/a 2, AC= +/a2 + C2, BC= /b2 + C2 A C + AB'-2 BC C a2 and cos z BA C =; 2AB. AC v/(a2 + b2) (a2 + c2) * //a2b + a c"' + b2 c2 sin L zBAC b- + V(a2 + b) (a2 + c2) and AABC= 1 AB AC. sin z B AC= /ab + a2b c + bec2; or (AABC)2 + j + ( = (A AOB)2 + (a A0C)2 + ( BOC)2o GEOMETRICAL PROBLEMS. NO. VI. DEC. 1835. 35 ST JOHN'S COLLEGE. DEC. 1835. (No. VI.) 1. IN 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 parallel 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 considerations 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 equimultiples according to Euclid's definition, that the magnitudes 4, 5, 79 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. A chord POQ cuts the diameter of a circle in 0 in an angle equal half a right angle; shew that PO0 + OQ2 2 (rad.)2. 5. Two circles touch internally; describe another of given radius (not greater than the difference of the radii of the former) so as to touch both. Prove that the locus of the centres of all the circles so described, is an ellipse whose foci are the centres of the two given circles. 6. If the lines bisecting the vertical angles of a triangle be divided into parts which are to one another as the base to the sum of the sides, the point of division is the centre of the inscribed circle. 7. Find the locus of the points of quadrisection of all parallel chords in a circle; employing the equations y2 = a - x; y = ax + 3. 2 36 GEOMETRICAL PROBLEMS. NO. VI. DEC. 1835. 8. If all the ordinates of a circle be moved through a given angle, the abscissa and magnitude of each ordinate remaining the same; what will be the curve, the origin of co-ordinates being the centre? 9. Eliminate (a) between the equations y = a (x - a); y2 = (a + ) ( - a - ). Explain the geometrical meaning of the result, and trace the change as (() diminishes and ultimately vanishes. 10. In the hyperbola, at a point P whose ordinate (BC)4 (Cs)' prove that Py = CS, Cy being a perpendicular from the centre on the tangent at P. 11. If about the exterior focus of a hyperbola, a circle be described with radius equal the semi-conjugate axis, and tangents be drawn to it from any point in the hyperbola, the line joining the points of contact will touch the circle described upon the transverse axis as its diameter. 12. If PSQ be a focal chord of an ellipse, tangents at P and Q will meet in the directrix. 13. In an ellipse the sum of the squares of two conjugate normals is constant. 14. S is any point in the diameter AB of a circle whose circumference is divided into 2n equal parts; if lines be drawn from S to all points of section, the sum of their squares = n (SA2 + SB2). 15. A sphere is described touching a face of any triangular pyramid and the other three produced, and three other spheres in like manner touching the remaining faces. If rI, r2, r3, r4 be their radii, and r the radius of the inscribed 1 1 1 1 2 sphere, -+- +-+-= -. r,1 2 r3 r4 r GEOMETRICAL PROBLEMS. NO. VI. DEC. 1835. 37 SOLUTIONS TO (No. VI.) 1. SEE Potts' Euclid, p. 50. Also Note to Prop. 6, Book i. p. 50; and Note to Def. 9, Book xi. p. 253. 2. Euclid, Prop. 11, Book v. Let the equimultiples of the first and third be 5, and of the second and fourth be 4; then the multiples of the first, second, third and fourth are respectively 20, 20, 35, 36; or the multiple of the first is equal to the multiple of the second, but the multiple of the third is not equal to the multiple of the fourth; and the four quantities are therefore not proportionals. Similarly, if the equimultiples of the first and third be 19, and the equimultiples of the second and fourth be 15, the multiples of the first, second, third and fourth respectively become, 76, 75, 133, 135; or the multiple of the first is greater than the multiple of the second, but the multiple of the third not greater than the multiple of the fourth; hence the first has a greater ratio to the second than the third has to the fourth. 3. Euclid, Prop. 19, Book I. If A: B:: B: C, then the ratio of A to C is that compounded of the ratios of A to B and of B to C, or of A to B and A to B, and is therefore the same of the ratio of A2 to B. 4. From the centre C, (fig. 52) draw CAN perpendicular to PQ; then since PQ is bisected in N, PO' + QO2 = 2(QN2 + NO ) = 2 (QN' + NC), (since z OCN = z PNC- z POC= 2 a right angle= z NOC);.. O" + QO2=2QC'. 5. Let C, 0 (fig. 53) be the centres of the two circles whose radii are a, b; P the centre of the circle which touches both circles; p its radius; then OP= b + p; CP = a -; hence if on the base OC a triangle OPC be described whose two sides OP, CP are equal to b + p, a - p respectively, the vertex P will be the centre of the required circle. Also OP + CP = b + p + a - p = a -+ b; 38 GEOMETRICAL PROBLEMS. NO. VI. DEC, 1835. hence the locus of P is an ellipse whose foci are 0, C; and axis major = a + b. 6. Let CD (fig. 54) bisect the angle ACB; divide CD in E, so that DE: EC:: AB: AC + CB, and draw CM, EN perpendicular to AB; then DE c DE EN c EC a+ b' DC CM a+b c' hence c (CM) = (a + b + c) EN. But if r be the radius of the inscribed circle, (a + b + c) r = 2 area of ABC = c. CM= (a + b + c) EN;.. EN = r; and the centre of the inscribed circle lies in the line CD, therefore E is the centre of the circle inscribed in the triangle ABC. 7. Transform the origin to a point X, Y in one of the parallel chords; then (Y + y)2 + (X + t)2 = a2, and y = ax are the equations to the circle and chord respectively. Hence at the points of intersection of the circle and chord, (r +a)' + (X + )= a. (). Now since the origin is in the quadrisection of the chord, if x' be one value of x derived from equation (1), - 3x' will be the other; therefore the equation must be of the form (x - ') (vo + 3x') = 0, or x2 + 2x'x - 3'2 =; but (1 + a') V' + 2 (a Y + X)x + X2 + Y - a' = 0; aY+X a,, (X+ Y2).. =; 3 x2 = - 1 + aX 1 + ra raF+Jr a2- (x, F2) or 3 - - _ - or 1 + a2 ] 1- a hence (4 + a') X2 + 6 aXY + (4a2 + 1) Y2 = (1 + a2) a' is the equation to the locus required, and is that of an ellipse whose centre coincides with the centre of the circle. 8. Let x', y' be the co-ordinates of any point in the required curve, 90 - a the angle through which the ordinates are GEOMETRICAL PROBLEMS. NO. VI. DEC. 1835. 39 moved, x, y, the co-ordinates of the point in the circle corresponding to the point x!, y'; then Xi '- y cota; y = y'coseca; and x2 + y2 a2;.'. (a! - y'cot a)2 + (y'cosec a)2 = a2, or 1_'2 2 Iy' cota + y'2 I1 + 2 (cot a)'} = a2, the equation to an ellipse, whose centre coincides with the centre of the circle. Let the curve be transformed to polar co-ordinates by putting a! = p cos0, y' = p sin 0; p2 COS2 - sin 2 0. cot a + (1 + 2 cot' a) sin2 0} = a, or P2 {1 +cot2 a (1 - cos 2 0) - sin 2 0 cota= a2; p2 cosec'a - cot a cosec a cos (2 0 - a)} = a'; 2 22a2 sin2 -a 2 sin2 a 2 and p2 =\ asn 1 - cosa cos (20 - a)a ( a CI- os a seC2- cos2 which is the equation to an ellipse whose axes are 2a-,/I - cosa, and 2a /1 + cosa, of a and the inclination of the axis-major to the axis of = 9. The equations to the two curves are y2 = a (v - a), and y2 = a (n - a) + Q (rv- 29a) 2; hence at their points of intersection S ((v - 2 a) - ~' = 0, or a= and x -aX2 82 y2 a (x - a) Hence for a given value of (s), every pair of parabolas whose equations are y2 = a (x - a), y2 (a + 8) (x - a - ~) intersect 2 2 - 82 each other in the hyperbola whose equation is y2 and whose axes are ~, 2. As 8 diminishes, the hyperbola y' = approaches to 40 GEOMETRICAL PROBLEMS. NO. VI. DEC. 1835. the two straight lines represented by the equations y2=4 or y = 4-; hence the parabolas ultimately intersect in the two 2'Y3 LIY: straight lines y and y The straight lines manifestly touch all the parabolas represented by the equation = a (;G - a) when every possible value is assigned to a. b" ~~~~~~b 10. y= - 1 2 - \/a2 + b2 a a 2+ b2 a2 b o r "I,_ _=_ _- - - Vd., + b 2 hence CP2=x2+y2 a2 + bV/aI+b2; CD2 = CP2 - a' + b2 = b2 + b Va2 + b2; a2C)2 CY2= PF2 b x,/a2~ba - b2; CD2 hence PY2 = CP2 - CY2 = a2 + b2 or PY = CS. 11. Taking the centre of the hyperbola for the origin of co-ordinates, the equation to the circle whose centre is Hii and radius b is (x + ae)2 + yl b2; and the equation to a tangent at the point x, y is x + ae or yy' + (x + ae) v'= y2~ +x (x + ae) b 2- aev - a 2e2 - a2 - aexv; now if this passes through a point h, k, ky + (x + ae') h = - a' - aex, or ky~+ (h + ae)v = - a" - a (i) which is the cquation 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, k. The equation to the circle described on the transverse axis is x2+y2 = a2; and the equation to a tangent at the point v,1 y, is axx1 + yy, = a2; and in order that this may coincide with equation (1), we must have XI h + ae G h + ae = or - a2 a24eh a ae + eh' and Y.=- or 1 -- a2 ab + aehf a a + eh c2' + (h + ae)2 (a + e h)2 a -C or 2 - (es - 1) 2 + a2 (e - ) = 0; 7h2 k2 hence - - 1, which shews that the point h, k is a point in the hyperbola. 12. Let tangents be drawn from a point whose coordinates measured from the centre are h, k; then the equation to the straight line joining the points of contact is hs + ky a2 b+ let this pass through the focus; therefore when x = ae, y = o; he a hence - = 1, or h -: which shews that the point h, k lies a e in the directrix. 13. Let PK, DK' be two normals at the points P, D whose co-ordinates are x, y and x', y' respectively; then PK2 = y + 2; Dr2 y'2 + b and,' ay, b' a 42 GEOMETRICAL PROBLEMS. NO. VI. DEC. 1835, 2 (1 +_ + - (a2 + b2), a2/ b a a2 " and is constant. 14. Let C (fig. 55) be the centre, PI, P2, -.. P2 the points of division of the circumference; LACP, =;.~. ACP2=0+- zACP,0+-+, &c. n n and SP2=SC2+ CP2-2SC.CPCcos = SC2 + CA2 - SC.CAcosO; sp2= SC' + CA 2 SC.CA cos (0+-; I 7r\ SP32 SC2 + CA2- 2 SC. CA cos ( +; 2(2n-l)7r SP, = SC2+CA2-2SC.CAcos 0-+ - )}; n j o. (SP2 + sP2 +. SP2,) 2n (SC2 + C42) -2SC. CA {c os+os + - +... +cos (0 + ( } {sin (27r -— +-) -sinO) - ) } n(SC2+ CA2) _ SC. CCA sin 2n =2n (SC2 + CA') = n (AS2 + SB') since AB is bisected in C. 15. Let V be the volume of the pyramid; 1,,, S3, S4 the areas of the four triangular faces opposite to the angular points A, B, C, D respectively; then GEOMETRICAL PROBLEMS. NO. VI. DEC. 1835. 43 = - (S, + S3 + S S); v =(S2 + S + S- 8); v= (S, + S3 + 4- S2); V = (s + S2+S 4-3); 3 V = 3 (8 + S + 3 - 4); l f2 3 r4/ 1 1 1 1 2 or - +- + + - - i r2 r3 }X4 1. 44 GEOMETRICAL PROBLEMS. NO. VII. DEC. 1836. ST JOHN'S COLLEGE. DEC. 1836. (No. VII.) 1. IN equal circles, angles whether at the centres or circumferences have the same ratio as the arcs 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 unequal circles, and 2a, 2a' be the angles which the two of each pair make with each other; then sin a R - sin a' R + r 4. A, B, C, D are four points in order in a straight line, find a point E between B and C such that AE.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 p and /' be the angles which this line and the chord joining the points of contact, respectively make with the real axis; then will tan (P. tan p' = 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 0 in the same direction, the loci of the intersections of each ellipse with its original position, are two straight lines whose equations are y = xtan-, and y - cot-. 2 2 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' = b2 + 12 where b and I 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 n times; prove that all the ellipses are similar, and find the sum of the areas of the n ellipses. 12. An ellipse has a square described touching it at the extremities of the minor axis: an ellipse upon the same axismajor circumscribes the square. This ellipse is dealt with in the same manner as before, and the operation is continued till there are altogether n + 1 ellipses; prove that if the original eccentricity = - the last ellipse becomes a circle. V/n+ 1 13. Given the equation Ay2 + Bxy + Cx2 + D = 0 to be the equation to the hyperbola; find the position of the asymptotes, 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 - 2_xy + 3X 2y - 4x - 3 = 0. 46 GEOMETRICAL PROBLEMS. NO. VII. DEC. 1836. SOLUTIONS TO (No. VII.) 1. EUCLID, Prop. 33, Book vi. 2. Euclid, Prop. 11, Book xI. 3. If C, C' (fig. 56) be the centres of the two circles; DD'T, ET'E' common tangents to the circles, meeting CC' in T, T' respectively; join CD, C'D', CE, C'E'; then if L CTD = a, L E'TC' = a', we have R- r., +r sin a sin a = -; CC' ' CC' sin a R - r sin a' R + r 4. Take any point F (fig. 57) not in AB; about the triangle ABF describe a circle ABFG, and about the triangle DCF describe a circle DCFG. Let the circles intersect each other in G; join GF and produce it to meet AD in E; then EA. EB = EF. EG = EC. ED. 5. Let h, k be the co-ordinates of the point of interk section of the two tangents; then tan qp =; and the equation to the line joining the points of contact, when tangents are drawn to the hyperbola from the point h, k, is hx ky =1; a 2 b= hb h b2.'. tan ' = - or tan ' tan -= - t a' k I a2 and if the hyperbola be rectangular, tan (p tan j' = 1. 6. Taking the centre for the origin, the polar equation to the ellipse is and the polar equation to to the ellipse is p2 = es cos2(P; and the polar equation to * - e2 cos2 <p) GEOMETRICAL PROBLEMS. NO. VII. DEC. 1836. 47 the same ellipse when the axis-major is twisted through an angle 0 becomes p2= 1 - ecos(- 2; therefore at the e1 - 2 cos 2 (4p - 0) points of intersection of the ellipses in the two different positions, cos2 (b = cos2 ((P - 0), 0 w- 0.*. - or - i.e. ) e=0-, or + 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 through the origin and inclined at angles - and - + - to the 2 2 2 axis-major; or the equations to the two lines will be 0 0 y = tan - y - o cot-. 2 2 7. Let S, H (fig. 58) be the centres of the two circles whose radii are r, r'; P the centre of a circle touching both circles: then SP- HP = SQ- HR = r - r', and is constant, or the locus of P is a hyperbola whose foci are S, H. If 2 a, 2b be the axes of the hyperbola, 2a = r - r', 2ae = SH = r + r'; r ___+ r_.. e= ---, and 2b = 2a/e l - = 2 /rr. r - r When r = r', 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; X, Y the co-ordinates of P; in PG take PQ = b, then the locus of Q will be a curve which by its revolution round AQ will form the inner surface of the vessel. Y Since the subnormal = 2 a, tan 0 =-; 2at o,. Y=2atan0, X=atan2 0; 48 GEOMETRICAL PROBLEMS. NO. VII. DEC. 1836. hence if x, y be the co-ordinates of Q, = X + b cos 0 = a tan2 0 + b cos 0, y Y - b sin 0 = a tan 0 - b sin0; 4ab.'. y - 4a = - —. + b2 (1 - oS20), cos 0 or b2cos 0 + (y2 -4ax -b2)cos0+ 4ab = 0; (1) and b cos' 0 - (xr + a) cos2 + a = o. (2) Multiply (2) by b and subtract from (1);... b ( + a) cos' 0 + (y2 - 4aa - b2) cos + 3ab = 0. (3) Multiply (2) by 4,b and subtract (1), or 3b2 cos2 0- 4b (x + a) cos 0 - (y2 - 4 a - b2) 0. (4) Multiply (3) by 3b, and (4) by (&x + a) and subtract;.. 3b (y2 4a - b2) + 4b (x + a)2} cos 0 + 9ab2 + (,i + a)(y2 - 4ax - b2) = 0. Multiply (3) by y' - 4ax - b2, and (4) by 3ab and add;.~. b (x + a) (y - 4ax - b2) + 9ab3} cos 0 + (y2 - 4 ax - b2) - 12 ab2 (x + a) = 0. Hence {9ab2 + (x + a) (y2 - ax - b2)}2 {3 (y2-4a-b2) ( +4 a)2 { (y2- 4a.-b2) -12ab2 (r +a) }, which is the equation required. 9. If Aa (fig. 60) be the axis-major, and a' be the semidiameter parallel to PSP', we have PS.SP': SA.Sa:: a': a2; or PS. SP'= 2. a 2. Similarly, QH. HQ' =. b'2; b2 P.. S. P' + QH. HQ' = - (a2 + b'2) = (a2 + b2) = b2 + 12. a2 GEOMETRICAL PROBLEMS. NO. VII. DEC. 1836. 49 10. Let E (fig. 61) be the middle point of the circumference AB; join AE, EB; and let / EAB = a, / EAD = (; then L EDA= z- EBA =a, L.-AED=r - (+a), LECB LEAC +,LAEC = w-;.-. ~ACE=q; sin (Q~ + at) and if AE = a, AP = AC = p, p = a sin (P 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 a2 -b! 2a'2= a-+b1, and sin a=; or Cosa= a2 + b2 a2+ b2 and the tangents at the extremities of these diameters will form a rhombus whose side = 2a'a and whose diagonals are I a a 4 a'sin-, and 4 a' Cos 2 2 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 a1, b be the semiaxes, then b" 2 1C-Acos a b 2 = tan _ — - a 2 1+ CoSa a2' or the ellipse will be similar to the original ellipse; hence if A, A1 be the areas of the two ellipses, a a Ai = '7ralb, = 47ra'2 sin - cos - - 2ra'2 sin a = 27rab = 2A. 2 2 Similarly, if A2, A,,...A, be the areas of the second, third, and n"th ellipses described in the same manner, A, = 2 Al = 22,, A3 =2A=2'A, &cand A, + A, +...~+ A"= (z 2 4- 2' +2'... + 2') A - (2+ I - o) A - (22+1 - 2) wab. 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, AE, BF its diagonals intersecting one another in C, AC =, BC '; then CP2 =CD2 = S2 + ',2 = AB2 4CD2 = 2 (a2 + b2), 3' = z 2^ACB =2 ABEF =2ab; X- m x y2... = '2 a, = /2. b; and 2 + + = ]1 2a~ ab 2b2 will be the equation to an ellipse circumscribing the rhombus, where m may have any magnitude less than unity. Let e' be the eccentricity of this ellipse; then 1 1 2 - e'2 2_a2 + 2b2 e'2 - 2a et /2 - e ' ( 2b2 2a') \abl 2-e'2 2 - e2 e V- /e4 +4m2(1 - e2) 2 2 hence - - 1 is never greater than - 1; or e' is never less e2 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 b1 be the semi-axis minor of the ellipse circumscribing the square, the co-ordinates of P are b, b, be b2 1 1 1 -* ~r ~~ = 1, or - 2 1 or-b2 1 2 a2 bl aSb similarly, if b2, b,...bn be the semi-axes minor of the 2nd, rd... th ellipses; 1 1 1 b^^b/" 'a2' GEOMETRICAL PROBLEMS. NO. VII. DEC. 1836. 51 1 1 1 b,2 b3, a2 9 1 1 1 b,,_ = a_; 1 n~+1 1 1 nn b2 b h2 = a2 9 and if b = a, the nth ellipse becomes a circle, or b 9 b2 a2 b2 1 and = 1- e=. = a2 n +1 n+1 13. Ay + Bxy + C = - D; y B /+Cw AC hence - + f - x 2A 4A2 442 and the equations to the asymptotes are y -B ^ /B2 - 4AC X 2A If a, a' be the inclinations of the two asymptotes to the axis of; B C tan a +ana tan a tana =;- and if x', y' be the co-ordinates of any point referred to the asymptotes as axes, X --, cos a + y cos a, y = sin a + y sin a.' (A sin2 a + B sin a cos a + C cos' a) x'" + (A sin2a' + B sin a' cos a' + C cos2 a') '2 + (2A sin a sin a' + Bsin (a + a') + 2 C cos a cos a') y' + D = 0o D 2 52 GEOMETRICAL PROBLEMS. NO. VII. DEC. 1836. Now the two first terms vanish;.2. 2Ccosacosa' tanatana'+ - (t anatan) +l fy+D = C 2C(-/ J or 2Ccosacosa 2 2AC) -- ' + D= 0. sin (a + a') B cos (a + a') A - C Now, = —; - -; a cos a cos A cos a cos a A 1 XV/(A - C)2 + B2 Cos a cos a A4 A D V /(4 - C)2 + B2 cos a cos a'B 4 B - 44C.- 4aC 14. The curve is an ellipse since 2s < 4. 1. 3. Let a, /3 be the co-ordinates of the centre, and let the origin be transferred to the centre by making aX = + 4- a, y = yy' +3; - (y'+/3)-2(2('+a)(y'+I3)+3('+a)2+2(y'+/) - '+)-=4(;+a)hence 2 3-2a + 2, - 23 + 6a- 4 = 0;.-. a= 2, --- —; and the equation to the ellipse becomes y'2 - 2fly' + $W2 - - = 0. 2 Let 0 be the inclination of the axis of the ellipse to the 2 y2 axis of ix, and let - + = 1 be the equation to the curve referred to its principal diameters; transform the axes through an angle 0, then (x cos 0 + y' sin 0)2 (y' cos 0 - ' sin 0)2 a2 b2= 1 cosO2 sin'O 6 2 + 2 9 a b 9 GEOMETRICAL PROBLEMS. NO. VII. DEC. 1836. 53 sin2O cos20 2 1 4 + - - and sin 20 --- = - a2 9 \2 b21 9 a 2 9) \a 9 1 1 8 1 \ 4 hence - + = -, and cos20 b2 - a2 b 9 \a2 or tan 20 = -1, and 20 = 135; 1 1 4\/2 I 1 8 b2 9a2 = b2 a 9; 2 8-4 42 2 9 9 (2+V/2)..2 - - ----- or a - a2 9 4-2/2 4v and b2 = (2 - /4 hence a = 3 cos - b = 3 sin -. 4 4 54 GEOMETRICAL PROBLEMS. NO. VIII. DEC. 1837. ST JOHN'S COLLEGE. DEC. 1837. (No. VIII.) 1. IF two triangles which have two sides of the one proportional 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 another. 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, ir, which touch first, the three sides of a triangle ABC, and secondly one side BC and the other two produced, touch AB in D1, D, AC in E,.E; shew that BD,. BD2 = CE1. CE, = R r. 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 4 - 2e2: 4 - e2. 9. If one of the co-ordinates of the centre of the curve GEOMETRICAL PROBLEMS. NO. VIII. DEC. 1837. 55 ay2 + bxy + cx2 + dy + ex +-f = 0, assume the form-, shew that the equation becomes bx+d 1 y=- -- /d - 4af, 2a 2a and explain the meaning of it. 10. AP is a parabola, vertex A, focus S; T the point where the axis intersects the directrix; join PT and produce it to meet the latus rectum in N; draw SPQ to meet NQ, which is parallel to ST in Q; and shew that the locus of Q is a circle. 11. If 0 be a point in the directrix of a parabola; and OA = a, OB =b, tangents at A and B; shew that the equation to the parabola referred to OA, OB as axes, assumes the form I/ = a6 b 12. Shew that 1+ =I, 1 is the equation to a a b 4a2 b2 parabola whose latus rectum = — (a? + b2)' 13. If in (11) any tangent to the parabola cut OA, OB in P and Q; shew that OP OQ + =1, OA OB 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 A, 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 - 2arcos 0+r (a-2 -r) where A 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 AC, 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. EUCLID, Prop. 32. Book vi. 2. Euclid, Prop. 20. Book xi. 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 from 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 A ADE = A ABE. Similarly A DEC = A BEC. 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 AA', BB' parallel to DE, EF respectively; these will be the two axes; and with centre B and radius = AC describe a circle cutting AA' in S, H; these will be the two foci required. 5. Let A, B, C (fig. 66) be the three given centres; find 0 the centre of the circle inscribed in the triangle ABC; draw Oa, Ob, Oc perpendicular to the three sides BC, AC, AB respectively; then Ab=Ac, Cb Ca, Ba = B; and the three circles described with centres A, B, C and radii Ac, Bc, Ca, respectively, will touch one another in the points a, b, c. 6. Produce HP, SY (fig. 67) to meet in V; then since L SPY = z YPV, 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; BD2=S-c; CE,=S-c; CE2=S-b;.. BD. BD, = (S - b) (S - c) = CE. CE,; and /(S- a) (S- b) (S- c) S 58 GEOMETRICAL PROBLEMS. NO. VIII. DEC. 1837. = /S. (S b) (S - c)RrH.= (S-b) (S - c). or BD.. BD2 = CE. CE2=Rr. 8. Let APQa (fig. 68) be half the hexagon, Aa being one of its diagonals; then if v, y be the co-ordinates of P measured from the centre C, PQ = 2x, AP= 2cr; and (a - x)2 +y2'= AP2 = 4iv, or 3x2 2+2a - a2 = (1 - e2) (a 2 2'VI); 2 -e2 (4- e')v2 + 2ac = (2 - e ) a'; or x= a 4 e2 4 - 2 e and 2x - a; and the side of the hexagon inscribed in 4 -- e2, the circle on the axis major =a; therefore the side of the hexagon inscribed in the ellipse: the side of the bexagon inscribed in the circle on the axis-major:: 4 - 2ee: 4 - e2. 9. Let a, 0- be the co-ordinates of the centre; transform the origin to that point by making V X + a, y = y+ 3; a(y ~/)2+ - b (v' ~+ a) (y& +3) + c (Jv + a)2 + d (y' + /) + e (x' + a) +f = 0; hence 2a/3+ba + d = O; b + 2ca + e = 0; 2ae - bd 0 a = b and when a assumes the form - 2ae -_bd=o; b2-4ac=0. In this case the equation to the curve becomes (2ay + bv)2 + 2d 2ay + bv) + 4af = 0, yz-Y — I ---6 —la — d 4-4af (1) 2- a 2a which represents the equations to two parallel straight lines; GEOMETRICAL PROBLEM S. NO. VIII. DEC. 1837o 59 and any point in the straight line whose equation is fbv + d' Y= \ 2a I 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 but any point whatever in the straight line y =will 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. 10o Let AM (fig. 69) x, MP = y draw QNA' per pendicular to TS, and let Z SN ' C', N'Q = y'; then y SN 2e MaP y 7 2c a o a -.G 2a f Y if f y 2 y yc y y y 2a V ( a +, a - x 2.z 4 a x all - -, -... y' y y y y 2ay 2a Hlence by multiplication 4r _ ce2 ', 2X y2 t The eqntuation to a circle whose centre is S and radius 2 =a = ST. 11 and 12, See Appendix, 1. Art. 19" 13, Let - e + - 1 be the equation to the parabola; antd i -- + 1, t-he eqLuation to the tangent; then, at the Dont in which the tangueni;t meets t;he cuIrve 60 GEOMETRICAL PROBLEMS. NO. VIII. DEC. 1837. y = b (1 - 2 N/a a) ( m) and since x has only one value, the quadratic equation /b n\ 2b 4- - or - -- v/ + (b - n) = 0 \a m must have its two roots equal; b2 (b n\ b2 bn b nn2 =(b - ( n) - + - - - - -; a a ml a m a O b b n m n or -- +-;.. — + 1. m a nm a b 14. First, let the plane of the paper which represents one of the given planes revolve round the point A (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 z BAB'= z PBP' = 0, since the angular velocities are equal; and if AP' = r, L P'AB' =, the relation between r and 0 will be the polar equation to the curve required. Let AB = c, PB = a, L ABP = a, then P'BA = (P + a, and AP'B = r - {(P + a) + (0 - 0)} = 7r - (0 + a);.'. 2= a2+ r2 + 2car cos(0 +a), which is the equation to a circle. Secondly, suppose the planes to move in opposite directions, then AB will move into the position AB1, and if Z P'AB1 = 0, we have P'AB =- + 0, and P'BA = + a; GEOMETRICAL PROBLEMS. NO. VIII. DEC. 1837. 61.. r in + -sin ( ( + a) = 0, and r cos ( + ) + a cos (/ + a) = c; or r cos (+ + 0) - a cos (p + a) = c- 2a cos (P + a); and by adding the squares of the two last equations r2 - 2ar cos ( - a) + a2 = l - 2a cos (P + a) 2 a 2 + c2 - r\) (a2 _r 2 ==c ---- C ---- 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 APO cut the sphere, the section will be a circle, and AP will be a tangent to this circle;.. AP2= A02 - OP2, 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 A e'= Ab', Cb' = Cb, Db = Dc, Be = Be';.. Dc + cB + Ab' + Cb'= 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. MENTION the principal methods that have been proposed 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 rightangled 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 AB, AC of a given triangle ABC, take two points M, N; and in the line joining them, take a point MB A AT MP P such that AM -. = -P; prove that if PB, PC be joined, the triangle PBC is twice the triangle AMFN. 8. In No. 7 the circle described about the triangle AMN will always pass through a fixed point, 9. Draw the straight lines represented by the equation (2y - X + C) (3y + -tc - 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 = mx + c = c'. 11. The circles represented by the equation (n + l) (2' + y') = a + b uy, when n 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 successively 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 = ~x + 2 /- - 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 ay2 + bxy + cx2 + dy + ex + f= 0; shew what the curve which it represents becomes when by - 4Eac = = C. 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. IX. DEC. 1838. SOLUTIONS TO (No. IX.) 1. SEE Potts' Euclid, p. 50. 2. Euclid, Def. 5, Book v. and Prop. 33, Book vI. 3. Euclid, Prop. 21, Book xi. 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;.. CD: CE: C: CB A:: BD: EA:: BF: AH:: CF: CH; hence the triangle CDF, CEH are similar, and DF is parallel to EH. Similarly, DG is parallel to EK,.-. z FDG = z KEH, and the segments FDG, HEK are similar. 5. Let x be the side of the square abed (fig. 74) inscribed in the triangle ABC, and having one of its sides ed on the hypothenuse AB; then Ca = x sin A, Ba = x secA, or (sin A + sec A) =BC =a; a cos A 1 + sin A cos A Next, let x' 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 x' tan A + x' = a; a cos A sin A + cos A4 Now (1 - sin A) (1 - cos A) is positive;.'. 1 - sin A - cosA + sin A cos A > 0, or 1 + sin A cos A > sin A + cos A; hence x is less than i'; 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 x and mx be the two sides; then (X' (1 + m2) (hypothenuse) = 2 (m + -); 2m 1 n-1 or I+ + m2= m2 -+;.. m = n Wn 2n and if v = 2 na, the sides of the triangle are 2n a, (n' - 1)a; and the hypothenuse = (n 2 + 1)a. Hence if AB (fig. 76) be the base, divide AB into 2n equal parts, and let AD, DB contain n + 1 and n - 1 parts respectively, and BE one part; then if BC be a fourth proportional between BE, BD and AD, BC = (n2 -1) a, and AC = (n'2 + 1) a; or ABC will be the triangle required. 7. Let AM (fig. 77)= a, MB =na;.. AB = (n +l)a; NC = b, AN - nb, AC= (n + 1) b; PN = c, PM = nc, MN = (n + 1) c;.A. ABC = - AB. AC. sin A = (n + 1)2 ab sin A; A AMN = 1 AM. AN sin A = 2n ab sin A, or AABC (n + 1) MN PN. NC 1 and A PNC = N AANM A a M V; MN. AN n. (n + 1) PM. MB n_2 A PMB = M A AMN= - A AMN; NM.MA n + I.. A PBCC AB - A AMN - APN C - PMB (n + 1) 1 1 A = 2A N A AMN A / AM-V. fn n(n + 1) n + a 9/ 8. LetAB= a, AC=4 3; ~.. AM= AN - +1 n 1 + 1n+ and the equation to the circle passing through A, M, N, referred to AB, AC as axes becomes xa + y2 - 2xy cos A = amx + b6y, E 66 GEOMETRICAL PROBLEMS. NO. IX. DEC. 1838. but when y = 0, xA = a=;.'. a, = n+o similarly, when M = 0, y b= b= AN, or b - =-, and the equation to the circle is x2 + y - 2xy cosA a= + -Y n +1 Hence if ax = fy; x2 y2 - 2xy cos A = a = y; from which equations two values of x and y may be found independent of n; therefore all the circles will pass through the two points so determined; hence they have a common chord a!3 / a2 whose equation is ax = /3y; and x = (-)-2; = (BC)2 are the co-ordinates of the common point through which they all pass. 9. The equations to the two straight lines are 2y - x + c = O, and 3y + x -c = 0; and by addition 5y = 0, or y = 0, and x = c; hence the two straight lines intersect in the axis of x 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 x; and 0 the angle between them;.~. m=, m' - and tan0= - -= - 1; 0 = 1350. 1 + 21 m' 1 -- 10. Let x cosa + ysina =p, (1) x cos a + y sin a = p', (2) x cos a + y sin a = p", (3) be the equations to three parallel straight lines, whose perpendicular distances from the origin are p, p', p" respectively; then if the second straight line be equidistant from the first and third, 2P p" p - GEOMETRICAL PROBLEMS. NO. IX. DEC. 1838. 67 and if cotan a = - m, the equations are P P P y= mx + -- y= m = +., sn sin sinna s a or the second equation is Y = + -( p +,P" -. sin a sin a Hence in the proposed example the equation becomes, y '= m + 2 {(c + ') + (C - C') = mx + c. 11. Equating the coefficients of n, we have x2 + y2 =by, and x2 + y2 =ax; hence the straight line whose equation is ax = by will meet all the circles in the same point, or the circles have a common chord whose equation is a = byo 12. Let C be the vertical angle, and z C= z A + z B = r - C;.L. z C - 900, and the locus of C is a circle described on the diameter AB. 13. Draw PV (fig. 77) parallel to AC meeting AB in V; and let AB, AC be taken for the axes of x and y respectively; then if = V = x, VP =y, (n+ l) AM= (n + ) x; AN = 1)y; AB = (n + 1) AM = (n + 1)2x = a; AC = n +1 I a \/- n /A n =.~. +....t- - + =1; a +b n+1 n+ 1 which is the' equation to a parabola referred to two tangents AB, AC. E"2 68 GEOMETRICAL PROBLEMS. NO. IX. DEC. 1838. 14. Let A be the vertex, S the focus, C AS =c= a(1-e);.. 1 +e = 2 -- and y2 = ( - e) (2ax - x) = (1 e) {2a (1-e)v - (1 e) 2} 2 - - 2cx - -ex). C. If a > c, 2 - - is positive; and as a increases the curve a continues to be an ellipse, having the axis of x for its axismajor until a = co, when y2 = 4cw, and the curve approaches to a parabola. If a diminishes until a = c; the equation becomes 2 2c 2 _ y = 2cw - i, or the curve becomes a circle when C moves up to S. C If a < c >, the curve becomes an ellipse having its axismajor in the direction of the axis of x. When a =, y =0, and the ellipse coincides with the axis of x. When a<- /2 = ( - 2) (2 -2cx), and the curve 2 a a becomes a hyperbola. fc 2 When a =0, y2 = (-) x2, or x = 0, and the curve co\ai incides with the axis of y. (This result is not mentioned in the problem.) When a is negative, y = (+ ) (ex2 + 2ccv), which is \a a \j still the equation to a hyperbola. When a - co, y2= 4ex and the curve again becomes a parabola. GEOMETRICAL PROBLEMS. NO. IX. DEC. 1838. 69 2 y2 y2 15. Let + -= 1, and - - 1 be the equations a2 6' a~ b~ to the ellipse and hyperbola; then CS2 = a2- b2 = a'2 +b'2, since the curves have the same foci and centre; and at the points of intersection -( - 41)2 + = 0; ~ a2 ab2 X2 a2a'2 b2 + b' a2a' or - = - b 2 = 2 ye b2b'":a2 - a b2bb" Now if 0, 0' be the angles which the tangents to the two curves at the points of intersection make with the axis of x, b2x h' tan =-,, tan ' =; a2 aZy b2b'2 x2.*. tan 9 tan 0' = -= - 1; a2a a' y'z 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 c; then its equation is y - mx = /b2 + a 2m'; and since it passes through the point h, k, k - mh = /b + asm2. Similarly, if a tangent be drawn from the point h, k to an hyperbola whose semiaxes are a, b', and ml be the tangent of its inclination to the axis of x, k - mih = a'2m2 - b; and when the tangents to the ellipse and hyperbola are at right angles m= - -- or km + h = \/a - bm; and k - h = and k - mh = Vb' + a4m, -; 70 GEOMETRICAL PROBLEMS. NO. IX. DEC. 1838. therefore adding the squares of these two equations, (1 +- m2) (h2 +?) = b2 + a'2 + (a2 - b'2) 2 = (l + m) (b2 + 2); hence h + k2 = b2 + a'. (1) But if x, y be the co-ordinates of the point of intersection of the ellipse and hyperbola,!2 2 al12 2 at Y2 2 y a2' y~ - bb; 9 ai = b b- or b;1 + bJ =b 1; 2 1b'2 bb2 b2<o b~ b2 a a2a hence y2 X= " =; a'. + 2 a + bI~ 2 2 a2'2 + bb',2 a'2 (a2 - b2) + b2 (b12 + a'2) and (2 b'2) + 2/ (a=2 + b'2) ___2 aa 2 + b+2 =(a2 + b2) a+- + b'2 which is the equation to a circle passing through the points of intersection of the ellipse and hyperbola: and from equation (1), h, k are co-ordinates of any point in this circle. 17. y- + 2 V/(2 - x) (5 + ). Let A (fig. 78) be the origin; draw the straight line BCD whose equation is y = + 2 by making AB = s, AC=2; then BCD is a diameter to the curve; and if = 2 or x = -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 symmetrical with respect to ED, having equal portions above and below ED. When = o, y = 2 = V/10; hence if AH = 2 + V/0, and AF = V/l - 2, the curve will pass through the points F, H, and it is manifest that r cannot be greater than 2 nor less than - 5, hence the abscissae 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, z BAD = z ABD= a; also AD = DC;.'. z DAC = DCA = 3; hence zBAC= a +; and 2(a +3) = r, 7r or a +3= —; but the angle AED which the circular section of the cone makes with AB = ACB =/3;.L..ADF = z AD EAD = /3 + a -; 2 or every section perpendicular to the axis will be circular. 19. When b" - 4ac = o, we must have one or more of the quantities a, b, c = co. (1) Let a be infinite, and b, c finite; then y2 = 0, and the equation represents the axis of x. (2) Let b be infinite, and a, c finite; then xy=O;.v. = 0, or y= 0; and the equation represents the axes of y and x. (3) Let c be infinite and a, b, finite; then x?= 0, and the equation represents the axis of y. (4) Let a, b, c be infinite, and d, e, f finite; b c then y2 + -xy +-x = 0, a a or b --- =-( (1) X 2a \2a a which may be finite or infinite, depending upon the values of b c -, and -. a a 72 GEOMETRICAL PROBLEMS. NO. IX. DEC. 1838. If b2 - 4ac be negative, equation (1) can only be satisfied by supposing b2 - 4ac b /c --- 0 or - = 2 V/ - a a a in which case = which is the equation to a straight Iv 2a line. Hence when b2 - 4ac = c co, the equation represents one or two straight lines. 20. Let ay' + bxy + czt + dy + ex + f =; a'y'2 + b''y 2 + c''2 + d'y' + e'' + f' = o, be the equations to two curves similar and similarly situated; then if ' = mx, y' must = my; a'm22y + b'm2~xy + c'mn2X2 + d'my + e'tmx +f' = 0 must agree with the equation ay2 4 bxy + ciz2 + dy + ex +f= 0; a a b b' c c — r=n2-M - -= mM2 ff A f f f' d d' e e M - = M f Y f Jn or a.f ' bf' cf' d2f'2 e2.f' a'f b'f - c 'f dl'f = e'2f2; a b c d2 f e2 f' a' bb' c' d'2 f e' f' which are the conditions required. GEOMETRICAL PROBLEMS. NO. X. DEC. 1839. 73 ST JOHN'S COLLEGE. DEC. 1839. (No. X.) 1. DESCRIBE a rectilineal figure which shall be similar to one, and equal to another given rectilineal figure. 2. Explain and illustrate the fifth and seventh definitions 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. 0 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 E, AD is divided into three equal parts in 0 and E. 4. ABC is an equilateral triangle; AF, BE are drawn perpendicular to the sides BC, AC intersecting each other in D; shew that if FG 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 iv. 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 CEF is to the triangle ABD as 3 + /5: 2, and that the triangle ABD is a mean proportional between CEF and BCD. 6. 0 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, Cc be drawn parallel to them from the angles A, B, C to the opposite sides, then OD OE OF a + Bb Cc = 1. 7. Find the polar co-ordinates of the points of intersection of the straight lines p = 2a sec (- ), p = a sec (~-) 9 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 L 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 I 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 SC and SP, 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, L, 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 = Fv are included in the equation k (y - b) cos 0 - k (,v - a) sin 0 = F I (y - b) sin 0 + k (x - a) cos0}. 14. CP, CD are any semiconjugate diameters of an ellipse; join 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 1+: 2 1w. V/2 15. SY, HiZ are perpendiculars from the foci of an ellipse upon the tangent at any point; find the locus of the point in which HY, SZ intersect each other. GEOMETRICAL PROBLEMS. NO. X. DEC. 1839. 75 16o AB, AG, AD are three edges of a given parallelopiped inclined to each other at angles a, /, y respectively, and to the vertical at angles 0, P, + 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 in a triangle that each touches the other two and two sides of the triangle; prove that the radius of that which touches the sides AB, AC is 1( B CA 1 + tan - 1 + tan - 1 + tanbeing he radius of the circle inscribed in the triangle. T being the radius of the circle inscribed in the triangle. 76 GEOMETRICAL PROBLEMS. NO. X. DEC. 1839. SOLUTIONS TO (No. X.) 1. EUCLID, Prop. 25, Book vi. 2. See Potts' Euclid, p. 162, and Euclid, Prop. 8. Book v. 3. (Fig. 80) AF = DC: and A,4D= AC2- CD'2= 4CD2 - CD2= 3 CD' = 3AF2 = 3AE. AD; AD.AD= 3AE, or AE.= 2 AD also ED = AD-AEAD AD.OE,=and OD= 3 3 4. Since the angles at E and F (fig. -81) are right angles, a circle may be described round CEDE; also since AB, BC are bisected in G and F, FG is parallel to AC, and -BFG-= 600;.-. zAFG = 300 = LDCF; hence FG is a tangent to the circle. 5. Let AB = a; AC=BD=2asin = a = CD; 10 2 5 + 1 also CF=a+Ac= -_ -a; AC. CF = a2; and BC=a-_C=a_=a and CE.CD=BC.CF=(V51 a!; CE =a; GEOMETRICAL PROBLEMS. NO. X. DEC. 1839. hence A ECF = 1. CE. CF. sin ECF, A ABD =!-AB. BD. sin ABIJ, A BCD = BC. CD. sin BCD; AECF CF CF CE AB ~ABD aABDBCD CD BCBC A BC'D' hence (A ECF) (A BCD) = (A ABD)2, and AECF: AABD:: AB: BC:: 3 + /5: 2. OD 6. (Fig. 82) A BOC = a ABC, A a OE OF AsAOC= A ABC, AAOB= AABC; Bb Cc therefore by addition AABC~/ OD OE OF\ C a Bb Cc OD OE 01 or + =. Aa Bb Cc 7. 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 (0 - a') be the equation to another straight line, the angle between them equal the angle between the perpendiculars = a - a'; hence the angle between the two given straight lines = - - 2 63 At the point of intersection 2a sec (- a sec 0 - 6 2co,(0 - ) sin 0 or 0 =-: 2/ and p = asec (o8ZE - = 2aup2 78 GEOMETRICAL PROBLEMS. NO. X. DEC. 1839. 8. Let A (fig. 83) be the origin, AB the axis of x; and the equation to the circle x2 + y2 = ax + by; therefore when = 0, x = AB= a; and if x', y' be the co-ordinates of any point P, tan PAB = —, tan PBA = Y; Xy a - O.. tan APB = - tan (PAB + PBA) y y Y+-. _ Xa a - ay ay a y, x + y2 - ax by' b a x' -'2 ax - hence Z APB = tan-' -, and is independent of the position of P. 9. Generally let QSq (fig. 84) be any chord of a parabola 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 q. 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 L be the latus rectum, PM2 pm2 AM - = = am; L L therefore Pp = Aa; and if a be the angle which Qq makes with the axis, L PQ. - Pp; s pQ. i n a hence area of segment Qpq Pp.PQ sin a GEOMETRICAL, PROBLEMS. NO. X. DEC. 1839. 79 APPpvIL. Pp- vL (Aa)3, and is constant. 11. Let SQ =pl; (fig. 86) zCSQ 0; iPSC= 20, and Sp a (I-e2) 1- e Cos 20' 2 ((~\=a'2e (I1- e2) p(a ej (SP) 1- -ecos20' or p'2 -1 e(cos2O-sin'20)} =a'2e(1 -e 2); +Y y2 -e (av'2 - y2) =a'2e (-e'2), or (1e)_2~ +(1~+e) y'2a 2e~ -e) hence +, which is the equation to an ellipse whose centre is S, semiaxes a Ve (1 + e), and a v/e(1 ~-e)~, and eccentricity \/_e 1 +e 1. Since the hyperbola is rectangular, CP = CD (fig. 87); and if CL, DP be joined to meet in E, L. DEL is a right angle; and E will be the centre of the ellipse which passes through the points C, P, L, D. Let CA4 be the semiaxis of the hyperbola =a; /2 1~ 2 CP = a'.. a sin 6O= a'2 or a" a a DE-;EL =a' and the equation to the ellipse is EL +ED-' a/ where m is a constant to be determined; 2 3a' a a 80 GEOMETRICAL PROBLEMS. NO. X. DEC. 1839. and since the tangent at D makes an angle of 150 with DL, its equation is X 2y or.-=1- a' 4, ([42V/3\ a2 a a 4,7?2 r 4-2fi/3 I2 4-2vl3 \.'. + 1 X — + mX 1 — 4 =1 3'0 (.Y a - ) ( a- ) and since in this equation x must have only one value, the coefficient of x will vanish; (4 - 2/3).. m- -o = 0, a and the equation becomes 4 V2 4y2 (4 - 2 V3) 2a'y = 1. a,- -- + '- + t2 "oY 3 a" a a 13. Similar curves are such that if two lines can be taken for the abscissa, and any two lines equally inclined to them for the ordinates, if the abscissa be taken in a constant ratio, the corresponding ordinates will always be in the same ratio. If y = Fx 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 0; then the equation becomes (y -/3) cos0 - (x- a) sin 0 = F I (y - ) sin 0+ (m - a) cos0; and the equation to a similar curve is found by putting kx and ky for x and y respectively; and if a = ka, /3 = kb, the equation becomes k(y - b)cos0 - k (- a) sin0= F{k (y- b) sin0+ k (- a)cos}O. 14. Let the ellipse be referred to two conjugate diameters x Y CP, CD; (fig. 88) then the equation to PD is - +j = 1; aL GEOMETRICAL PROBLEMS. NO. X. DEC. 1839. 81 and the equation to CP' is - + - 0; hence for the coa b' ordinates of P', y2 yf b' -2 +- or- a'b' sin a ab and APUCP' CP C y sin a Y= -2 2 V/2 2\/2 ab also A PCD = a'b' sina = - 2, trapezium CDPP' ab (1 + - - 15. Draw the normal PG (fig. 89); then SG SP PY SY S'Q GH PH PZ HZ QZ9 therefore Q the point of intersection of SZ, HY is in the normal PG. PQ YP SG QG Also O HZ YZ SH HZ;.o PQ = QG; and PG is bisected in the point Q. Hence if x, y be the co-ordinates of P; Xt y' the coordinates of Q; 2 yt = CG + 2 GMn, 2 ^ - (_ - e2) X =;,-P1M y 2 yP - -9 2 ' y= 2'; 2 2 1 ^+ e2,( 2t }2 4 y2 or {a (1 e) + y - 1; which is the equation to an ellipse whose centre is C, and axes a (1 + e') and b respectively. F 82 GEOMETRICAL PROBLEMS. NO. X. DEC. 1839. 16. Let AP (fig. 90) make angles a', P', y', with the three edges of the parallelopiped AB, AG, AD respectively; then a ' h osaahcosa -cos3 acs a + h + kcosy COS a AP Cos AP f acos/3 + hcosy + k Cos 7 = AP hence a', /3, y' 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, 0 respectively;.'.ab=a, ac =, be y;pa = a',pb =f',pc Oa = O, b =,Oc = #; cos 0 - cos j Cos in A Oac, cosOca= si/i sin 3 sin r COs a -cos/ cosy' and cosacp i sin,.. z Ocp is known; and cos Op = cos +cos y' + sin xfsin -y'cos Qpc; which determines the cosine of the inclination of Ap to the vertical, or the sine of its inclination to the horizon. 17. fLet 'I, 12;, r, be the radii of the three circles which touch the sides terminated at A, B, C respectively (fig. 92) 01, 02, 03 their centres; then AO,, BO,, CO. bisect the angles A, B, C; and if 02 a,, O,a, be drawn perpendicular to BC, we have 0203 2 + r.; alap = V"(2 + 13) @2 - r)2 = 2 V2r,; B C BC rBC C ao=t r 3cot-ot2 ot - + Cror,; 2 2 2 2 2 GEOMETRICAL PROBLEMS. NO. X. DEC. 1839. 88 0- -11 C / ---- I cot C\ Similarly, r, cot - + r3 cot - + 2 r = cot + cot-, 2 2 2 2/ rc cot _ + tr cot+ 2 cot 2 2 2 2 2 A C cot- r - cothence r 2 2r 2 A C C A C cot- + cot- cot - + cot - cot- + cot2 2 2 2 2 B C cot - r cot - 2 2 2 — r 2 2 2 2 + +2 B C B C B C cot - +cot- cot - + cot - cot- +cot - 2 2 2 2 2 2 A AC A C C. A cos - sin - sin - sin- cos- sin - 2 2 2-r 2 22 9 2.'. r, - + V/r3 +r3 —3 B B B Cos - COS - COS - 2 2 2 B C B C C B co - sin sin - sin - cos - sin2 2 2 2 2 2 + 2 V-2-3 + 32 V/r2r - - + Cos - Cos - Cos - 2 2 2 4A A A C. A A. r cos- + 2 /r-r3 sin -os cos+ 3 cot- sin - cos2 2 2 2 2 2 B - B B C B B = r cos2 - + 2 V/rr sin - cos - + r3 cot - sin - cos -; 2 2 2 2 2 2.A / — A A A 2 2 2 2 B -- B B B = r~ cos" - + 2 x//r3 sin- cos - + r3 sin —. or vr, cos - + v3. sin - = / cos - + V/r sin-. F2 84 GEOMETRICAL PROBLEMS. NO. X. DEC. 1839. Similarly, /r3coS- \/rl sinC r cos - + /r, sin -; - /. 4 B - C -f. C. B vr cos — + r3 sin —si =\/r3cos- + /r sin —sin2 2 2 2/ or V/l tcos- - +sin sin /r3cos + sin - -sn A B.C.B /. 2 + B. 27r- B Now cos - + sin -- sin - 2sin- sin + sin 2 2 2 4 4 4. B. r+ A C = 4 sin -.sin ---. cos - 4 4 4 B. 7+ A C.B. +C A - 4 - sin s cos - r =4 sin -.sin.cos4 4 4 4 4 4 C 2 rl 4 rs hence r cot +(. cot-+- 2 1 + tan-/ tan - 4 4 r cot - + cot - l from which by reduction 1 + tan - )(1 + tan - r ( 4) ( 4) rl = - -- 9 1 + tan - 4 and similarly the values of re2, r, are determined. Next let AB, 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 -r - B, r - C, and - A; and if p, be the radius of the circle which touches BD, CE, pl= / 3- a7r - 7r - C\ R1 + tan ) (l + tan- C 2 A 1 - tan - 4 2R1 1 + tan - 1 + tan - ( - tan ) 4/ 4/ 47 86 GEOMETRICAL PROBLEMS. NO. XI. DEC. 1840. ST JOHN'S COLLEGE. DEC. 1840. (No. XI.) 1. IF two triangles have two sides of the one equal to two sides of the other, eaclh 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 triangles have to one another the duplicate ratio of their homologous 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, A.Q 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 0 is a circle. 8. In two given straight lines drawn from a point 0, 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 sec - sec e0 GEOMETRICAL PROBLEMS. NO. XI. DEC. 1840. 87 the focus being pole, and O' being the spiral angle of the point of contact. 10. Determine the position and dimensions of the Conic Section 3y2 -ay + '2 - 2 V3s y + 8aw /3 0, and trace the curve y4 + 2y = a2 (2 -y2) 11. Two cones whose vertical angles are supplementary, are placed with their vertices coincident, and their axes perpendicular; 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 AP, 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. 88 GEOMETRICAL PROBLEMS. NO. XI. DEC. 1840. SOLUTIONS TO (No. XI.) 1. EUCLID, Prop. 24. Book I. 2. Euclid, Def. 10. Book v. and Prop. 19. Book vi. 3. Euclid, Prop. 11. Book. xi. 4. Let A (fig. 93) be the origin, and the diameter AB the axis of x; then the equation to the circle is y2 =2a - ax 2; and if y = mx be the equation AP, at the point P 2a 2am m2cV = 2ax - x2;.. x =, and y =; 1 + m2 1 + m2' hence the equation to BP which passes through the points 2a 2am 2a, 0; and, 2 becomes, 1 + m2 1 + m2 1 Me+ 2a (x - 2 a), 1 + m 2 or y = - - ( - 2 a) 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 DAB ( - A); DBA = ( - B); A+B _r z. ADB= - - - ( - A) + 2 (r - B) - _ 2 4 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 A O, whatever be the lengths of AB, AC. Next let O' be the centre of the circumscribing circle; BC then radius AO'- 1; and if BC and L A be constant, = sin A AO' is constant; and the locus of O' is a circle whose centre BC is A, and radius s 2 sin A GEOMETRICAL PROBLEMS. NO, XI. DEC. 1840. 89 7. PQ (fig. 96) passes through the centre;.-. PQ = 2a, PO=2na; and CO = (1 -2 n) a which is constant; hence the locus of 0 is a circle whose centre is C, and radius = (1 - 12n) a. 8. Let OPQ, OP Q' (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 a? y - +- i a b and the equation to P'Q is cX Y a b hence at their point of intersection \a l cr ( - ` 1 a I Y = 0 1 1 1 1. but X- = Y -=. a a b b' hence the intersection will always be found in the straight line whose equation is x - y = 0, which bisects the angle POQ'. 9. Let r = p sec (0 - /) he the polar equation to the secant; hence p = r cos (0 - 3); and if 0', 0" be the polar angles of the points of intersection of the secant and curve, r'1, r1 the corresponding focal distances, we have p = r'1 Cos (0' - /3), p =r"cos (0"- 3); 2a r" 2 a 1 cos0 I+ cos 0" cos (0'-) os (0" — 3) I+cos0' Ic + os0O" or (1 + cos 0") (cos 0' cos 3 + sin 0' sin /) = (1 + cos 0') (cos 0" cos /3 + sin 0" sin /3); hence (cos 0'- cos 0") cos/3 = - sin 0" -sin 0' + sin (0" - 0') sin /3; 90 GEOMETRICAL PROBLEMS. NO. XI. DEC. 1840. e, fit [" fQ t - fOe\'.*. sin- cos -- + cos --- — tan 2 \ 2 2 0' 0" = 2 cos - cos - tan; 2 2 hence tan p= tan - + tan - * 2 9 and when 0', 0" approach to one another, 0' 0' tan = tan-, or =-; 2 2 0' 0' 0' 0' also p= ~ cos ( - ) = ' cos - = a cos sec2 co sec-; 2 2 2 hence the polar equation to the tangent becomes 0 0 r =a sec- sec (0- -) 2. 2 10. The equation is that of a parabola. Transform the origin to a point a, /3 in the curve by making = + a, y =y' /3;.-. 3 (y' + 3)2 - 2 v/3 (m,' + a) (y' + ~) + (v' + a)2 -8 a (y' + f) + 8a V2/ (' + a) = 0; hence 3y'2 - 2 V\/3 'y + '2 + d'y' + e'' = 0, where, (a, /3) = o; 63 - 2 /3a - 8a = d', 2a - 2 +/s + 8 a/3 = e'. Again, transform the equation to polar co-ordinates by putting ' = p cos 0, y' = sin;.'. (V/3sin 0 - cos 0)'p + (d' sin 0 + e' cos 0) = 0; d' 1 7r and if - = =tan = tan; e ^/3 6 4 sin2 (0 - ) p + e' sec. cos (0 - ) = 0; which is the equation to a parabola whose axis makes an L 30 e sec with the axis of x, and the latus rectum = - 4 GEOMETRICAL PROBLEMS. NO. XI. DEC. 1840. 9 91 6f3 - 2 V3a — 8a 1 Nw2a -2 /3) + 8 a = ' 2. 3 N/3 -a2a V\3; and e' =8a \1-2 (2a /3_)= 4a '3; 4-aV\3 2 latus rectum = -. _ = - 2a; b~a/3)= 3322 Va/3+ a -8a/ + 8a V' 3. a = 0; 8 a (/3 - '\/3 a) =(//3-a)2 12 a, or 2 (/3 -V'3a)=3a; and 6/3 - 2V3a =12 a, 4/3 =9a; and 9a 4 also 2 V'-3a = 2/3- 3a = 9 3 hence aa, / are the co-ordinates of the vertex of the parabola, and the negative sign of the latus rectum shews that it extends indefinitely on the negative side of the origin. (2) To trace the curve y=+~vy a2(2-y) let the equation be transformed to polar co-ordinates by putting 'V = pCos 0, y sin 0; 2 'a"2ct'-) and when 0 = 0, p = co:also p2 sin.2 0 = a 2 COS 20= a2: hence the value of the ordinate = -- a when iv is infinite, or the curve has two asymptotes parallel to the axis of iv at a dlistance a fromt it, 92 GEOMETRICAL PROBLEMS. NO. XI. DEC. 1840. 7" As 0 increases from 0 to -; p diminishes from infinity to 0, and the curve passes through the origin. Since by putting - x for x, or - y for y, the equation to the curve is unaltered, it is symmetrical both with respect to the axis of x 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; 0 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, 2b the minor axis of the section, then 4b2 = BD. CE = 2p sec (O - a) sin a {2p sec (0 + a) sin a} 4p2 sin2 a 4p2 sin2 a o ( )cos ( + a) cos (- a) cos 0 - sin2 a Similarly, if 0' be the angle which AP makes with the axis of the second cone whose vertical angle is 2a'; 4p sin2 a' sin a' - Cos2 0' and when 2a' = r - 2a, and 0' = + 0; 4b'2 - cot2 a; 4b' which is the ratio required. 12. Draw PM (fig. 83) perpendicular to AB; then since the angles AMP, BMP are right angles, the semicircles described 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 AB. 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, ac, be; then ab = ac= 600, be =90~; cos 90 - cos 60. cos 60.'. cos z bac = =- ~ sin 60. sin 60 3 and z bac which measures the inclination between two triangular faces = cos- (- -) = 7r - cos-1 1 aV 2y 2 y2 14. Let - + 2 _ 1, and - + - 1 be the equations a2 b2 a2 b'2 to the two ellipses; then the equation to a tangent at a point a', y' of the first ellipse is IV'V y'y +. (1) Let a pair of tangents be drawn from a point X, Y to the second ellipse, then the equation to the line joining the Xxa Yy points of contact is -,- + -b =1; and in order that this may coincide with equation (1) we must have x X y Y 2 y\2 - = -,- = and 1; as2 a'~ b2 b'2 b1a b (aX 2 (bY 2. ) + = b - 1, which is the equation to an ellipse whose semiaxes are at2 b2 a b 15. Transform the origin to a point a, /3; then the equation to the hyperbola is -x C- ( — 1; l \ a b 94 GEOMETRICAL PROBLEMS. NO. XI. DEC. 1840. and if y = mx be the equation to a chord passing through the origin, when the chord meets the hyperbola (V + a)2 m + =3 \2 a2 b and the values of x in which the chord meets the conjugate hyperbola will be determined from the equation (X+ a) 2 m (m + a32 but when the origin is the middle point of the chord x'= -; /I m 2 /2 2 ma 9m a2 O2 a2 __+2 a2- b2 a b2 and -, -2)- + a - +1 = 0;,d2 6b2 a h2 2 / +2 -a - 2. a2 bS2 )-= and by addition V1 ta2 a) 2 2t hence by eliminating.x, /1 m /a m/2 a2 09 g.( + aa2 =b2 which gives a relation between a and 3, and is the equation required. N2 '2 16. Let the equation to the ellipse be - + 1; and let x, 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 Xx' Yy' a'2 b'2 - + 1; and = 22x, -7=2y; a' b~2 B y GEOMETRICAL PROBLEMS. NO. XI. DEC. 1840. 95 a\2 /2 b' 1 or a'2y2 +J b'2e = 4x~y2, and xy = m2;... a'2y2 + b'2l2 = 4Mm4 which shews that the point x, y lies in an ellipse whose semi2m2 2mi2 conjugate diameters are - and -, or the ellipse is similar bf a a 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 xm, y; but the two equations a'2y2 + b'2x2 = 4m4, and my = m2 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 a1y2 6 2a'b'x y + b'2,2 = 4m4 =- 2a'b'm', or a'y 4. b'x = = mn \/4 m2: 2a'b'; 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. THE rectangle contained by the diagonals of a quadrilateral 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, N, P, Q the bisections of its sides, prove that AC' + BD2 = 2 (MP2 + NQ2). 4. Three circles, whose radii form a geometric progression, 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 r; shew that the locus of the centre of the circle touching these three semi4r 2r circles is an ellipse whose semiaxes are -, and 3 A/3 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 A 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. diameters including an angle a, and B that contained between two semi-diameters at right angles to the first, then (a b\ 2A 2B +- - cot a s cot - + cot -- \b a/ ab ab 10. The straight lines drawn from any point in an equilateral hyperbola to the extremities of any diameter, are inclined 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, b and the curve in p; thenpaccpb. 12. A, B are two fixed points, P, Q any two points in the line AB or AB produced, such that PA QB QB PA m n n m find two other fixed points All,, 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 produced 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. G 98 GEOMETRICAL PROBLEMS. NO. XII, DEC. 1841, SOLUTIONS TO (No. XII.) 1. EUCLID, 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 L AOB = r - (OAB + OBA) r - (-A)r +- r ^B) } - a Similarly if O' be the centre of the circle touching CD and the two sides BC, AD produced, zDOC C + _ 2r - (A + B) _ + B AOB D^O'C = = -~ - =r - AOB; 2 2 2 hence z AOB + z DO'C = r. Also if P, P' be the centres of the circles touching the sides AD, BC respectively, z APD + z BP'C = 7r; and the opposite 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 -; also QM and PN are each parallel to BD, and equal to..; hence MNPQ is a parallelogram whose diagonals are MP, NQ;.. MP2 + NQ = MN' +2 QM 1 AC2 + LBD2;.A. AC2 + BD2 = 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; a b 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. X1I. DEC. 1841. 99 centres are A, C; be 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 abc will touch the three circles with the sides be, 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 r,, 2*2 r. be the radii of the three circles whose centres are A, B, C; LCfe = O, L Aec =; then r2 - r3 sinO0 r3 + 'P, -9, sinA db = I 4- sin 0; r1+r2 4 + 2 3 LABC - LBfc =w- Labe -LAdb; or Labe wrr - ABC=7I - -B. Again z ead ~ L eAd L Adb + LAec = 80+, or Lcab=0+ -z/CAB;. L aeb = vP - Irr - B + 0 + -44 1 A +- B - (O ) ) 'It~~ut sir, -r 3 but sin 0 =I sin r; BC=3Or, AC=5r,, AB=6r,; r, 1+ 3 5 V214 20 4 / 140 or sinA-; sin; sinC= 95 g 15,2 / 4! V4.. 12VI L a= sin'4+sin'*-s11 15 __ 12V" B 14~ L c =sin V- + sin sin -'I sin-' A 15 9 3 5 5. (1) Let 0 (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, CP are IV IV y +-=I;, -+ =1; S-c c a S-a and the co-ordinates of N are a C - (AN), and - (CN), b b a C or (S - 0), and (S - a); b b hence the equation to BN is y c (S - a) X a (S - c)' and for the points of intersection of AM, BN we have a (S - C) c (S - a) S S which are the same as the co-ordinates of the point of intersection of BN, CP; therefore AM, BN, CP meet in the same point. (2) Next let BC be the axis of x; BM =,, MO =y; BC = a; BA + AC = s; then B - C B C cos 1 + tan - tan - b +c sin B + sinC 2 2 2 s a sin (B + C) B+C B C a cos 1 - tan - tan - 2 2 2 B C s-a an B tan tan - and cot 2 2 s + a IV 2 - tC e cx - x2 s - a) =colt - - enceeI a- - IV y Y" s + aj GEOMETRICAL PROBLEMS. NO. XI1. DEC. 1841. 0 I 01 and y2 _ (ar -IV2) a which is the equation to an ellipse having the base for its 8 +t a axis minor, and axis major = \/8_+ a a. 8 - a 6. Let AB (fig. 105) be the diameter of the given semicircle whose centre is 0, and radius r; C a 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, 0, D; draw PM perpendicular to AB, and let AM = v, MP=y; t' y' the co-ordinates of D; then (Solutions of Trigonoinetrical Problems. Ex. 18. No. 15) v r +v r a?j P P P and when the point D is in AB, y'= 0; r v-v' y -=2; p rv-' p and ( - v)2 + y (p ~ ')2,... c/ 2/a? +y' p2 +2p1.'; andL P; P.. a = 2y (22"v' P) 2v(x - p); or X 2 2 y1=2r (qV -); 4 \ 2/ oY2 hence # 21v + _+ +vy = 0; 4 0 r (e- - r)+ t -- \ 3' + 3 g? which is the equation to an ellipse whose semi-axes are and 2r 102 GEOMETRICAL PROBLEMS. NO. XII. DEC. 1841. If P be the centre of the nth circle inscribed between the circles whose centres are C, 0, D; xny its co-ordinates; we have IV r + r' y y y, n^ = =2n2n.; or =2n, since y'= 0; p, r-r pn p pn and (x - r')2 + yn = (p,, + r)2;.. 2r7. + y = pn t+2r' p,; and <2 + - p = 2r' (x + pn) = 2r (xn - p); 2. 2 __]_ ( y \2 (4n2- ) 2 r.'. x2n + 4~ y "- 2rxn + - yn - O; /4n2-1 I 2n 2 4 2.'.i(%-t )" + (~n~ - - 1 (x ^ -2) - t4n 2 4-2 - 4n-1; which is the equation to an ellipse whose semi-axes are 4-n2 / 4?z2 4n2 s 1 9 4n2 - 4 - 2?z and co-ordinates of the centre r, and -- r. 4n2 - 7. The equation to a tangent at the point ', y' of an X'x y'y ellipse is - + 1; and if tangents be drawn from a a b point X, Y without the ellipse, a'X y'Y -2+ -- = 1. Let m be the tangent of the angle which one of the tangents makes with the axis of x; then b2a' b2 = 9 and Y r- z X = a"y Y <' am y am2 y'2 but -- = - -;.-. + ~, a b b (f7 J ' GEOMETRICAL PROBLEMS. NO. XII. DEC. 1841. 103 or 7 2 Mb+ a'm'; hence Y - mnX V62 + a 2m', or m" (2 ( a2) -29XYm + Y2 - b=0; (1) and the product of the two values of m = a _ _ _2 _ - 2 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 2 XY values of m be constant = /3 which is the equation to a hyperbola. 8. Taking the polar equation from the vertex, if LzBAC (fig. 106) = 0; LB'AC = ', AB =p, AD = p'; 2a, aa' the major-axes, and e the eccentricity; then 2a (1 - e2) cos, 2a'0(1 - e') cos 0 P~i 1 - ecos~g 0 -1 e2 COS All p a AB'-a A~p= '- Similarly -4D p a J D' at AB A4B' AB A4D or AD AD" oABAD" hence BB' is parallel to DD'. 9. If ', y, x', y' he the co-ordinates of P, P' in the ellipse (fig. 107); then area PCP' area A CP' - ACP ab Xs( ' IV Co2 - - Cos'-) 104 GEOMETRICAL PROBLEMS. NO, XII. DEC. 1841. ab [I ay ay\ or A =- abtan- tan-; 2 \ b bxbt ay' ay 24 b x' b x a.'. tan - ab 1 +. Y b d7 XI (tan O' - tan 0) 1 + - ) tan 0 tan 0' b where z PCA = 0, Z P'CA = ', and ' - 0 = a; tan O' - tan 0 ' + tanOtanO'- tana, 1 + tan 0 tan 0' 2A b or cot = - cot a ab a 1 2B b Similarly, cot - - cot a ab a 1 () tan 0 tan' b + tan 0 tan ' J a)cot 0 cot 0' b) C + cot 0 cot 0' J b b, c 1t{(-Y+ tan 0 tan 0' a - co t an -- t a 1 + tan 0 tan O' 2A 2B b a\ a b. cot-+ cot = - cot a 1co + 2 = + - cot a. ab ab a \ b a 10. The straight lines drawn from any point of a conic section to the extremities of a diameter are parallel to two conjugate diameters; and in the equilateral hyperbola every pair of conjugate diameters is equally inclined to the asymptotes; hence the straight lines drawn to the extremities of a diameter are equally inclined to the asymptotes. 11. Let the asymptotes (fig. 108) be taken for the axes; a, t3, al, /13 the co-ordinates of P, Q respectively; a', 3' the co-ordinates of p; a, 3", a", /3 the co-ordinates of b, a respectively; then the equation to pba is GEOMETRICAL PROBLEMS. NO. XII. DEC. 1841. 105 y - /3= m (x - ai), and /' -/3: = m (' - a), 3" - 1 = im (a p,=- a-)" = t (a - a); and pb T' -f m(a' -a) andpa —T ' /3' - /3'-3 Now a(3= a2, a't3' a2, from the equation to the hyperbola; a2 a2 a2. 3 - 1 =-7, - = (a' a); a a aa pb aa hence - mpa a2 a a a but /3'- t3 a-= m (a'- ai); a ac a2, pb a m =- -; hence -=a al pa al which is independent of m, or pb cc pa. 12. Let AM (fig. 109) = ma, BN= n/3, PM = mx, QN= ny;... k ( + a) (y + 13) + I (y + 3) - g (x + a) - h = 0, or kxy + (ka + ) y + (k3 - g) +J + kaf3 + I - ga -h = o; and if a, 3 be assumed so that kac + 1 3 -ga - h = O, and ka + I = - (kt3 - g);.'. k= =g-l-ka; hence -ka(ka + ) + I(g - - ka) -hk = o; 2. a2 + 2kla + 12 = gl- hk; or ka + l= - V/gl - hk, and kixy + (ka + l) (y - )= 0; 1 1 k hence - = _ = K. y r ka + I V/gl- hk 106 GEOMETRICAL PROBLEMS. NO. XII. DEC. 1841. Also AM =ma=m -k-(g Vg hk) and AN -= AB - BN is known. 13. Let AB (fig. 110) be the common side of the quadrilateral ABCD; produce AD, BC to meet in E; and AB, DC to meet in G; then EF is the chord of contact of two tangents drawn from G, a point in AB (App. II. Art. 61); but if pairs of tangents be drawn from any point in the straight line AB, the chords of contact will all pass through a fixed point; hence EF pass through a fixed point. GEOMETRICAL PROBLEMS. NO. XIII. DEC. 1842. 107 ST JOHN'S COLLEGE. DEC. 1842. (No. XIII.) 1. WHAT are the objections to Euclid's theory of parallels? State any other theory that you recollect. How did Legendre escape the difficulty by an analytical process. 2. Similar triangles are to one another in the duplicate ratio of their homologous sides. 3. If a straight line be at right angles to a plane, every plane which passes through it is at right angles to that plane. 4. If two circles intersect, the common chord produced bisects the common tangent. 5. If A be the vertex, S the focus, PSp a focal chord in a parabola, the triangle PAp varies as V/Pp. 6. Let the three perpendiculars from the angles of a triangle ABC on the opposite sides meet in P; a circle described so as to pass through P and any two of the points A, B, C is equal to the circumscribing circle of the triangle. 7. APB is a segment of a circle. Take any point P, and let PC be drawn dividing the angle APB in a given ratio; shew that the dividing lines all pass through a point, and find its co-ordinates. 8. The locus of the foci of all ellipses described with their major-axes parallel to the base of an isosceles triangle, and touching its three sides is a circle. 9. Given a circle and two points A, B exterior to it, find a point X in the circle such that if XA, XB be drawn cutting the circle in P, Q; PQ shall be parallel to AB. 10. If r be the radius of the inscribed, r,, rs, 3, r the radii of the escribed spheres of a triangular pyramid, 2 1 1 1 -- - - + -r, r 9' 9'l ',z 9a) 9:4 108 GEOMETRICAL PROBLEMS. NO. XIII. DEC. 1842. 11. If a tangent be drawn to the interior one of two similar concentric ellipses whose axes are in the same straight lines, meeting the exterior one in P, Q, and at P, Q tangents be drawn to this latter intersecting in R, the locus of R is a similar ellipse. 12. CP, CQ are conjugate diameters of an ellipse; on PQ describe an equilateral triangle PQR, and find the locus of R. 13. Find the area included by normals to an hyperbola, which pass through the foci of the conjugate hyperbola. 14. From P, the point of concourse of two tangents to a parabola, PQ, PQ', draw PABC meeting the curve in J, C, and the line QQ' in B; then PC is divided harmonically. 15. Two equal ellipses which have the same centre, have their axes inclined at a given angle, find the angle between the curves where they intersect. 16. A plane is drawn through a tangent to the circular base of a right cone cutting off an oblique cone whose volume 1 is - th of the original cone, if 0 be the inclination of the n cutting plane to the side of the cone, 2a the vertical angle, sin 2 a cot 0 = n - cos 2a. 17. If a hexagon be described about a circle or ellipse, the three lines joining opposite angular points intersect in a point. 18. Give the geometrical construction of the following equations: (1) y2 - 4,y + 5,c2 - oy + 5 = 0. (2) y2- 2y +2 - 2y - 2 - 3 = 0. 19. If a conic section be touched by four straight lines, the locus of its centre is a straight line. GEOMETRICAL PROBLEMS. NO. XLII. DEC. 1842. 0 109 SOLUTIONS TO (No. XIII.) 1. SEE Potts' Euclid, p. 50. 2. Euclid, Prop. 19, Book vI. 3. Euclid, Prop. 18, Book xi. 4. Let PQ (fig. 111) be the common tangent, AB the common chord; produce AB to meet PQ in T; then TP = TB. TA = TQ9, or TP = TQ; hence PQ is bisected in T. 5. Let ASB (fig. 112) be the axis; LzPSB = 0; then 2 a 2a 4 a SP= *5p..Pp 2= 1- cos$' 1 +cos 0 sin'2 0',Q a2 P PO 2PJ and A PAp = AS. Pp sin 0=~ =a2/ V-P. - H,,. r "' -sin 0) a 6. LAPB (fig. 113) =w7r - (z PAB + L PBA) -7r - B+ —A D A) A+B C; therefore the radius of the circle circumscribing APB AB AB 2sin APB 2sin C the radius of the circle circumscribing the triangle ABC. Similarly, the radii of the circles circumscribing the triangles APC, BPC are each of them equal to R. 7. Let L PAB (fig. 83) =; LAPB = (n1)a; LAPC = a;._. LPCB = 0a; and if a be the radius of the circle, AP = 2a sin ABP = 2a sin 10 + (n + 1) a}; hence the co-ordinates of P are 2a sin {0 + (n + 1) al cos.0; 2a shni{0 + (ii ~ 1) a) sinO; 1.10 GEOMETRiICAL PIRODLEM11S, NO. XITII DEC. 1842. and the equation to PC is y 2 9 a sin 1O + (n + 1) a} sin 0 -tan (0 + a) [x - 2 a sin 10 + (n +- 1) a} cos 0]; or ycos(O +a) +2 asin 10+ (n+ I) a~sin a=Sin (0+a) x; hence ycos (0 + a) + 2 a sin a.- cos nza sin (0 + a) + 2 a sin a sin flacos (0 + a) = sin (0 + a),e; which is always satisfied by making x =2a sin acos na; y- -2a sin asiu na; hen-,ce PC always p asses through a fixed point whose coordinates are 2 a sin acos na, 2?a sin a sin fl a. 8. From the vertex A4 (fig. 114) draw AB perpendicular to the base of the triangle; let C be the centre of one of the ellipses which touches the base in B, and one of the sides in P; S its focus; draw EN perpendicular to AB; then if L PA4B a, CN =y, PN = r,, AC= X, CS = Y; w~e have X+ b =AB c; AC - =X; tan L PAB = U=tana; Y2=CS2-~ b2 X a cota a a + - ~= 1, or - (b2+ a'cota) =1 N2 b + a2 cot'a; Y2 = a - b or N2 _ Y2 cot2a =b2 cosec a =(c -XN)2 CoSeCI2; (X2sinl) OSa-Yco_ ac-,N~ hence (N Y)cs' cX+ c2 = 0 or (N _ C secQ a)2 + Y2 = (c seca tana)',,which i's the equation to a circle the co-,ordinates of whose centre are c see' a, 0; and radius e sec a tan a. GEOMETRICAL PROBLEMS. NO. XIII. DEC. 1842. 11 If EF be drawn perpendicular to AE meeting AB in F; the circle described with centre F and radius FE will be the locus of S. BX AX 9. Since PQ (fig. 115) is parallel to AB, QB- From A, B draw AT, BT' tangents to the circle; then AX.AP = T; BX.BQ=BT'; BX2 AX2 or B A,= 2 T;.'. B X A T' * AT. Hence divide AB in D so that AD: DB:: AT: BT':: AX: BX; and AE: EB:: AT: BT':: AX: BX; then the angles AXB, BXP will be bisected by the straight lines XD, XE; therefore z EXD is a right angle, and if a circle be described upon the diameter DE it will intersect the given circle in two points X, X' either of which will satisfy the conditions of the problem. 10. Let 14, A2, A3, A4 be the areas of thle four triangular faces of the pyramid; V its volume; then if C be the centre of the inscribed sphere, the volume of the pyramid is equal to the sum of the volumes of the four pyramids whose common vertex is C and bases A,, A,, A, A, respectively;.-.= - (A, + A, + A, + A,). 3 Also if C, be the centre of the sphere touching A,, and the planes A,, A3, A, produced; the three pyramids whose bases are A,, A3, A and vertex C,, will exceed the original pyramid by the pyramid whose vertex is C, and base A,; or V = (, + A3 + - A,); $ similarly, V = (A, + A, + A, - A); 3 112 GEOMETRICAL PROBLEMS. NO. XIII. DEC. 1842. V = 3 (Al + A2 + A4 - A3); V = _4 (A1 + A2 + A3 - A4); 3 V V V V 2TV... - +- - + _r- - - (Al + AX + A3 + A4) - 7,1 72 r3 724 r 1 1 I 1 2 hence +- + - +-=-. ',1 r2 23 7,4 r 11. Let a, b; ma, mb be the semi-axes of the interior and exterior ellipses respectively; then the equation to a tangent at a point x, y of the interior ellipse is a + = 1... (1). a2 b2 Again, let X, Y be the co-ordinates of R; then if tangents be drawn from the point X, Y to the exterior ellipse, the equation to the chord of contact is x'X y'Y =; -- 1; m2 a m b2 and in order that this may coincide with equation (1) we have X Y X/ \2 [y2 - 2 = 2; and - + + =1; (m m andM2 which is the equation to an ellipse whose semi-axes are m2a, m2b, and therefore similar to the two given ellipses. 12. Let CP= a', CQ = b'; (fig. 116) z CPQ =; z PCA = O; z QCa = 0'; z PCQ = r - a; X, Y the co-ordinates of R referred to the axes of the ellipse as axes; then X = a' cos 0 - PQ cos (60 + - 0); Y= a' sin 0 + PQ sin (60 -- 5 - 0); GEOMETRICAL PROBLEMS. NO, XIII. DEC. 1842. 113 also PQ cos ( = a' + b'cos a; PQsin b = ' sina;.. = a'cos 0 - (a' + b' cos s (60 - 0) + b'sin a sin (60 - ) = {cos 0 - cos(60 -)} -' cos(60 - + a) = a' sin (30 - 0) - b' sin (30 - 0'); Y =a' sin 0 + (a + b' cos a) sin (60 - 0) + b' sin a cos (60 - 0) =' sin 0 + sin (60 - ) } + b' sin (60 - 0 + a) a cos (30 - 0) + b' cos (30 - 0'); now a' cos 0 =,a 'sin = y; a b b' cos0' = -y, b'sin O' - y; 6' sin 0'=-; b a X A /3 a b - a + b/ y 2 2X 26 2a 2 a b) __3 y / 3 a b b + a V/3 (V Y.....+- + = -- - - 2 2 2 b2 2 \a b) 2X \2 12Y 2 (v y\ (ora+bv) +( y) +2 +- )2= + 2 which is the equation to an ellipse whose semi-axes are a + b \3 b + a \/3 v/2 9; o 13. Let S' (fig. 117) be the focus of the conjugate hyperbola; S'PK a normal; x, y the co-ordinates of P; then the equation to the normal S'PK is - 2y and when ax = 0, 62 y' + Y= CS' = l 2/ + b2;. Y=V/a2+ b2 H 114 GEOMETRICAL PROBLEMS. NO. XLII. DEC. 1842. also when y'=O, av,= CK ( + 2) X22 2 a2+ 2b + - 1 + and 2 =1 a'b V (a 2~ b2) (a2+2b2) a and the area included between the four normals = 4 zA CS'K = 2 x'y 2 (a2+ b2) /a 2+ 2 b2 14. Let PQ, P4~ (fig. 118) be taken for the co-ordinate axes; PQ = a, PQ' 4 b; then the equation to the parabola is a b and the equation to QQ' is + - = 1 also let y = mx be the a b equation to PABC, a?', a?2 m"' the co-ordinates of A, B, C respectively; hence X maX ma? a? ma? 1 I) I a* =1, -+ =1; a b a b a b But PA, PB, PC are proportional to a?', a?', a?11 d C s d har c 15. Let CA, CA' (fig. 1 19) be the major-axes of the two ellipses inclined at an angle a; their polar equations referred to the fixed line CA are PlC e' c os' 1 - e2Cos 2(O ) GEOMETRICAL PROBLEMS. NO. XIII, DEC. 1842. hence at the points of intersection cos 0 = cos (0 - a); a w a a 3 v a e2sin 0 cos 0 a but cot TPQ = -- where -; e' sin a ~ 2LTPQ= 2cotV 2 ( - e2 os2 a At the second point of intersection 0 = + - therefore r22 2 the angle between the curves e2 sin a.= 2cot - 16. Let AB (fig. 120) be the slant side of the cone; BC the diameter of the base; BD the axis-major of the elliptic section; draw DE parallel to BC, and AP perpendicular to BD; then the axis-minor of the elliptic section = VCB. DE, z CAB = 2a, z ABD = 0; volume of oblique section of the cone the anvolume of the cone A]. P x area of elliptic section AB cos a x area of circular base 1 AB sin 0 x 7r.BD. e CB s DE or - = n AB cos a x 7ot BC2 sin0 BD / DE sin 0 D /AD H 2 H:2 116 GEOMETRICAL PROBLEMS. NO. XHLI DEC. 1842. henc L =sinO0 COS a sinO0 fl cos a in (2 a +) sin (Qa +O) -~sinO0 ~sin (2 a +O) 2 sin (2 a +O)_ fl3= - sin 0 -Sn2a Cot0+ICOS Qa. 1-7. See Appendix, Art, 75. 18. (1) y - (4x +2) y~+(2x+ I)' +x2 — 4,v+ 4=O; which equation can only be satisfied by making v~- 2=0, and y - 2,v-I=O0;.,. x = 2, y = 5; or the locus is apoint whose co-ordinates are 2, 5. (2) Transform the origin to a point a, /3 in the curve, by making aw = x! + a, y = y'+ /3; (y'+/3)2-2Qv'+a)(y'+3) +(x'+a)'-2(y'+3) -2Gtv/+a) -3 =0; or (y' - x!)2+ (2/3 -2a -2) y' +(2a -2/3- 2)!= 09 and (/3-a)2-2/3-2a-3=0. Let j3-a I= I = tan ~;and transform the equation a -/3-1 to polar co-ordinates; hence a /= -0 p sin' (O -)V cos (O -) 0; which is the equation to a parabola, the co- ordinates of whose vertex are a = /3 = —;the latus rectum =V;and axis inclined to the axis of av at an angle =45. See fig. 12 1. 19. See Appendix, Art. 31. GEOMETRICAL PROBLEMS. NO. XIV. DEC. 1843. 117 ST JOHN'S COLLEGE. DEC. 1843. (No. XIV.) 1. SIMILAR triangles are to one another in the duplicate ratio of their homologous sides. 2. Draw a straight line perpendicular to a plane from a given point without it. 3. Shew that the equation to the ellipse referred to any pair of tangents as axes, is y-k — ~ + cxy c =l, h, k being the portions of the tangents intercepted between the ellipse and their common point of intersection. Find the corresponding equation to the parabola. 4. Prove analytically that if from a point without a circle two straight lines be drawn, one of which cuts the circle and the other touches it, the square of the line that touches the circle is equal to the rectangle contained by the straight line which cuts the circle and the part of it without the circle. 5. The portions of the tangent at any point of an equiangular hyperbola intercepted between the curve and asymptote, are equal to the distance of the same point from the centre. 6. Find the locus of the middle points of all chords of an ellipse, (1) which pass through the extremity of the axismajor; (2) which pass through the focus. 7. Find the diameter of the circle whose equation is + y2 + 2 y cos w = a + by. 8. A straight line of given length slides between a circle and a straight line; find the locus of the middle point. 11 8 GEOMETRICAL PROBLEMS. NO. XIV. DEC. 1843. 9. Two straight lines revolve uniformly in one plane about one extremity; the one moving twice as fast as the other. Find the locus of their points of intersection, supposing them to begin to move together from the straight line joining their fixed extremities. 10. Find a point external to two circles in the same plane that do not meet, such that if straight lines be drawn through it cutting both circles, the portions of all such straight lines intercepted within the circles shall be proportional to their radii. 11. Hence draw a pair of common tangents to two circles; and determine within what limits a point must be situated, so that a straight line may be drawn from it cutting both. 12. Two straight lines include a given angle 2a, and from their point of intersection a straight line of given length is drawn bisecting the angle between them. Determine the locus of the middle points of all straight lines drawn through the extremity of this line to meet the other two. 13. Lines are drawn through the angular points A, B, C of a triangle through any point meeting the opposite sides in a, b, c; a circle is described through these three points cutting the same sides in a', b', c'; shew that Aa', Bb', Cc' meet in one point. Assuming that Ab. B. Ca = Ba. Cb. Ac and conversely, that when this relation holds the lines pass through one point. 14. If the angle between the focal distances of a conic section be constant and = 2a, the locus of the intersection of the tangents at their extremities has for its polar equation p (cos a+ e cos 0) = a (1 - e2). 15. Find the locus of a point from which if perpendiculars be dropped on three given straight lines their points of intersection shall all lie in a straight line. GEOMETRICAL PROBLEMS. NO. XIV. DEC. 1843. 119 16. In an ellipse if two focal distances r and r' include an angle 2a, and T be the intersection of the tangents at their extremities; then (1) the angle between the focal distances is bisected by ST, and (2) ST2 = r sin b - rr sin2 a Hence shew that for the parabola ST2 = rr' always; but in the ellipse only when a = 0. 17. Find the locus of the centres of all circles which cut off from the directions of two sides of a triangle chords equal to two given straight lines. Hence describe a circle that shall cut off from the direction of three sides of a triangle chords respectively equal to three given straight lines. 120 GEOMETRCAL PROBLEMS. NO. XIV. DEC. 1843. SOLUTIONS TO (No. XIV.) I. EUCLID, Prop. 19. Book vi. 2. Euclid,.Prop. 11. Book xr. 3. Hymers' Conic Sections, Art. 228. 4. Let S (fig. 122) be the given point without the circle; C the centre; P any point in the circle; SC= c; CP = a; SP = p; PSC = 0;... a2= p C2 +C 2 - 2cp or p - 2os oOp + C2 - 2 = 0; and if SP be produced to meet the circle in P', the two values of p are SP, SP'; hence SP x SP' = c - a2 and is independent of 0; but when SP is moved until P and P' coincide in T, SP and SP' become equal ST, and ST becomes a tangent; hence ST2 = c - a2; or SPSP' = ST. 5. Let TPt (fig. 123) be a tangent to the hyperbola at P, meeting the asymptotes in T, t; then TP= Pt; but PT= CD; and in the equiangular hyperbola CD' - CP2 = BC2 - AC2 = o;.'. CD= CP; hence TP Pt= CP. 6. (a) Let A (fig. 124) be the extremity of the axismajor; P any point in the ellipse whose centre is C; Q the middle point ofAP; zPAC=0; AP=p; AQ=p'; b2 p2 sin 0 = (2ap cos O - p2 cos2 0), 2 b or p ( - ecos2 0) =-cos 0; and p =2p'; a b2 ' (1 - e2 cos 0) = —cos 0; which is the equation to a similar ellipse whose semi-axes are a b 2' 2 GEOMETRICAL PROBLEMS. NO. XIV. DEC. 1843. 121 (/3) Let PSp (fig. l25) be any chord passing through the focus; Q its middle point; SQ = r, PSC 0; then SP - Sp a (1- e) a (1 - e2) 2 2 1 -e cos 0 1 + e cos 0' a (1 - e2) e cos 0 or r= 1 - e~ cos' 0 which is the equation to an ellipse whose vertex is S, eccentricity e, and axis-major 2ae. 7. The equation to a circle referred to oblique axes inclined at an angle w is ( - a)2 + (y - 3)2 + 2 (,v - a) (y - 3) cos w = r2; and comparing this with the given equation, we have 2 (a + 1 cos w) = a; 2 (3 +- a cos ) = b; a2 + 32 + 2 a/ cos w = r; a - b cosw c b - a cos w 2 sin' w 2 sin2 w aa + bB a2 - 2abcosw + b2 and r2 = = 2 4 sin2 whence the radius, and co-ordinates of the centre are determined. Any line passing through the centre will be a diameter to the circle, and y - / = m (x - a) will be its equation where m is arbitrary. To determine the diameter which passes through the origin, its equation is 3 b - a cos w y =- x, or y x. a a - b cos w 8. Let C (fig. 126) be the centre of the circle; draw CA perpendicular to the given straight line AB; and let AC, 4AB be taken for the co-ordinate axes; suppose PQ any position of 122 GEOMETRICAL PROBLEMS. NO. XIV. DEC. 1843. the line whose length is 1; draw PM perpendicular to AC, and let AM iv, MP = y, AQ = y'; AC = c, CP r; X2~ + (yy)l2, y2+(c-_)2 = -2.; and if X, Y be the co-ordinates of the middle point of PQ, a? __.y '/1 4X2; Y Y- =y2' y+y'=2y; or 2y = 21Y+ V"1 - 4X2 = 2V'lr2 - (C - 2X)', which is the equation required. 9. (i) Let A, B (fig. 127) he the two fixed points; P the intersection of the lines PA, PB in any position; then if L_ PAB = O, and z PBC = 20, we have LAPB =0, and PB = AB; hence the locus of P is a circle whose centre is B. (2) If LzABP = 20, AP = tp, AB= a, then a sinD20 2a cosO0 2 a cosO sin 30 3 - 4 sin2O' OrP COS2 O -_ hec y {( a 2 a2} 4,v-~m - (X2 + y'?) av2a, or y2 -y = 2 Q(: a a which is the equation to a hyperbola whose semi-axes are a a = 3 and eccentricity 2. 3 38 B is the focus of the hyperhola, and A the vertex of the exterior hyperhola. 10. Let PQRS (fig. 128) he a straight line cutting the two circles whose centres are A, B, and intercepting the chords PQ, RS respectively proportional to AP, BR. Produce BA to meet SRQP produced in T; and draw AM, BN perpendicular to PS; then PQ _ RS lip _ RN = - ~ I or -- zIP BR AP BR' GEOMETRICAL PROBLEMS. NO. XIV. DEC. 1843. 123 therefore AP is parallel to BR, and AT AP AT AP BT BR' ' B BR - AP' hence AT is constant; and every line passing through the fixed point T and cutting both circles will intercept chords PQ, RS proportional to their radii. If T' be between the circles, it may be proved in like AT' AP manner, that A- = BR +; 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 P and Q coincide in P', R and S will coincide in R', and TP'R' will be a common tangent; also L AP'T is a right angle; hence if AT be taken a fourth 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 Q', Q" and T'Q', 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 D'T'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, PSp any line passing through S; Q the middle point of Pp; z PAS = a;,Z PSB = 0; SQ = p; then a sin a a sin a sin ( a sin( a) sin (O - CA) sin (0 + Ca) 124 GEOMETRICAL PROBLEMS, NO, XIV. DEC. 1843. SQ~~P=~(SP~SP)=asm s{n (-a) sin (O +a)} a Sin2 a. cos 0 sin2 0 - sin2 a y/ Sa~t + y') = a sina x, or y COS a - x2 sin2 a = a Sif ax; ( a ) -2__ _ which is the equation to a hyperbola whose axes are a, a tan a' and whose Centre is the middle point of AS. 13. Fig. 130. Ab.Ab' =AC. Ac'; Be. Bc'= Ba. Ba'; Ca. Ca'= Cb.- Cb'; (Ab.Bc. Ca.) (Ab'. Bc'. Ca') = (Ac. Ba. Cb.) Ac'. Ba'. Cb'; but A b. Be.Ca = Ac Ba.Cb; A b'.B c'. Ca' = Ac' B a'.C b'; or Aca', 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 T; zLPSC =0; L PSQ =2a;.'.LzPST a; LSPT-=.c/; LTSC=xf; ST=p; ~SP sin (P SP sin + a) -cos a + sin a cot ( and ~e sin 0 a a(1- e') GEOMETRICAL PROBLEMS. NO. XIV. DEC. 1843. 2 125 a (1 - e2) ' (i - aecosO) cos a+ esinOsina-' a ( - e2) a (I - el) or d O os a - e Cos (0 + a) cos a - e os# I Cos2 a - 2eCos a Cos (O~+ a) + e' cos' (O + a) ('3) Also - = I-ecoseO 1 e cos (O +t 2 a) and if SP= r, SQ r'; 2) T ac ( - e" r~ a 0I - e2 1 1 -eecosO+cos(+2a)}~e2{cos2 (O + a)-sin2a) rr la (1 - e') 52 1 1 sin'a (I - e') hence2 T hec =rr' a2 (1 - e2)2 1 sin'a b2r r' o r__ _ P2 2_ _ __ _ _ __ _ 2_ _ _ r r b2 orp b2-rr'sin2a' which shews that in the ellipse p2 cannot = rr' unless a = 0; but in the parabola b is infinite, p2 r' 15. Let the base AB (fig. 132) be taken for the axis of iv; A the origin; x, y the co-ordinates of the point P from which the perpendiculars Pa, Pb, Pc are drawn upon the three sides of the triangle ABC; then the co-ordinates of the points c, b, a respectively are c, 0; (x cosA.4 + y sin A) cosAJ, (x cos A4+ y sin A) sin A; c-{(e-x)cosB+ysinBkcosB; {(c-x v)cosB ysinB sinB; but when three points XI,, y,; I V, Y2; 2,, y3 are in the same straight line, Y2 - _ Y - Y' hec i,bb3 - X I hence if a, b, c be in the same straight line, 126 GEOMETRICAL PROBLEMS. NO. XIV. DEC. 1843. (x cos A +ysin A) sinA4-O 0 (c-x)cosB~ysinB~ sinB-O (vecosA ~ysinA)cosA -x - c- {(c -x)cosB +ysinBi cosB -C or cosA + ysin A = (c -x) cos B + ysin B.ry cosA 4 -x sin A (c -x) sin B- ycos B' wV(c-x)(sinAcosB+ cos.AsinB) -y(c-x)(cosAcosB- sin,4sinll) -y2(sinA cosB + cosAsinB) -xy(cos-4cosB - sinlsinB) '=0; (c (V x2) sin (A +B) cy (cosA +B)-_y2 sin (A +B) =O; or cx - Cy cot (A +B)-_y2 =O hence v'+ y2 _-x - Cy cot C =o; (Xil)+(y - cot<=( cosec which is the equation to the circle circumscribing the triangle. 16. See No. 14, j3 1-7. Let 0 (fig. 133) be the centre and r- the radius of the circle which intercepts from AB, AC the two portions,cc', bb' = 8~ respectively; draw 0OM~ parallel to AC; and let A4M =x,MO y; 9..1-2y sin2A. L 4 j2 Similarly, r2 - 2' sin2 A = 2 2 _2 __ __2 I-V ' 4 sin.2A' or the locus of 0 is an equilateral hyperbola whose centre is A, having two conjugate diameters in the directions AB, AC. Again, let ON be drawn parallel to BC, and let BN =, ON= y' then if aa' =a, 2 a2 X 2 'Y2 Y -a -y sin2 B GEOMETRICAL PROBLEMS. NO. XIV. DEC. 1843. 127 and if two hyperbolas be drawn with centres A, B, whose equations are Y2 - (32 2 - d2,-2 _y2 7_ 2 '2 _ y 2 -a I 4 sin2 A " — 4 sin2 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 produced. 128 GEOMETRICAL PROBLEMS. NO. XV. DEC. 1844. ST JOHN'S COLLEGE. DEC. 1844. (No. XV.) 1. IF 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 given line, so does the extremity of the other. 3. A series of triangles are constructed on a given base, their vertices being in a line parallel to the base; shew that the perpendiculars through the extremties of the base to the sides of these triangles will intersect in a parabola whose latus rectum is the distance between the lines. 4. CP, CD are a pair of semi-conjugate diameters in an ellipse whose foci are S, H (S being the nearer to P) prove the following properties: (a) If the ordinates at P and D be produced to meet the circumscribing circle in Q and E; then QCE is a right angle. (3) The normals at P and D meet in the line through' C perpendicular to PD. (y) The sum of the squares of the perpendiculars from P and D on any fixed diameter is constant. (S) The perpendiculars from P and D on diameters drawn respectively parallel to SD and HP are each equal to the semi-axis minor. 5. Straight lines are drawn from the extremities of a given diameter of a circle whose radius is (a) to the extremities of a chord which always subtends an angle a at the centre; GEOMETRICAL PROBLEMS. NO. XV. DEC. 1844. 129 shew that their intersection traces out a circle whose radius a = a sec -, whether the diameter and chord are joined towards the same or opposite parts. 6. If a circle cut a conic section the chords joining the points of intersection are equally inclined to the axis. 7. If a triangle formed by a pair of tangents to a conic section and the chord of contact be of constant area, the vertex traces out a similar and similarly situated conic section. 8. Two focal chords of a conic section are drawn, and the lines joining their extremities are produced to meet in two points P, Q; shew that SP is a perpendicular to SQ: and if the focal chords be produced to meet PQ, each of them as well as PQ will be harmonically divided. 9. If a series of chords to an ellipse pass through a fixed point, so do the chords of the corresponding conjugate arcs. 10. A pair of tangents are drawn at the points P, Q of a conic section, and another tangent RST meeting them respectively in R and T; shew that for every position of this RP TQ. latter tangent the ratio RS: TS is given; and RT subtends a constant angle at the focus. 11. Find the axes of the curve y2 + x*y + x' = 1, the angle between the co-ordinate axes being 45~. I 130 GEOMETRICAL PROBLEMS. NO. XV. DEC. 1844. SOLUTIONS TO (No. XV.) 1. EUCLID, Prop. 16, Book xI. Let A (fig. 134) be the given point, AB a perpendicular to the plane, AC perpendicular to CD in the plane; join CB, and draw CE parallel to AB; then since AB is perpendicular to the plane BCD, CE is perpendicular to the same plane; therefore CE is perpendicular to CD, or CD is perpendicular to CE and CA, and therefore perpendicular to the plane ACB; hence CD is perpendicular to CB. 2. Let A (fig. 135) be the fixed vertex; BC the straight line along which the extremity of one of the sides moves; draw AB perpendicular to BC; take any point M in BC; join AM, and make L MAP = a, and AP = n (AM); let AB = a, AP=, - BAP =;.A. M = a sec (O - a), and AP = p = na sec (O - a); hence the locus of P is a straight line making an angle 90 + a with AB, whose perpendicular distance from the origin is na. 3. Let AB (fig. 136) be the given base, C the vertex of one of the triangles; draw CM perpendicular to AB; and A a perpendicular to CB, meeting CM in P; then P is the intersection of the perpendiculars Aa, Bb upon BC, AC respectively; let AM = x, MP = y, CM = p, which is constant;.*. y = x tan PAM = M cot B; and CM BM. tan B, or p =(e - ) tan B; /C2 c\2.'. = cX - x2 or p ( - = -, which is the equation to a parabola the co-ordinates of whose v Ca vertex are -, - and latus rectunm H= p. 2 4p GEOMETRICAL PROBLEMS. NO. XV. DEC. 1844. 131 4. (a) Let QPM, EDN (fig. 137) be the ordinates through the points P, D; then QM a PM CN sin QCM= - c -- os ECN; CQ bCQ CE.. ECN + z QCM =-, or ECQ =-. 2 2 (,3) Let x, y; x', y' be the co-ordinates of P and D respectively; then the equation to the normal at P is a1f2Y r- -Y IY (X- x,), b2Cv similarly, the equation to the normal at D is Y= ---X+ 1 -^ '; (2) multiply (1) by y', and (2) by y and subtract; therefore at the point of intersection of the two normals (y - y) Y = - ('- x) X; but the equation to a line through C perpendicular to PD is I - X Y1 - -, --- W1; y -y therefore X, Y are co-ordinates of a point in this line; or the normals at P and D intersect in the perpendicular drawn from C upon PD. (y) Let a diameter be drawn making an angle a with the axis-major, and let PM', DN' be perpendiculars upon it from P, D;.'. PM' - sin a + y cos a, bx ay. DN'= ' cos a + x sina a = cosa - sin a; a b 132 GEOMETRICAL PROBLEMS. NO. XV. DEC, 1844. hence (PM')2 +(DN') (a2 sin a + b2 cos2 a) + a2b =a sin2a + b2cos' a; and is constant for every pair of conjigate diameters. (~) Let CR be drawn parallel to SD; then the perpendicular from P on CR = CP. sin PCR = CP. sin SD T -cP.V' b2 Pb SDk~. HDb P Similarly, the perpendicular from P on the diameter parallel to HD = b. 5. Let AB (fig. 138) be the diameter of the circle; PQ a chord subtending an angle a at the Centre; join AP, BQ meeting in B, and AQ, BP meeting in B'; then '7 ARB APB= LAPBP - BQ -_; 2 2 and is constant; hence the locus of R is a circle whose radius Ar-4 a = a sec -. a ~ ~ s Also LARB~F'Z = zAQR~ z-L PBQ W a —t I? sin - _- + - 2 2 therefore the locus of B' is a circle whose radius = a sec - 2sin(-+ —) 2 6. Let P, Q, B, S be the four points of intersection of the circle and conic section; Y - Mix - C, = y - m2a7 - C2 0; Y - M3 tV - C3 =O; Y - M4V - C4 -; GEOMETRICAL PROBLEMS. NO. XV. DEC. 1844. 133 the equations to PQ, QR, RS, SP respectively; then the equation to the conic section is (y -,m -c) (y - mv - c) c+(y- ma m-c) (y-m, v-c) =0; (1) and the equation to the circle (y-mm - cl) (y - m3o-c3) +X'(y - mxv-c2) (y-m x-c) =0; (2) where X, X' are constants. Now when the conic section is referred to the axis-major, the coefficient of xy vanishes; m'. + m3 + X (m2 +m4) = 0, and in the circle mI + m3 + X' (mn + m,) = 0;.m. 2m + m4 = 0, since X and X' are different constants; hence m4 = - m.; and m +- m., -, or m3 - mn; hence PQ, RS are equally inclined to the axis, and also QR, SP. If equation (2) represents any conic section whose axis is parallel to that represented by equation (1) the coefficient of my will vanish in equation (2); and we shall still have M3a - -- m1, /i4 = - 2; hence if two conic sections whose axes are parallel, intersect one another; the straight lines joining the points of intersection will be equally inclined to the axis. In like manner it may be proved that PS and QR are equally inclined to the axis. 7. Let Qq (fig. 139) be a chord of an ellipse whose centre is C; QT, q T two tangents meeting in T; join CT meeting the ellipse in P and Qq in V; let CV = x, QV =y CP= a'; then a'2 '2 a2 y a a aT, T CT --, V- - IV b' ) i V 134 GEOMETRICAL PROBLEMS. NO, XV. DEC. 1844. and AQTq = QV. TV sin TVQ a a2 y 2 ab f - = - Y ab Mb/\ ~~~~~ ab~( abr; a'2 a hence isconstant n; and if CT p, p -=-; therea ir 'n fore the loCus of T is an ellipse similar to that of P, and its a b semi-axes will be - respectively; where n f m4 (1 -n2)' ' ' a' b2 8. (a) Let RSr, R'Sr' (fig. 14o) be two chords drawn through the focus S of a Conic section; take SR, SR' the axes of x and y respectively; let SR = a, SR'= /; Sr a', Sr' = /': then the equations to r, R' r'R respectively become 1; and 1; /3 a /3 therefore by subtraction (1+ e) rP - + y 0; 1 1 1 1 but + -t + - + 05 a afP P and if rR' r'R intersect in P, the equation to SP is y -. = 0. Again, the equations to RR', rr' are - + - 1, + a/ a/ therefore by addition (+ 1I) + + = 0; aa a ~ GEOMETRICAL PROBLEMS. NO. XV. DEC. 1844. 135 therefore X + y = 0, which is the equation to SQ; hence SP and SQ are at right angles. (3) If iv', y' be the co-ordinates of P, and X", y"' the coordinates of Q, then the equation to PQ is yf -yIf Y - Y', (, -XIv'), bntu y'Iv', y" =M" - iv -x Y - X (X~W - XI let this meet the axis of y in V; Sv = -X ++; SV V' I re 1 ST iv i y y Also by combining the equations to rR', SP we have 1 1 and by combining the equations to rr', SQ we have 1 1 1d 1 1 1 1 2 1 1 1 or XT pi ' p p SR Stherefore the axis of y is harmonically divided. Similarly, the axis of x is harmonically divided. If SR meets QP in U, then UP, UQ, UV are proportional to y', y" and SV; 2 1 1 U V UP UQ therefore PQ is harmonically divided. 136 GEOMETRICAL PROBLEMS. NO. XV. DEC. 1844. 9. Let PP' (fig. 141) be any chord; DY the chord of the conjugate arc; m, y, aI, y' the co-ordinates of P, P' respectively when the ellipse is referred to its principal diameters as axes; then the equation to PP' is Y -y = YY - X) V r( - V) a -ta and the co-ordinates of D, D' are - ay bx -ay' ba/ _b -a hb' a therefore the equation to DY is bix a a 9Y - -- ~ ~ Y fay' ay\Y b b) oil Yn =I) a a2m b b2 b my.ml+ -X=_(-( MX-y); a a and if PP' passes through a point a, y = m (a - v) m~x - y =ma -s b2 b or mY+-X=-(ma a a which is satisfied independently of m, by making b a Y = -al, X = — f; a b therefore DD' always passes through a fixed point whose coordinates are f, ~ b a GEOMETRICAL PROBLEMS. NO. XV. DEC. 1844. 137 10. Let a, a' a" be the semi-diameters parallel to PR, RT, (fig. 142) TQ respectively; then Rp2 a2 TQ2 a"2 RS2 a'2 a TS2 a' RP TQ RS:: a: a and is constant for all positions of RT. Also, if H be one of the foci, and RH, SH, TH be joined, we have z RHS - z PHS, z SHT = z SHQ; therefore by addition z RHT -- Z PHQ, and is constant. 11. Transform the equation to polar co-ordinates by making sin (45 - 0) sin 0 sin45 Psin 445 *.. p2?(cos - sin 0)2 + a/2 (cos 0- sin 0) sin 0 + 2 sin' 0 = 1; or p{1 -sin20+ sin 20 - (1 - cos 20) + or 2 I 1 - sin 20 + -)-) 2 + 1 - cos20} =1; { 2/2 - ( /2) sin o- 1- cos20 =1; I /2 9 /- - I hence p 2 2/ -(sin 20 + cos 0)}= + /2. 3 { + /2- /2 cos (2 0 - 45) 2 + /2; 3 2- 8 ) w\- 7 or p +[1+ -- {2cos 0 ) -]=v2+1; V2 v 138 GEOMETRICAL PROBLEMS. NO. XV. DEC. 1844. hence p2(3 + 2\1,) -2V2 Cos2(0_ )}2~V 2; p21 - V2 /(3 - 2'\,/2) Cos'( 0 i} (2 - V which is the equation to an ellipse whose eccentricity =\/2Q V-2 (3 - 2 V/2) \/2' (2 -\/Q semi-axis-minor = v'2 - V 2 and semi-axis-major \ /2- V2 2 (2 (3 C2,\2 9-V The axis-major, is inclined at an angle - to the axis of x, and 8 bisects the angle between the co-ordinate axes. GEOMETRICAL PROBLEMS. NO. XVI. DEC. 1845. 189 ST JOHN'S COLLEGE. DEC. 1845. (No. XVI.) 1. IF two triangles have two sides of the one equal to two sides of the other each to each; but the angle contained by the 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 similar rectilineal figures, and shew that parallelograms about the diameter of any parallelogram are similar to the whole and to one another. 3. Explain briefly the advantages gained by the application of analysis to the solution of geometrical problems. If a line of given length in a given direction is represented by a, shew how a line of the same length, inclined at an angle 0 to the former may properly be represented, and apply the method to express the cosine of an angle of a triangle in terms of the sides. 4. Through any point of a chord of a circle other chords are drawn; shew that lines from the middle point of the first chord to the middle points of the others, will meet them all at the same angle. 5. A tangent at any point P of an ellipse meets the majoraxis produced in T, and perpendiculars upon it from the centre and focus in Y, Z; shew that TY has to PY the duplicate ratio of TZ to PZ. 6. The diameters of circles described about and within a semi-ellipse bounded by the minor axis are D, d; shew that D is a third proportional to AC and AB, and d a fourth proportional to AC, BC, and SH. 7. An ellipse and a pair of conjugate hyperbolas are described upon the same principal axes, and at the points 140 GEOMETRICAL PROBLEMS. NO. XVI. DEC. 1845. where any line through the centre meets the ellipse and one of the hyperbolas tangents are applied; find the locus of their intersection, and determine all its points of contact with the ellipse. 8. ACP, A'CP' are two diameters of a circle; tangents at P, P' meet the diameters produced in B, B', and two parabolas are described touching them in A and B, A' and B' respectively. Shew that the quadrilateral formed by the two diameters and the axes of the parabolas may be inscribed in a circle. Find also the smallest possible value of the diameter of this circle, and the condition that it shall equal that of the original circle. 9. Trace the curve whose equation is 3V2 -_ IOy - 8y2 + 4v - 2y + 1 = 0. 10. Two cones having their vertices coincident, their axes at right angles, and their surfaces in contact, are cut by a plane parallel to both axes: find the condition that the section shall be a pair of conjugate hyperbolas. 11. A straight line is drawn through a fixed point 0, meeting a curve in the points Pi P2... &c.; and in this line a point Q is taken, such that OQ-~" = OP-n". Determine the nature of the original curve, that, when n is any positive number, the locus of Q may be a similar curve. 12. The double tangents of the curve whose equation is;;+7 + (4+ 4+ y (+ ) + 22 y 1 - a b — 2' (I + 2) - 2y1 + + )+ 2 = 0, touch it in points, all of which lie on a circle whose centre is the origin of co-ordinates. 13. Parabolas are described touching two lines at right angles to each other; shew that if the chords through the GEOMETRICAL PROBLEMS. NO. XVI. DEC. 1845. 141 points of contact are parallel to one another, the locus of the vertices of the parabolas is a straight line. 14. P = o, Q = 0, R = 0, are the equations to three curves, all of which pass through the same point. Shew that of the curves whose equations are P = a. Q. R, P = b.Q.R, the former will in general touch the curve whose equation is P= o, at that point, but the latter will not, a and b being constants. Exemplify this when P=y-, Q=y- m, R=y + mx. Why must the words in general be used? 15. With the asymptotes of a hyperbola as conjugate diameters ellipses are described touching the hyperbola. From any point of the hyperbola a pair of tangents is drawn to one of the ellipses; shew that tangents applied at the two points where the chord through the points of contact meets any other of the ellipses, will intersect in the same hyperbola. 142 GEOMETRICAL PROBLEMS. NO. XVI. DEC. 184s5 SOLUTIONS TO (No. XVI.) 1. EUCLID, Prop. 24, Book i. 2. Euclid, Def. 1, Book VI. and Prop. 24, Book vi. 3. See Peacock's Algebra, Vol. II. Arts. 823-831. 4. Let A (fig. 143) be a point in the first chord whose middle point is B: D the middle point of any other chord passing through A; C the centre of the circle; join CD, CB: then angles CDA, CBA are right angles, because D, B are the middle points of the chords; therefore a circle described on the diameter AC will pass through B and D; and L ADB = L ACB is constant for all positions of AD. 5. Let C be the centre, S the focus (fig. 144); draw the normal PK meeting CST in K: then a2 TY: YP: TC: CK:-: e2x:: a2: e:v2 a2 and TZ: PZ:: TS: SK:: - ae ae - ex a:: a: exr;.. TY: YP: TZ2: PZ2. 6. Let 0 (fig. 145) be the centre of the circle described about the semi-ellipse, then OB = OA: let OC = w;.'. /b2? + 1 = (a -,), or b2= a2 - 2ax; a- b2 a 2+ b2 *.. = -, and AO a -: 2a9 2a a~ + b2 AB' or D0 - a AC GEOMETRICAL PROBLEMS. NO, XVI. DEC. 1845. 143 Again, let o be the centre of the circle inscribed in the semi-ellipse; draw the normal o P and PM perpendicular to AC, then oP = o C: let CM = ', MP y'; b2.. oM= - (1 - e) x', and Co =e '; a2 hence e4x2 = (1 - e2)2' + y'2 = (1 - e2)2x2 + (1 - e2) (a2 - ') - b2- e2'x2 + e4xV2; b.. '=-, and Co = e2' = be; e b. (2ae) BC. SH or d= 2be = a 4AC 7. Let y = mxn be the equation to any line CPQ (fig. 146) drawn through the centre meeting the ellipse in P and the hyperbola in Q; then the equation to the ellipse is 2 y2 +_- 1; a2 &, Wx yy' and the equation to the tangent at P is -- + - 1, abm a2 a- - IV' =+ -- b \/b= + 2 ~A2 (1) also the equation to the hyperbola is 11 2 x2 y2 - - = I, a 2 b and the equation to the tangent at Q is x y 1 m2 1 -x' Y: 1, where xc 1 a b2 a2 =b2; a~ _ -- 144 GEOMETRICAL PROBLEMS. NO. XVI, DEC, 1845. *... a m y, =: -* = - - or - o - /a 2 -, (2) b2 iV b 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" ( 2 + 2 ~-, am- y 2) = b2 + a22; o + " b 4: + =b m+; ab2 ( 2 a m a4 Y y ) b2 n; a 2 x'Y W + 2 4b2 2 therefore by addition, 2b2 a4, a4, a4m,2 a ( f+ b4 Y) =b2, or + =2 a; by subtraction, 4m' y' = 2a m2, or a2m = 2x'y';.'. (1)+ b4 ) a-. (3) Again, using the negative sign in equation (2) b j2 ( t y +af 2) = b2 + a2m2; 2 a2 m\ IV ^r y + 2 a2m2 -b2; a V,'+ therefore by addition, a42 a4m2 a4m 2 2IV + ( T Y 2a m,2 o a m2+ 4 by subtraction, 4 'y' 2 b2; 1 a a7b2 a) b s4.,~ +4 s=, 2,/2, or y2 1+ — =; (4) 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 CPQ meets the hyperbola and its conjugate respectively. If -(1 + 4 ) = 1; when the curve meets the ellipse (i y1) C-^ ) ' 1 5 01 y ( 1 ) 2=0; (5) a/2 -l + = 1, or - = o; (5) y'=O, or y'=/ and the corresponding values of v' are a ' == a, and x' = = —_; /2 similarly, the curve y 1 + - ) = 1 meets the ellipse when x'= 0, and x' = = - 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 r' = 0, ax = a, and I =1 a. V/2 8. Draw AQ, A'Q perpendicular to AC, A'C respectively (fig. 147); join A'B', 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 A'C 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, AQ is the direction of the axis of the parabola which touches AC, CB: and since the angles CAQ, CX'Q are right angles, the circle described on the diameter CQ passes K 146 GEOMETRICAL PROBLEMS. NO. XVI. DEC. 1845. through the points A, A', and the quadrilateral CAQA' may be inscribed in a circle. a Also, if L PCP' = a, CQ = AC sec - the least value of which is AC: and if CQ = 2AC, sec - = 2, 2 a.. cos- =, and a = 120. 2 2' 9. (10 y - 4> 8Y2 + y- 1 '. - - (Y- 4) (y - 2)2 49 y2 - -1y + 3 I 3 9 B 5y 2 7y -1 or xi- - ---- 3 3.3. S =.y - 2 (7y - 1) = 12y- 3, or - 2y -; therefore the equation is reduced to two straight lines whose equations are 4y - = 1, and 2y+3x + 1 = 0. 10. Let the two axes of the cones which are at right angles be taken for the axes of x 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 Ax, 7r - -a will be the semi-vertical angle of the cone whose axis 2 is Ay: and if x, y, z be co-ordinates of any point in the first surface, V', y', A' co-ordinates of any point in the second surface 2 2 2 + z = X tan2 a or - - = 1; (1) '~+z'2 — y2" cot2 a o IV'2 Y/2 '2 + 2 = y, ~= - tan2 a. (2) Xor cot' a ' GEOMETRICAL PROBLEMS. NO. XVI. DEC. 1845. 1147 Now by making z' = z constant, equations (i) and (2) are the equations of the sections to a cone made by a plane parallel to both axes, at a distance (z) from their plane: and if a they manifestly become I2 Y -— =1 2 2 z I2 2 z2 2 the equations to two conjugate hyperbolas. 11. If the equation to the curve be referred to polar co-ordinates it will assume the form Pmf2 (O) + pnmzIfnI (O) +... ~p2f2 0 + pflO f +o= 0. When p is the radius vector, and fm(0), f(,;1-,)0... 110 homogeneous functions of sin 0 and cos 0 of mn, (m - i) &c. dimensions respectively, and fo a constant; 1 f () 1 f1Q9) 1 fr (0) + 4+ ----- "~...+ = 0; pi fo P f ptm2 fo.f (0) is a homogeneous function of sin 0 and cos 0 of a dimension; 2 f ((0) I 2f2 0 160 fO is a homogeneous function of sin 0 and cos 0 of 2 dimensions; ~-3 f 0p-2 _ - 3f3(0) 4b fo ~~~~f a homogeneous function of sin 0 and cos 0 of 3 dimensions. Similarly, Yp-'t = F, (0) a homogeneous function of sin 0 and cos 0 of n dimensions; therefore the equation to the locus of Q is -= a2F? (0) = a-" (cos0) F(tan 0), K2 148 GEOMETRICAL PROBLEMS. NO. XVI. DEC. 1845. where F (tan 0) is a rational and integral function of tan 0 of n dimensions; and in order that this curve may be similar to the original curve, the original curve must have for its equation ) cos" (0 - a) F tan (0 -a). p 12. The equation may be put under the form ( 2 2 J -2 _ 2 ( 2 2) + _ 0, or ( - 1 +Y2 ( - 2 — 2 (- +- +1 = 0 o2 a12 \a 2 + \a b2 or -+ + +- 1 =0. Now ( 2)+ 2-1 ( Y +1= - a2 (y b )2} {a2 (Y 2)2 \a b a b b a b 2+y2 \2 y \ y \ /x y \ Ix y \ \ c \a b r b a b \a b Ia b therefore the four straight lines e y XI Y + -+t 1 = - + — 1 = 0, a b a b so y o Y - — +1=0, - - 1=0 a b a b are double tangents, and meet the curve in the circle whose equation is m2 + y2 = C2. 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 PSp (fig. 148) a chord passing through the focus; A 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 PTA = 0, AM=x, MP = y; then tan 0 = Y = X. = m cot2, y = 2mcot0, SPQ zSTP = 0; and the perpendicular from 4 on PQ COS2 0 a = AT sinO= i sin 0 = m sin 0 Similarly, the perpendicular from A on Qp sin' 0 cos 0 hence M = tan30 = cot3 QpP. a Now a, / are the co-ordinates of A referred to the fixed axes Qp, QP; and if L QpP be constant, the locus of A is a straight line whose equation is 3 = cot3 QpP. a. 14. Let u = y -/3 - m (x- a) = 0 be the equation to a straight line passing through a point a, 3; and P = 0 the equation to a curve passing through the same point; then if /3 t- m (x - a) be substituted for y in P = 0, the resulting equation will be of the form (x - a) P' = 0, since one value of x is a; P' will be a function of x and m; and by putting s = a in the equation P' = 0, an equation will be determined in terms of n; and if the values of m found from this equation make P' a multiple of (sa - a), P = becomes of the form (x - a) P" = o, and two values of x 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 /3 + m (x - a) of the form (x - a)' P' = 0 for every value of m, which will always be the case if P be a rational function of x - a, and y - f3 of which no term is of less than two dimensions, u = o will not be a tangent except for those values of m which make P' a multiple of xa- a, since two values of x will only be made to coincide on this supposition, the equation (v - a)2P'= 0 indicating that two branches of the curve intersect when x = a; hence in this case in order that u = o may be a tangent to P = o we must have P = (x - a)3 P"= 0. Hence when u =0 is a tangent to P= 0, if u = 0 be substituted in P - a QR = 0, the resulting equation will be (av - a) P' - a (x - a) Q'. (m - a) R' = 0; or (v - a)2 (P' - a Q'R') = 0; or u = will in general be a tangent to P - aQR = 0, as well as P = o; or the curves will touch one another. If u = o when substituted in P - aQR = o gives the result ( -a ) (P' -a Q'R') = o for every value of m; the value of m which makes u = 0 a tangent to P= o, will not make u = o a tangent to P - a QR 0o, 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 u = 0 to the tangent to P = 0 when combined with P - a QR = 0 gives a result (xo - a)3R" = 0. If u = o be substituted in P2 - b QR = 0, every value of m will give the result (v - a) (P'2 - b Q'R') = 0; and the value of m, which makes u =o a tangent to P = 0, will not make u = 0 a tangent to P- b QR = 0, unless P'2 - bQ'R' be a multiple of x - a; or unless u = 0 when combined with p2 - bQR = 0 gives the result (= - ) 3R" = 0. Hence P = 0 does not touch the curve P' - b QR = 0. GEOMETRICAL PROBLEMS. NO. XVI. DEC. 1845. 151 (P) If y2 - m2-2 - a(y - a)) = 0; when y - =, we have (1 - m2) x2 = 0; or two values of x become =; and y - x = 0 touches the curve y2 - mo2 - a (y - x) = 0. If y2 - m2x2 - b (y - x)2 = 0, this represents the equation to two straight lines passing through the origin; for (n\ 2 / 2 ( - 2 - b - 0 (Xt) (.X) or (1 -b) () b -) + 1 -m2 =; therefore!- has two constant values; and neither of the lines will touch or coincide with y - = 0 unless the value of one of the constants is unity. 15. Let a', b' 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 x', y'; TPt the common tangent, C the centre; then because TPt is a tangent to the ellipse a2 b'2 CT =; Ct = -; x also because TPt is a tangent to the hyperbola, a2 b12 CT - 2' Ct=2y';.. =; -' 2y xo y or a'2 = 22/2; b'2 2y'2, and a'2b'2 -= 4ax' y' = 4m4;.~ a'6' = 2Q2; similarly, if a", b" be the semi-conjugate diameters of any other ellipse, a"b" = 2m2. Let tangents be drawn from a point h, k in the hyperbola to touch the ellipse whose equation is x y2 2 =1; a t B 152 GEOMETRICAL PROBLEMS. NO. XVI. DEC. 1845. then the equation to the straight line joining the points of contact is hx, ky — +-=1; (1) and if tangents be drawn from a point c,, y, to touch an ellipse whose equation is,2 ya2 a + b,-= 1 a"2 b the equation to the straight line joining the points of contact is xIox yly ae + b =21; and in order that equations (1) and (2) may be the equations to the same straight line, h x1 k y, x,yI hk x y1 m2 a 2 a":2 b ' b"2 * * a2"2 = a' b' o 2m2 =- 2; therefore x2,y = m; which shews that x, y1 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. MAGNITUDES 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 + b, a + c, 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. In the circle, of which AB is the diameter, take any point P; and draw PC, PD on opposite sides of AP and equally inclined to it, meeting AB in C, D. Prove that AC: 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 0. Prove that OP = OQ. 7. The circle described from an extremity of the minoraxis 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 1 /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, b) of two tangents to a parabola at right angles, are connected by the equation ca b3 1 b a- (s latus rectum)3 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 S on PQ, PR, QR: the feet of these perpendiculars are in the same straight line. 12. Two 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. Two tangents are drawn to a parabola making angles 0, 0' with the axis. Prove that (1) if sin 0. sin 0' be constant the locus of the intersections of the tangents is a circle whose centre is in the focus; (2) if tan 0. tan 0' be constant the locus is a straight line perpendicular to the axis; (3) if cot0 + cot0' 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 similarly situated, take two points P, Q whose abscissas are as their major-axes; and P', Q' two other such points. If '04'~' be the angles which the tangents at PP'QQ' make tan 0 talln with the axes; prove that ta =tan P tan 0' tan (p' If the curves be likewise confocal, prove that PQ' = P'Q. 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 hyperbola, whose asymptotes have the same direction as their axes, in QQ'Q"...: prove that the extremities of the diameters respectively 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. EUCLID, Prop. 15, Book v. Potts' Euclid, Def. 5, Book v. and note to the definition, p. 162. 2. Euclid, Prop. 9, Book xi. 3. (1) Let ABCD (fig. 150) be a quadrilateral figure touching a circle in the points a, b, c, d; then Aa = Ad, Ba = Bb, Cc = Cb, Dc= Dd;.. Aa + Ba + Cc + Dc=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 + c; 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 E, then ABCD will be a parallelogram; from E draw EF perpendicular to AB, and produce FE to meet CD in G; then z EGD = z EFA= a right angle; and AB CD FB= =-= DG; 2 2 hence FG 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 = b, or the parallelogram is a rhombus. GEOMETRICAL PROBLEMS. NO. XVII. DEC. 1846. 157 4. Let A, B (fig. 152) be the given points; draw AM, 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 AP, meeting MN in Q; and draw BQ; then tan z AQM tan L PQN PN =__ _ _ _ = _ _ _ _ _ _= =n. tan z BQN tan z BQN BN tan z AQM _ AM If tan z BQN BN teAM PN then BN-B or P= AM; hence in this case make PN = AMJ. 5. Since angle CPD (fig. 153) is bisected by PA, CA: AD:: CP: PD; and since angle APB is a right angle, PB bisects the exterior angle of the triangle CPD;. CP: PD:: CB: BD; hence CA: AD:: CB: BD, or AC: BC:: AD: BD. 6. Let A (fig. 154) be the common vertex, CD, C'D' the semi-diameters of the two ellipses parallel to PQ; CB, C'B' the semi-diameters parallel to AO; then since the ellipses are similar CD C'D' PO CD C'D' QO CB - CB AO B t C'B';.. PO= QO. 7. The equations to the circle and ellipse are a2 X2 9y + (y + b)- e, and - - 1; e:! a2 b2 hence at the points in which the circle meets the ellipse, a2 b2 a2 a2 ~ y +y +2by+ - b2,, b" e 158 GEOMETRICAL PROBLEMS. NO. XVII. DEC. 1846". or 2 ab y-as 1 - - e2 - b and since has onl one value, the ewill in e ] eby + b2 = 0; - e2 - eay b2 b3 b ea e a" and since y has only one value, the ellipse and circle will in general touch one another in the two points in which e a If ea < b, or e < V/1 - e2, and therefore e < the values \/2 meet the ellipse. If ea = b, or e; the two points coincide in the extremity of the axis-minor, and there is only one point of contact. 2 yi2 x'2 y'2 8. Let -- b = 1, - -- =1 be the a2 b26 a1" b2,v' x y' y two hyperbolas, and let - -; then b- =-; a V' b be the co-ordinates of the middle point of the the points V, y, x', y'; we have equations to and if X, Y chord joining v -F aL' or X= 1+ 2aX 2bY 2aX 2b Y.'a a --- + by a +a" b5+5' y 1 +y and - - = I; a b6 GEOMETRICAL PROBLEMS. NO. XVII. DEC. 1846, 159 {f(a+ }' - {h(b+ b')} 1+ which is the equation to a hyperbola whose axes are a + a', b + b'; which are arithmetic means between 2a, 2a' and 2b, 2 b' respectively. 9. Let PSp (fig. 148) be any focal chord making an angle 0 with the axis; PQ, Qp two tangents intersecting at right angles; Qp = a, QP b; latus rectum 4m; angle PSMV = 0; 0 m zSPQ=-, and SP= sin - 2 m m Cos' - sin' -Cos' - 2 2 2 hence a= Pp sin 0 2 sin - cos - 2 2 b = Pp C 0 m Cos -sin 2 2 2 sin'-0 Cos30 a 2 b or -, si20 Cs20 sin - +I cos1_4+ -f U3 a3 ma m 10. The equation to the circle is (,v - ae)' + y' = b and to the asymptotes y = - vr; hence at the points in a which the circle meets the asymptotes (X - aCe)2 + (e - 1) x' = a' (e2 - 1);..e2 X2-2aexr +a' = Oor (ev - a)2 = O; 160 GEOMETRICAL PROBLEMS. NO. XVII. DEC. 1846. which shews that the circle touches the asymptotes in two a b h points, the abscissa of which is -, and the ordinates - and -- e e e respectively; or at the point in which the directrix meets a them, since for every point in the directrix x= - 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 S is a point in the circle circumscribing the triangle PQ1R, 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 in Q; AM=x, MQ=y; AM'= x', M'P'==y'; then x+ ' =2a; y y /a / a y2 4a and - = 2 -- =2 V --; *.* c G X x 2 a -- x I 2 a - x is the equation required. When x 0, y = 0 or the curve passes through the origin, and limit y = = /2; as x increases y increases, and when x. = 2a, y is infinite, or the ordinate at a distance 2a from the origin is an asymptote. When x> 2a, y is impossible. When x is negative, = 2a+; as x increases y increases, and when x is very large, the curve approaches to the parabola y2 = 4a (x - 2a) as the asymptotic curve. The curve is that traced in (fig. 156). 13. (1) Let y2 = 4aix be the equation to the parabola; then yy'= 2a (x + x') is the equation to the tangent, and if 2a aan)n y a \a y \ + _ GEOMETRICAL PROBLEMS. NO. XVII. DEC. 1846. 161 2y a or m2 - - m + =; (1) but when sin sin 0' = a; cosec" 0cosec2 =-, or ( + + ) = 1 L 1 1 a2 M ] m22 aI, hence + -- 1m /I 1 2 1 12.'. -+ + -1 =-; m m m mm a 1 1 y 1 v ' and - +- =-;,-; and-+ an n a ) n a m m a mm a (y'\\ (P7- a\2 1 as hence (x' - a)2 + y'2 = a2 which is the equation to a circle whose centre is the focus, and radius = - = a cosec 0 cosec 0'. a a a (2) Let tan 0tan 0'=;.'. mm' =-; hence x'=which is the equation to a straight line perpendicular to the axis at a distance a cot 0 cot 0' from the origin. 1 1 y (3) Let cot 0 + cot ' = y; ' -- =; or -; m m a which is the equation to a straight line parallel to the axis at a distance ay from the axis. (4) From equation (1) 1 y' V/y'2 4ax' m 2a 2a 1 y' v/y2 - 4 a ' m' 2a 2a L 162 GEOMETRICAL PROBLEMS. NO. XVII. DEC. 1846. 1 1 _ 4a ' - - = cot 0 - cot = m m a. y,2 4, ax =a2; hence y'2 = 4a (' - which is the equation to a parabola whose latus rectum = 4a, a3~ and vertex is at a distance - from the origin. 4 14. Let P, Q, R (fig. 157) be three points of a parabola whose ordinates are y, y', y"; 0, ', 0" the inclinations of the tangents at P, Q, R to the axis of the parabola; then since the tangents are in harmonic progression, cot 0 + cot" = 2 cot 0'; or Y+ - = Y 2a 2a 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 R intersect in T; then the tangent LQM is parallel to PR; TQ = 2 TV; and the triangles LMT, RPT are similar;.. AMT - = R?PT I A PQR- Q V. R sin QVR; 4a' but i- QV= RV2; and RV. sin QVR = Y y sin2 QVR 2 1 QV= (y"-y)2; 16a __ y"-y (y"- y)3 and A LMT = (y _ y) y - 16a 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 x in harmonic progression. 15. (1) Let x, y, v', y' be the co-ordinates of P, P', and X, Y, X', Y' the co-ordinates of Q, Q'; then GEOMETRICAL PROBLEMS. NO. XVII. DEC, 1846. 163 bx b2 x' tan 0 =-; tan 0= a-y a"y tan 0 X y' tan ( X Y',tan = 7 By '; similarly t-a n x- -; tan O' \y v tan (P. y; Xx Y y but -;. a a b' X' X' Y' y' also -- = -;.. a a b x X y Y hence =;, - tan 0 tan p tan 0t tan ' 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'= - x, 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 x, y be the co-ordinates of Q, so that xy = m2; X, Y the co-ordinates of D; then X= y; Y -bv; b a the positive or negative sign being used according as the series of curves are ellipses or hyperbolas; '. XY = =I= y = m or the locus of D is the hyperbola, or the conjugate hyperbola, according as the series of curves are ellipses or hyperbolas. L52 64 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 2b be the axisminor of the elliptic section, SY. HZ = b2; and S and H are on different sides of PQ, therefore PQ always touches a hyperbola whose foci are S and H, and conjugate axis (b), and P is a point in this hyperbola, therefore the locus of P is a hyperbola whose foci are S, H. If a', b be the semi-axes of the hyperbola; S', H' the foci of the ellipse, then SH2 = a2 + b2 = a2;.'. a'2 = a2 - b = CS'2 = CH'2;.-. a'= CS' =CH'; 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', H'Z' 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 = (altitude of cone x triangular base); and the area of the triangle which touches (a + b + c) the circular base = r, ---- ) where a, b, c are the sides of the triangle, and r the radius of the base of the cone; this will be least when a + b + c is least; but + --- — =r r cot- + cot cot2 2 2i b and if A be given GEOMETRICAL PROBLEMS. NO. XVII. DEC. 1846. 165 B+C A 2 sin - 2cos - B C 2 2 cot - + cot- = 2 2 B-C B+C B-C A' cos --- -os cos - sin2 2 2 2 B-C which will be least when cos is greatest, or when B= C. Hence the triangle will be least possible when A = B = C. In this case the area of the triangular base = 3 r2 (tan 60) = 3 /3. "2; hence the volume of the pyramid - i {3 V/32 h = v/3rh. where h 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, b, a, whose co-ordinates are a3, b3; a,, b3; a1, b, respectively; let the centre O be assumed for the origin, and let x,1 y,; 2,, y2; a,, y3 be the co-ordinates of A, B, C, and x, = a cos 01; therefore y, = b sin 0; similarly, let x., = a cos 02, x3 = a cos 03 therefore Y2 = b sin 0,, Y3 = b sin 03; and the equation to AB is b (sin 0, - sin 01) y - b sin 0, (sin sin ( - a cos0); a (cos 02 - cos 0,) and if tan 0- = t,; tan 2 = t,, tan - t3, 2 2 2 b /I - tt b (1 + tlt2) = -- + ( + a J t + t2 and since AB passes through a point whose co-ordinates are a3, b3, we have ab3 -b (tl + 2) = - a) + ( 1 + t1t) or (a+ a3) tt, - b3 (tl + t2) + (a - a3) 0; (1) similarly, a (a + a2) tt3 - b,2 (t, + t,) a - ag = 0 (2) (a + a1) t23 - b. (tt + t3) + a - ca, c 0. (3) b b /a + a3\ b /a - a,\ b (a + a.) b (a - a,\ -- =m, 2 - o-. =,29 ac b, a x a APPENDIX I. 167 b a + a,) b a - a1 _ -- — = 'M1i - = nl; Ca b I 1 a bl hence m tl t - (t1 + t2) + n3 = 09 m,2t t3 - (tx + t3) + n2 = ~, m1,t, - (t2 + t3) + n, = 0; tl - 23_ t2 - n1 _ 1 3 '.3 t3=, and t2 = m2,t - 1 m, t - 1 m3,tl- 1 ts - (1 - n (1 M13) l + n - n3 0 -or - ml t2 - 1 (Mi - m3) t + 1 - nmitn t1 - n2 (1 - nm,) tl + (n, - n3) m2st - (- t + (1 -,) (1 -m 3) hence (ml - m2 - m3 + nl^2ms) tt2 + }2 - mI (n2 + n3) - n, (in2 + fi3) + n23m + M2n,3 t + n2 - 2 - n3+ mln2n3 = 0; (4) from which equation the two values of t1 or tan 1 may be determined; which will give two positions of the point A, and by drawing AB through c, and AC through b, 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 A. Let the coefficients of equation (4) be A, B, C; then 0, A tan2 + B tan - + C =0, 01 01 01 01 or A si sin- in - cos - + Ccos - = 0; 2 2 2 2, A (1 - cos01) + B sin 0, + C(l + cos0,) = 0; 168 GEOMETRICAL PROBLEMS. hence (C- A) cos 01 + B sin 01 + (A + C) = 0 (C - A) + B. + ( + C); a b or {(n, - m,) - (e n- 2,) - (n, - m3) + mln2n3 - n,m2m3} - + {2 - m, (n, + n3) - n, (m2 + 4 3) + n2m3 + m2n3} Y + { n, +M l-(n2 + m) - (i3 +.3) + mln2,n3, + m, 3 = 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 inscribed in this circle whose sides shall pass through the three points a', b', c' whose co-ordinates are a a a a, - b; a2, - b.; a3, b3; b b b then using the same notation we shall obtain equations (1), (2), (3) for determining the values of t,, t2, t3 in the circle; hence the values of 0,, 02, 03 are the same as in the ellipse; and if AA', BB', CC' be drawn perpendicular to ED to meet the ellipse in A', B', C', the three sides of the triangle A'B'C' will pass through the three points a, b, c whose co-ordinates are a, b; a2, b2; a3, b3 respectively. 4. To inscribe a triangle in a hyperbola whose sides shall pass through three fixed points. Using the same notation as before, let (v, = a sec,,.'. y1= b tan O0; similarly let X2 = a sec 02, 3= a sec 03;. y = b tan 0,, y = b tan 0O; and the equation to AB is y - b tan 0b =b (tan 02 - tan 0,) 0 a (sec 02 - sec 00) APPENDIX I. 16 oy= b I ~tj t2\ _ _ b ______ a hence -4b3 (tl + t2) = (I + t tj) a, - al- tj t) b b or M3 tlt2-(t] + 2) —n3 ==O similarly n2 tI 13 - (tI -+- t,) - n,= 0; Ml Gt t3 -(12 ~ t3) - n= 0; hence (MIn - in2 - in, - n1rnkM3) t12 ~ {2 +m, (n2~+n3)+ n, (Mrz+ M3) -n.rn,- m n;3 ji -(n, - n- n- minn2n) = 0; from which the two values of t1 may be deter-mined. 5. To find the equation to the straight line passing through the two positions of A. If At,2~+Bt, +C = 0; we have (C - A) cosO1, ~BsinO1, + (A + C) = 0, or (A + C) secO,1 + B tanO ~1 (C - A) = 0; (A~ Cj I +B Yi+(C A) 0; a b is the equation to the line joining the two positions of A. 6. To inscribe a triangle in a. parabola, whose sides shall pass through three fixed points. Let ABC be the required triangle, whose sides AB, AC, BC are to pass through three points c, b, a whose co-ordinates' measured from the vertex are a3, b3; a., b.; a,, b1; let tv,, yi; IV2, Y29; X3, y3; be the co-ordinates of A, B, and C respectively; then the equation to AB is YV2 -V or Y 2- Ya1;1'2- I but y, = 4ax,; y2= 4 ax2; 170 170 ~~~GEOMETRICAL PROBLEMS. 4a Y1 Y2 Y1 +Y2 Y + Y2 and since this passes through the points a,, b,; 4a Y1 Y2 Y1 +Y2 Y + Y2 or YI y2- b Y + Y2) +4 a a3=O (1 similarly y, y,- b,(y,~+y3)~+4 aa. =; (2 y2y, - bi(Y + Y3) +4aaiO=; (3) henceY;3-b2y, - 4aa, by2 - 4aa, b3y - 4aa, bly2 -4aa, (b, b; - 4aaly,) i~4a (a,b3 - a~b) b2YI-4aa., y_9-b I (b, - bl)y, ~b, b3 —4 a a3 - 1- or bi b3 - b2 b3 + b, b2 - 4 aa, jy12 + E4a{I(a, + a2) b3 - (a,,+ a3) b, + (a, + a3) b,} - 2blb~b,]y, + 4a {a2 (bib3 - 4aa.) - b2 (alb3 - a~b,)} 0; (4) from which may he determined the two values of y, which will give the two positions of the point A. 7. To find the equation to the straight line passing through the two positions of A. Since yI' 4ar,; substituting this for y,2 in equation (4) we have 4a (b,b3 - b~b3 + b,b2 - 4aa,) XI + [4a J(ai + a,)b3 - (a2 + a3) b, + (a, + a3) b,} 2blb~b.,] y, + 4a {a2 (b~b3 - 4aa3) - b, (alb, - 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. 35) he the three points; draw any line mnab through rn; through b drawn ben, and through c draw pcd; join ad and let it meet the line mnp in q; then abcd APPENDIX I. 171 is a quadrilateral figure, three of whose sides ab, be, 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 abed becomes a triangle, and qda becomes a tangent from the point q. Hence from the point q draw the tangent qB (Appendix II. Art. 66); join mBA, 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, n, 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 a3, b3; a, b2; then from equations (1), (2) Art. 1, we have n3tlt2 - (tl + t2) + n3 = 0; m2tt3 - (t, + t3) + n2 = 0; and the equation to BC is ay (a + at) tt3 - (t3 + t3) + (a - x) = 0; t 11 - U3 tl - i2 hence 2, t -1 mail - 1 mt1 - 1 (m + m3) t - (2 + mn3 + nm3) tl + n2 + n3 or t, + 3 ==; or + 3 (mnt,- 1) (mt - 1) t2 - (n2 + n3) t + n2T3 (2 3 (, - 1) (m3t, - 1) and substituting these values in the equation to BC, (a +,) {t1 - (n, + nz) t1 + nn3} - y (m 2+ 3,) t12 - (2 +4- 2'31 + n 2m3) 1t + n2 + q2} + (a - }) minm3t,2 - (m2 + m3) + 1 } - 0; 172 172 ~~~GEOMETRICAL PROBLEMS. or (I+ +m^m)a~ (1 -M2M,)xV- a (M2 + m,)y} t12 b + [JM2 + rn3 - (n. + n3) } aV + -(2 + m2 n3, + n2, in) y - (in2+ mn3 + n2 + n3) a] t, b +(i + n2n) a-( -n~nv a n3)aY7- b or u t,2 + vt1+ w=0; where u, v, w are known linear functions of x' and y; and to find the curve to which this line is always a tangent, we must make Qwut, + v = 0;.-. 2=4uzw; which is the equation to a conic section having two tangents u=0, w =O; and v=O0for the -equation to the straight line joining their points of contact. The same proof will equally apply to the hyperbola and parabola. 10. If a polygon of n sides be inscribed in a conic section, and (n - 1) sides taken in order pass through n - 1 fixed points, the remaining side will always touch a curve of the second order. Let AB, BC, CD, &c. pass through -the points a,, b1; a., b.;... a,,-,, bn-I; then we have MI tl t (t, + 12) + n1 =-0 m'n21t2 3 - (12 + t3) + n2 =-0 MnInIn- (t,1 + tn) + n,,.1 = 0; and the equation to the last side is (a 12= t t -a 1 +tn a x -1'- 13=mt2 - 1 or 13=( - MIn2Q) t1 + (n2- n1) _t atl +/33 where a., /3 3, ~3, are known quantities. APPENDIX I. 173 t3 - n3 a4 tl + /4 Similarly, t4 = - = m,3t- 1 74 tl + 4 where 0a4 /34, Y74) 4 are known; and 7a, t + On where a,, 3^, y35 3 depend only upon a,, bl; a2, b.;.. a,_1, bn_-; and are therefore known; but from equation (1) CO\ Y t Y 1 + - t - t, +1 t- =o; a] b a b i (l +) t1 - } (ant +3) + (1 - - b t) (y? tl + S2) ~ 0 a a bJ a b or ut12 + vtl + w = 0; where u, v, w are known linear functions of w and y; hence as before v2 = 4w 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. m3tlt2 - (t1 + t2) + n, =3 0 mlt2t3 - (t2 + t3) + n = 0; tl - n3 2.. t, ---- and (mt3 - 1)t2 - (t- n,) 0; m3At - 1 or (m, t3 - 1) (t, - n3) - (, - ) (t, - n) = 0;.'. (mn - m3) tit:3 + (n,m, - 1) t +- (1 - n3m) t, + n3 - n, = o; but in order that this may pass through a fixed point a,, b, we have (a + a2) tlt3,-,b, (t + t3) + a - a2 = 0; or the coefficients of t1 and t3 must be equal; n'. nm3 - 1 = 1 - -nam1 or nma + n:m = 2, 174 GEOMETRICAL PROBLEMS. b a+ a, b a - a Now m1=- =. -- 9 a bl a b6 b a +a 3 b a-a3 m3=-. 9 3=-. -; a b3 a b3 b\hen 2 t a + - a3a a a a -a a-a1. 2 hence - be a-) = I 2;. alas b2 If two tangents be drawn from a point whose co-ordinates are a,, bi; a,, b3 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 c, 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 bAC be made to coincide with bA'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 A', 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, t = -;1 (mt - 1) t - (t3 - 2) = 0; mtl - 1 APPENDIX I. 175 (m3t4 - 1) t3 - (t4 - n3) = 0o (m, - m,) tlt + (,m, -1) t, + (1 - nm) t +, - nO = 0, or t,{(m - m)t, + (i - m) + (nm, - 1) t1 + (, - n2) = 0;. (M,4 - 1) {( - n,2) t + n2 - n} - (t4 - n,) {(m, - mi) t1 + (1 -n, r,) 0, and when the fourth side passes through a fixed point the coefficients of t, and t, are equal, hence m, (n, - n,) - (1 - nlrml) = n3 (m, - mi) - (1 - m1n,), or ml (n2 - n,) + m. (n3 - n,) + m3 (n, - n2) = 0; 2b2 /^ - a2 Now nIl, - mn2 = 2- b - -; a 1b b2b b92 ba, - l a3 _ " 2 a - a3 a \ bb, b3b2 bb, =J; hence (a, - a,) b3 + (a - a2) b, + (a, - a3) b2 = 0;.. (3- a) b, + ( - a,) b - (a, - a~) + (a( - a3)},3 = 0;.-~ (-) - b3) = (a - a3) (b3 - b,); which shews that the three points mus be in the sae righ which shews that the three points must be in the same right line. 14. If n - 2 sides taken in order of a polygon of n sides inscribed in a conic section pass through n - 2 fixed points; find the position of the point through which the n- 1th side must pass, in order that the remaining side may pass through a fixed point. Using the notation of Art. 10; t_ a-1,t1 + P~, _ (1) ere a_, t -l since they nly where a,,,,,-,, n-19 n-i are constant, since they only 176 GEOMETRICAL PROBLEMS. depend upon the co-ordinates of the points through which the first n - 2 sides pass; and (mlaltn - 1) tn-,: = 4, - n,-1; (2) for convenience we will represent equations (1), (2), by -4 t, + B - Ct1 + D and (ut, -1) tn.1 = (t, - v).I. Ca tn - 1) (At, + B) (tit V) (Ct, + D); or (AMIx - C) tj t, - (D - BIAu) t -(A- Cv) t +(D v - B)= 0; and in order that the nth side may pass through a fixed point the coefficients of t1 and t,, must be equal; D - Bu = A - Cv; and if a,,-, b.-I be the co-ordinates of the point through which the n - 1th side passes, c( a + vJajb a - an1) or (D - A) bn-1 B (a ~ an,) - C(a -ajc1), or (B + C) a,,-,= - (D -A) bn-1 (B - C) a; hence an-!, bn-l lies in a straight line whose equation is (B + C) v -,(D - A)y + (B - C) a = o; (3) and if the (n - i)th side pass through any point in this line, the nth side will pass through a corresponding fixed point. 15. To find the position of the fixed point through which the nth side passes. Writing for convenience M', v' instead of mn, n,, we have f - ( t,) + V, = 0; and making this equation coincide with equation (2) Art. 14; AMu - C Dvy-B v'; andD - BMu = A - Cv; D - Bu D - Bu APPENDIX I. 177, ABM - DCv.. Bat' - Cv' ID - Bl AB D - D (A - D + B/A) D -— h. Tg^ -B- D -A; or D - Bm' = A - Cv'; and as before, (B + C) a,= (D - A) b, - (B - C) a; hence a,, b, is a point in the straight line in which a,_,, ba_, lies. Upon the whole we conclude that if a rectilineal figure of n sides be inscribed in a curve of the second order, and n - 2 sides taken in order pass through n - 2 fixed points, there is a certain straight line, through any fixed point of which if the (n - 1)tl 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 respectively. If (n - 2) sides taken in order pass through n - 2 fixed points, and the n - 1th side passes through a fixed point which does not lie in the straight line determined in equation (3) Art. 14, the nth 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 AD, DE, EF pass through three fixed points in a straight line, the remaining side FA will pass through a fixed point in the same straight line. M 178 GEOMETRICAL PROBLEMS. 17. Hence generally, if 2n - 1 sides taken in order of a polygon of 2n sides inscribed in a conic section pass through 2n - 1 fixed points in a straight line, the remaining side will pass through a fixed point in the same straight line. 18. If 2n - 1 sides of a polygon AA,... A2^+ 1 of 2n + 1 sides inscribed in a curve of the second order pass through 2n - fixed points in a given straight line AB, and the 2nth 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 + 1th side will pass through a fixed point in the straight line AB. The chord A, A2, will pass through some fixed point A in the given straight line AB (Art. 17); and if A2, 2An+ pass through any point in the chord of contact of pairs of tangents drawn from A, the side AA2,,+1 will pass through a fixed point in the same chord of contact (Art. 11); and if it passes through 0, the straight line A4A,,, will pass through a fixed point in AB (Art. 12), which will be the point of intersection of the chord of contact of a pair of tangents drawn from A 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 DbE be the chord of contact of a pair of tangents drawn from c (Appendix ii. Art. 66) intersecting AB in b; through c draw any line cA'B'; join A'bC'; draw B'C'E meeting DbE in E; then AE, BE are tangents at the points A and B. If B'b C 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 ay2 + bxy + cx2 + dy + ex + f= 'p (X, y) = o be the given equation; transform the origin to the point a, /3 by making x = x ' + a, y y 3; a. (y' + (3)2 + b (' + a) (y' + 3) + c (x'+ a)2 + d (y' + 3) + e (' + a) +f 0, or ay'2 + bCy' + + cx'2 + d'y e'' + / (a, /3) = 0, where d'= 2a3 + ba+d; e'= b3 + 2ca + e. Next, transform the equation to polar co-ordinates by putting ' = p cos0, y'=psin;..(asin20+bsin0cos0+ccos20)p2+ (d'sinO +e'cos) p + ( (a,) = 0, 1 Q: V/Q2 - 4P() or Pp+ Qp+(=o;.. = - p 24) but Q2 - 4Pq) = (d'2- 4a () sin2 + (2d'e'- 4b4()sin0cos0+ (e'2 -4c)) cos2; and if a, 3 be assumed so as to satisfy the conditions d - 4a) = e'2 - 4c(D; 2d'e' - 4b -= 0o; then Q2 - 4P4D = e'2 - 4c(D, and 4- = 4c t1- 12 (e' cos + d'sin 0) p 2'4 ^es } e'2 -4c ) e' sec d =/eh2- 1 -l-r ' cos(0) - if =tanc =..(1) D ( \/eI c_ 4 c(I) e l M 2 180 GEOMETRICAL PROBLEMS. which is the equation to a conic section from the focus, having its axis-major inclined at an angle S to the axis of x, e' sec ~ its eccentricity =............ (2) 2(1 and semi-latus-rectum L --.e.................(3) /Ie" - 4c(? the two signs corresponding to the equations from the two foci respectively. Now d'2 - e'2 = (a- c), and d'e' = 2b; d' e' 2(a- c) 1 2 (a - c) or a- 2-= cot23 = e d' b g b 1 _ 2/(a- c)2 + b2,o + —Q(;.................. (6) u bb e sec ~ +. =. ve_ = b //; /C^w@w@*@@......... (7) b b'V 2 c ffi (a _ C)2 + e2 ad'e' - J h - b = ba + d) (b/3 + 2ca + e) v/e'2 - 4 c4 ~ 2c b i - 2b (a24 + a' ca2 + d + ea +f) = 2b ab/322 + a(b + + be) a + (2ae + bd) 3 + de - 2b (a32 + baf + ca2 + ea + dft +f) = - (b- 4ac) af3 + (2 d - be) a (2 ae - bd)/ - (2 bf- de) = 0; APPENDIX II. 181 or a3-ka-h3 + g = O............... (9) 2ae - bd 2cd - be 2bf - de where h=; k= and g= 2 -ac 4ab2 - 4ac b - 4ac Also d' - 4 a = (2a/3 + ba + d)2 - 4a (a/32 + baf3 + ca2 + da/ + ea +f), or d'2 - 4a4D = (b2 - 4ac)a2 - (4ae - 2bd)a + d2- 4af. Similarly, e'2 -4c4< = (b2- 4ac)/32- (4cd - 2be)/3 + e2 - 4cf; and putting b2 - 4ac m, 2a 2c d'2 - 4ac = r ((a - h)2 J+r b (hk - g)}; e'2 - 4c( = m {(/ - k)2 + 6 (hk -g); 2 2c (a - h)2 + 2 (hk - g) ( 3 k)2 + 2 (hk - g), b b or (a - h)2 - (3 -)2 = ) - ( - c) (h/k - g); (10) and (a -h) (- k) = hk - g; from (9) a - (3- k- 2 (a - c) 1 hence - - =- -; l- k a- h b 6 - h.a.-h =t, and p (a - h)2 =hk-g; a-h pt. (11) /3=ic v/"a(hk-g) Hence hk - g and g must have the same sign, and since 2(a - c) b 182 GEOMETRICAL PROBLEMS. the two values of tx have different signs; and only one of them will make a and,3 possible. It will afterwards be proved that in the ellipse where m is negative, b ( +-) is negative; and the negative sign alone of equation (6) can be used. Hence in the ellipse 1 2 /(a - c)2 + b2 2 /(a + c + m, + - b b 1 2(a - c) -- = -; b 1{(a - c + -/(a + c)2 + mn ~ ~ ~b — f)A b '.1 + 2c 2/(a+ c + m + c2 a ml a ln - - - --- b 1 2c a + c + cv/(a + c)2 + m a+ ~V/a'2 + m where a + c = a', or E 2 (a ")- +?) - 2(a a'/ a ' + m )2 =Ei = =1. (12) mn9 m A2E2 (a-h)2+2 (-k) =( +-) (hk -g) = - 2V/a' +2 ( —L-).(13) A2 ( h ) V ' (9 / )2 h () 9kk-g\ t / += - (a + Va'' + m) (14 1) 4E APPENDIX II. 183 BL = A' (I E) 2 (a'- V/a' + m) (hk-g (-) L- = -A(i - )T (a-. (16) And the co-ordinates a,, 3 of the extremities of the axis-major are found from the equations 1 a + Va 12 + m hk(a, - h)2= (a )- h a = / + ) ( 17),2 Va'm +m (3 - Ik)= ( - ) (/3 c)- a' + -g) V.+ (18) If a', a"; /3', /" be the two values of a and 3 respectively determined from equations (11), a' + a" = 2/h; 3' + /" = 2k; but a', 3'; a", /3" are the co-ordinates of the two foci; therefore h, k are the co-ordinates of the centre. 3. To find the values of d', e', and q5 (a, /). e'= (bt3 + 2ca + e) = k (b + 42he) + ) +/ + -j /_v k-g; or e' (b:, + 2c) N/hk -g.k d' (2af + b) h/h - g And Q2h O (a, ) =d'e' - e'" (bh + 2c)2 (hk - g). 4. To deduce from the above expressions the co-ordinates of the vertex, and latus rectum when b' - 4ac = 0. In this case the values of h and k become infinite; let 2cd - be K, 2a -bd - H, 2bf- de = G; (a - c) - v/( + c)' + w then... b 184 GEOMETRICAL PROBLEMS. 1 (a - c) + V/(a + c)2 + M ~(b and expanding to the first power of m, 2c m, 2c [ m /A= - 1 + 1 _ 2a m 2a / m - h-b 2 a'b b 4 a- 'a) now hk~- g=-b(ae2 bde + cds + mf) now hk-g — 2 b --— 2 -; and if M = d2 - 4af, N = e2 - 4cf; mh2 - M (ae2 - bde + cd2 + mf) = 4a 2 ----4-; m m2 mAk2 - N (ae2 - bde + cd2 + mf) = 4c -- qn I/g2 hk-g mh - M mk2 - N 2b 4am 4cm hence a - /(H - mM) 1 + m m 4aa H H ( m mM\ M H or= - 1 +- _ (19) m m, 8aa' 2H2 2H 8aa ' '1 K ( 2 M, N K /3 = A (KK - M N) 1+ =,...I (o0) L -- m m m K m / 4ca') 2K 8ca' (20) _ 2 j" K2- mNa_ K2 / mN\ Sa' \ 2cmN2 1 -6ca' 3 K2 K / mN 4or L ='-........................ * * (a) K and when m vanishes L= -.................. (21 4 ' APPENDIX I8o 185 /E=:+ +; ~ 8~=+ (p3) 1+ 4 2 ' E 8 t2..............(. ) which gives the eccentricity of the ellipse when m becomes very small. 1 f H I m mM\~ 1- \ Also a, =h h ( - h)= - + 8, - ) 1 r 7 7 m m aa 2HpA 8 a12 M H(a'- a) M cH or a, = 2H..."7 - H - 8aa (22) 2H 8a a'2 2H 8a2a'....... Similarly, 3, = k + (3 - A); KT K ( m mN\ N( mr N aK or6= — + - i --- 11- a'( ) -- = m 1 + Sea 2K) 1 8a2 2K 8cxa (23) 5. When b2 - ac = 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) 3 - (2b f- de) = 0; o + GK G b G or Ka + Jr3 = G;.. 3 + l a =- or 3 - =; (A) also the two values of A, derived from the equation 1 2 (a - c) 2a 2c are - and —;, b b b d' 2a/3 + ba + d 2a (2ae - bd) but g == e = -b e b3 + 2ca + e b b (bf + 2ca + e) 2a and cannot therefore = b 2c b hence the only admissible value of 3A is - - b 2a Again, d'-4a = -(2Ha-M); and e'24 c4= -(2Kf3-N);,. T, ' -. K M=K -:N; o.. oOO.e..... (B) 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 (2Ha) + (2 K3)= G; K ~ (2Ha(2 K - KN) - G- HM- - N 211 2K 2H 2K iK H K H or ) (2K3 -N) =2hf - de+ c (d - 4a) 2K~3N=2HaM-(2 + (KK/H -=-N cc"d + ae - bde K Hb 4be 4ab' K2 H2 2. -i N 2H- 2 - = - 4c(a + c) 4a (a + c) N K 32K 8c(a + c). (C) M H a _........(D) 2H 8a(a+c) d' 2a3 +ba + d b Again, -7)= bf-e +2Ca+ - d+ e - b 2ca + e 2a (4a2 + b2) 3 + 2b (a + c) a+ 2ad + be =- 0; b 2ad + be or +-a + —= 0,a- M........(E) 2 a 4 a (a + c) b /2ad be\ H, and e'=b + 3+-a + e=e - - K --- ) 2be (F) 2and b\ a+c c 2 (a + c) but d'e' =2b = me2= - e2;.. e12- 4a(n; L 24)cosS e' cosS H cos4 4(acnd - E ea 2 4+ (a + c APPENDIX 11. 187s H K L= 0O.. G 4x/a(a + c) 4\/c(a+ c)i 6. To find the equation to the axis of the parabola. b Since tan $ = equation (E) is the equation to the 2a) principal diameter. 7. When the general equation of the second degree is referred to oblique axes, to find the polar equation from the focus. Let ay" + baxy + cxi + dy + ex +f= f (x, y) = 0, 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, /3 by making x = x' + a, y = y' + /3; ay'2 + bx'y' + cv'" +d'y'f+ e'i&v' + (a, /) = 0, where d'= 2a)3 + ba + d; e' = b13+ 2ca + e. Next let the equation be transformed to polar co-ordinates by making I sin (te-O ) sinO0 tv-,ine P; y P;l) sin sin w a sinO2 + b sin O sin (w -0) + csin2 (w -_0)} p2 + sin o I d'sin0 + e' sin (w -0)} p + sin2w/ (a, /) = 0, r ~ (a - bcosw + c Cos2 W) sin2 0 + (bsinw - csin2 w)sin Ocos 0 + csin2w cos20} P2 ~ sinw {(d' - e' cos w) sinO0 + e' sin w cos0} p + sin2w 0 (a, /) = 0; and representing this equation by Pp2 + Q sin wp + sin2w4 = 0, we have 1 Q 1 P i Q 24d 208 9inw(B, p sin" te, (I,' p ~2 sinw2 p sin w4 Dp sin 2 p 2 s i n 188 GEOMETRICAL PROBLEMS. Now Q2 - 4P = {(d' - e' cos w) sin 0 + e' sin w cos O '2 - 4 (1 (a - b cos w + C cos2W) sin20 + (b sin w - c sin 2 w) sin0 cos0 + c sin2w cos2O}; and assuming a, / so that the coefficient of sin0cosO may vanish, and the coefficients of sin20 and cos20 be equal in the expression for Q2 - 4Pq, we have (d' - e' cosw)2 - 4~ (a - b cos(e + c cos2o) = e'2 sin - 4c~ sine2 o,................. (1) and 2 (d' - e'cos) e' sin c) - 44( (b sin w - c sin2 o) = o,... (2) or (d' - e' cosw) e'= 2 (b - 2c cosC),......... )(3) and d'2 - 4 a~ - 2d'e' cosw t e" cos 2w = 44( (c cos 2 w - b cos w) = - 4 ( - 2 (d' - e' cosw) e' cos w;.'. d'2 - 4a(I)= e'2- 4c~,............... (4) and d'e' - 2 b6 = cosw (e'2 - 4c)). d'- e'cos w n d e COSt Hence, putting e' = = tan,............ (5) e sin t I - e' sin o secc cos (O - 6): sin o V/e',' - 4c(b p 2 sinw)t 1 /e'2 -4c D) w e' sec S cos ( - ) or - =............... (6) p 24) which is the polar equation to a conic section from the focus, 4~) whose latus-rectum =-........................ (7) V/e'2 - 4c( e sec e eccentricity E........................ (8) a/e 2 4c ( and inclination of the axis-major to the prime radius =. Also from (3) e' (d' - e' cos w) = 2~ (b - 2 c cos w);.e e'2 sinwou - 2 i4> ( 2 c cos wo), APPENDIX II. 189 /_ I Fo+ 1 and E = s A/ 2csinto 1 - A b - 2 c os w................ (9) S. To find a, we have e'2 sin wp = 2 (b - 2 cos o) (; - u sin Co + cosw, and d' - e' 4 (a - c); e ' e'2 {(2 - 1) sinw s + k sin 2w} 4 (a - c); (,2 - 1) sinS a, -+ sin s 2 (a - c) hence =9 bI sinw - 2ccosw 1 2 (a - b cosw + cGos 2w) or = - - cot 2... (10) tu b sin o - c sin 2w and,+-= 4+ ~Ik \ M /It 1 2 V/(a - b cos w) + c)2 + (b2 - 4ac) sin2w or + -= - ( 1) ju b sin w - c sin 2 t It will afterwards be proved that in the ellipse and parabola the negative sign only can be used in the expression I for u + -. 9. To find a, 3, and L.. From (Art. 2) d'2 - 4a4 - nm l(a- h) + (hk- g), e'2-4cq = m (-3-)2} + (hk -g), d'e' - 2b = -m {(a - h)(3 - k) - (h - g); but d'2 - 4a4' = e'2 - 4c@, and d'e' - 2b1 = (e'"- 4c4) cost (a h)2 (/3 k)2 2 ( (h-g)......(1... (a - ) - ( - )=-(- (hk - g))...... (1 b 190 GEOMETRICAL PROBLEMS. coswt, (P - k)2 + (a - h) ( k - k) b (hk - g). (13) a - h sin (w -.) Also.- k - sin for dividing (12) by (13) and 3 - k sins a-h putting = X (3-k X2 1 2 (a-c) _ (2) sino + 2 cosw cs b -cosww + X b-2ccos1 \ f ' sin o sin (w -) from which X = - cosw = s y sin s( sin (w - b) - 2ccoswc_.' (3 k)2 cs i (hk - g); Sin % b hence (/ - 2 b 2 cos tan (hk - g);..... (14) b sin w b - 2a cosw similarly, (a - h)2 = b -. tan (w - ) (hk - g), (15) 6 sin w sin2co 2 (b sinco - c sin 2z t) and (AE)2 = (13 - ) Sin2 = sin 2 - (hk - g) sinw = b sin 21 or(AE2)-cn ( 2 )(hk-g)....... (16) b \ u Let a - b cos + c = a', b - 2c cosw = b', and (b2 - 4a c) sin2w = m' therefore in the ellipse 1 2 \/a'2 + n' 1 2 (a' - 2c sin2w) / + - = —. b - - =.- -,u bbsin A b' sin to 1 -( a' - 2c sin /a'2 + \/a'2 + or - ---, an b sin ftcI 1 2c sin2 fa +,t/aa'2 +m and --. _- --;, b sin \ b sinc o APPENDIX II. 191 2 /a2 ml 2 (_ 't2 + m) -2 a / a + s hence E2 a + /a2 + n' m (a' - /a'2 + mT')2 orE2J=l+,;. (17) m [ \ fhk - g\ 1 A2 - sinw b' ( + b) x Es; A2 _ (a + \/a' + i') v(h ),............ (18) (a' - Oa + m') B2 = A2 (l E2) = A2 ' -+ a; a + / a'2 + m'... B - (a ) (k b ) (19) L2 = A(1 E )2 = B2(1 - E2); (ffl f/b^T~yAAb ** - - ( a'- _'- /'** (20) J2E2 = 2 \/at2 _ ^/ -.......... (21) b - 2aCOSWa coh ^k- g\ (a - )2 — =oi tan(w_)h -..). (22) b-2c cos w {tanc g (hk (3) ~(ft -ky== —: — tan b-;...... ( (2 b ) b - 2 c cs 2 hk -g o~~r (1)b' sin )v b sinw o a('- $('2csinb2 ~7/a'2 + m' (hk -g) or (/ - k)s ==.- s ---- 7-in --- - - w. (/3 - k)2 = (a - b o+coscos w - + \/a + ) (m, g \c/ sins;t/ and (a - h)2 (c - b cos o + a cosw - v/a'2+ m') - k g\ And if ai, /3 be the co-ordinates of the extremities of the axis-major; 192 GE1OM ETRICAL PROBLEMS, (3~)1 - ( ) (- 2 / +, - (k)- k) or (/3, - fr)2 fmn + 2c (a + + 4 m (') h - (2 4) or (/31 - k).. (. ) 29I/a;? -kn b Similarly, m(a a (a +m\/a2 ' hkg (25) (.,- 5)} f )......(~5) [ 2 v/o"2 b 2 And (area of the ellipse)2 = 2A' 2 = -72 m' ( - g)2; thereforearea of the ellipse = - 7r/4 a c - b.sin w (b (26) It may be observed that as in the former case hk- - /h2 - \ m k2 - N b \ 2cam ) 2cm 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 b2 - 4ac vanishes, a - 2 c sin2 (o - v a' + m In this case we have u =, b' sin o m m' 2 a 2c sin w ^ __________ _= _ _ ____ _ 1 + - -- b' sin Co b' 4 ac sin2 ) 2c sino b ( { m\ -N o - l+ - sn......................... (1) Hence (3- k)= n. /'( ), sin \ 2cm APPENDIX II[ 193 / m \ K2 m or /3 -V + - 1 MN in \4ac m \K2 K K = 1 A 1T m - l m+ m, / I; Also 0, - - E (k - k), and E2 + 1 t,;:I mn m sin (c N K M H M H Simi larly, 8a - a.... (3)a I/ Jff 1 \m' / msin2mnc 13 m 8 a la a 2 ) 1Na 2K' 82c(a~ -bc co(4) M.,, cosH similarly, a '2ai- - - - (c - b cos wo + a cos2 w)..~<..e. (5) 1, /..,. m c sin w And from equation (1) AL = - 2c s; e 2 csin2 cv 2 a-bosw)+ccos2Ctw.. sin -= - cos2 4= - - -; a a VN Kr cos2c oE 8 ' '2K 8a'c; 2H 8 a'a N GEOMETRICA L PROBLEMS. 2 (a' - aV/n + 3/A2 L,a -' ) - m \ 2m 2 r m2 (K2_ - mN\ ( f2 - mN = 8a" 2cm / sn S lGa3 "; sin2 ws / m wV or L = -S7 _- - — K) when m is small; 4 ca' \ 2K2! and when m vanishes, L= — = --... When m is very small, B 2 - (4ac - b~) sin2wo 2 E 4a' - 4 a'2 or as b2 - ac diminishes, the ratio of the axes of the ellipse approaches to /4 ac - b'. sin (o 2 (a - b cos o + c).. () 11. When 2 - 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'2 - 4a4) = - (2Ha - M), e'2 - 4e = - (2K 3 - N); and d'e' - 2bt) = Ka + H/3 - G;. 2Ha -M = 2K3-N, and Ka + H3- G =- (2K-N) cos; K H or (2 Ha) + l(2Kf3) + (21 /3- N) cosw G G; ' -* ( +-+cos (2K 3N) = G G- -M- N; 2H 2 K os t 2iH 2 K APPENDIX II. 195 hence- ( c — cs) (2K3 - N) = 2bf- de +- (d2 - 4a f) + (e - 4cf) b b cd2 + ae - bde K2 H2 b 4bc 4ab N K N H 23 K Sca' 2K +4ba' M H M K similarly, a = -a'... (2) 2H 8aa 2H iba d' Also -, = (cot + ) sin u; and since b2 - 4ac = 0; e equations 10 and 11, Art. 8, give b - 2 cos w - 2 c sin,u - --. nor 2csin, b - 2c cos b bcoso - 2c.. cot o =. or s.i-; 2 c sin w sin w (b - 2 c cos w) d' b b cos w - 2c and — or e 2c b - 2c cos d' 2a3 +ba + d b (2cd - be) b K but - + -- + e' bf +2ca e 2c 2c (b/3 + 2ca + e) 2e 2ce b which cannot = -; hence the only admissible value of u is - 2c sin.t 2c sin c — Cs; and sin s = b - 2c cos w x/(b - 2c cos wo)2 + 4c2 sin2 w Vc sinw \/a d' b K b cos - 2 c also - = - + = e 2c 2ce' b - 2c cos w N 2 196 GEOMETRICAL PROBLEMS. K bcosw - 2c b 4bccos w - 4ac - 4c' 2a' 2ce' b - 2 cos o 2c 2C (b - 2c cosw) b' Vb'K hence e' - -, 4ca' L 2(i cosS e" sino tan.cos o e' sin o sins and L=....; E, e e' ' (b - 2c cos aw) b -2c cosw, sj 1sin sin2K R but sin $ -;... L......(3) va/ 4v/e a. L sin N K Ksinwo 2 sino 2K 8 c a' + a';...... L sin (to - ) M H H sin2 a, a -,. (5) s2 sin o 2H 8 aa 8H2 a'..(5 b' K Again e'=3-b+2ca+e=- b, 4ca sin /2 c\ and sin ( S) b b'K.. b3 + 2ca +, + e =0;............... (6) 4fca is the equation to the principal diameter; and the intersection of the two straight lines 2Ha - M= 2 K - N; K2 COS t( and Ka + H/ - G = - cos o (2K3 - N) =....(7) 4ca will be in the focus of the parabola. Let A be the vertex, S the focus, ASa (fig. 163) the axis of a parabola, and let the equation be transformed so that Z PSa = 0 - ~; then when the axes are inclined to each other at an angle o, d'e' - 2bq) = (e'" - 4c4)) cos tw;.. (d' - cos co) e'= 2 (b - 2c cos o) (P; APPENDIX II. 197 ~. but - 2c sin w or Le2e sin w = 2 b(; but ut --: b c sin'. e'2 b,2 and is negative since a and c are positive. Hence in order that the latus rectum of the parabola may be positive we must take the equation J a/e/2 -' 4 l ea sec$- c -- _. - -- 1 + cos (0 - )>; ~P 2(1~) (,/e'2- 4 c therefore e' sec $ is negative; K sin w K sin w but e'tan K= s;.. e sec=S sin 2 a' 2 a' sin c which is negative; hence K and sin S have different signs. From equation (6) if b be positive, CD (fig. 164) is the direction of the axis, and the parabola will assumre the position AJ P2 or A4P4 according as sin ~ is positive or negative; i.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 AP, or AP, according as sin ~ is positive or negative; i. e. according as K is negative or positive. This equally applies to Art. 6, where the axes are rectangular. 12. If any relation among the coefficients of the given equation should make d' and e'= 0; since d'- e' = e 4(a - c) 4), we have ( = 0; and the equation when transformed to polar co-ordinates becomes Pp2 =, or { a sin2 + b sin 0 sin (-j - 0) + c sin2 (w - 0) p2 = 0, and since p is not always = 0; this equation determines two values of (0), or the equation represents two straight lines passing through the point a, f. 198 GEOMETRICAL PROBLEMS. 13. When d and e are each = o, the origin is the centre; and d'e' - 2bb = - ma3 - 2bf, d'2 - 4a4D = ma2 - 4af, e'2 - 4c6 = m/2 - 4cf;.. m(a2 2) = 4 (a -c)f, and cosw (m32 - 4cf) = - ma - 2 bf;...2 =f (a + Va'2 +m'), m B3 =2 (da- va' + M), 2 (b - 2a cos) 2bf a= - tan (t - c) b sin w m 2 (b -2a coso) t f == -—. --— ^ --- tan (w -),< sin o m 3'2= e2 (b - 2c cosw)tan f sin w m 14. To find whether u = o or co, when b - 2ccosw = 0. a' - 2c sin2w - V/a'2 + m' b' sin w a= a +c-bcosw9 b'=b-2cos orb=b cos o =b;.. a'= a - cos - c cos 2 o - b cos 0, and a' - 2c sin2w = a- c - ' cosw also a2 + m' = a2 - 2accos2w +ccos2 2 -2b'coso (a - ccos2sw) + b'2 cos2 + (b'2 + 4b'c cosw + 4c2 cos2W) sin2w - 4ac sin2st = a2 - 2ac + c2 - 2b' coswo (a - c cos 2 w - 2c sin2w) + b'2 ( - c)2-b'cos (a - ) +b'2 (a - c - b'cosw)2+b'2 sin2+. Now if a>c; (a - c - b' cosw) - /(a - c - b' coswo) + b'2 sin'w b' sinwo b' sin wo - - h s O, when b' = 0; 2 (a - c - 1 cos W) APPENDIX II. 199 and when a < c, a - c - b'cosw + (a - c - b' cosw) = --. ~: b' sin w hence the axis-major is parallel to the axes of x or at right angles to it according as a > or < c. When a>c A2= - (a'+ V/ ' +m ) k - ghkc-g hk -g = - a- ccos2w+(a -c)} b - 2(a-ccos'w) b- g B - (a'- _ Va2 + ') hhk-g c in2w (hi -g - a - ccos2w - (a - c))} - - 2csin b \ b When a< c, A2 = - h1C - g 2hk - g A2 = - {a -ccos2w) (c - a) -g=- 2 csin2e b hk-g hk-g B2 - {a - c cos2t -(a -c) = - 2 (a- cos2'o) — g b b (a) When a = c, - - 2coto= - 2cot 2c; '. =-. (/3) When b2<4ac and hk-g= 0, A = B= 0; which is the condition that the equation of the second order may be a point. (y) When b2 is not less than 4ac, and hk -g = 0, the equation represents two straight lines. (Q) To determine which sign ought to be used in the 1 expression for M +-. /x (b - 2c cost) tanc (hk -g We have (p sino \ b (b - 2 cosw) tan m (3/M -H)2 Si I1 (o a O Mt' 200 GEOMETRICAL PROBLEMS. /,. rM- Hz /m (x + h)2 +mM- I — bx + d m and y=-: -- 2a 2a therefore when m is negative, mM - H2 must be negative, otherwise every value of x would make y impossible; hence (b- 2ccosw)tan is negative, since a is supposed positive; and (b - 2 cosw) + i ( b - 2- cosw) [ (+ is negative; but - 2 1 v/a 2 +mI (b - 2c cost) l - = ==. A\ a/sin 1 o. when m is negative, the negative sign only of v/a'" + m' is admissible. When m is positive, and a positive, the positive or negative sign 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 x; for if m M> H2, every value of x will make the quantity under the radical in the value of y positive, or y is possible for every value of x, and the curve is unlimited from x = + co to = - co. If mM< H', the quantity under the ralical = /m (,V + h) 2-_p, and for all values of x between p - h, and - p -h, y is impossible; or the curve does not extend indefinitely on both sides of the axis of y. 15. (1) If ay2+bxy+ cm2-+dy + ex +tf=c /(x,y) =0 be the equation to a curve of the second degree; and - 4ac = 0, in which case the curve represents a parabola; to transform the origin to the vertex, the axes being rectangular. Transform the origin to a point,19, f3 in the curve by making x = v' + al, and y = y' + /3;.'. ay' + bx'y' + cx' + d'y' + e'x = 0; where d (a, 31) = o, d'= 2 a,3 + ba + d, and e'=b,+ 2ca+ -e. APPENDIX II. 201 Next let the equation be transformed to polar co-ordinates by putting v' = p COs 0, y' = p sinO; (a sin2O0 + b sin 0 cosO0 + ccos2O0) p + (d'sinO0 + e'cosO0) = 0, or (b sinO0 + 2c cos O)2p + Ut (d' sin 0 + e' cos 0) = 0; d' 2c and if - = tan 3; we have e b2 sec2(3 sin2 (0- ()p + 4ce' sec S cos (0- () = 0, e'cos (3 or sin2'(0 - i)p + e a cos(0........ (1) Now the polar equation to a parabola, whose latus rectum is 2L, having its axis inclined at an angle ( to the axis of x, referred to the vertex as the origin, is p2 sin2QJ -(3) =2Lpcos (0 -), or p sin2(0- (3) = 2L cos (0 - (3); hence equation (1) is the equation to a parabola from the vertex, whose axis makes an angle (() with the axis of v, fcos (3 and latus rectum 2L = - And p (x,, y)= 0 is the equation to a parabola, the co-ordinates of whose vertex are a2,,3; having its axis inclined at an angle ( to the axis of x, e'cos (3 and latus rectum = - _ _ a (2) To find a,, 31,7 L. d' 2a/3i + ba, +d 2C We have e 0, +f 2 ea, +e b b 2 (a + c) (b3, + 2ca,) - (bd + 2ce), or 2 (a + c) (b31 ~ 2Ca, 4- e) =2ae - bd = H, and bf3 + 2cal + e = el 2 (a + 0 202 GEOMETRICAL PROBLEMS, also 4cq5 (al, 3i) = 0, and e'2 - 4cp (a., 31) = - 2KT31 + (e2 - 4cf); '2 HH or e'2 = - -2K/3 + N; 4 (a - +c)2 N H 2 N aK and, = -.......() an l 2K 8 K (a + c)' 2K 8c (a + ec)2 Similarly, 2 (a + c) d'= K, and d'2- 4aC (al, (3) = - 2Ha1 +,l; M cH 2H 8a(a+ c)2 ' e' os Hcos And L =- - 2a 4a (a + c) and cos ( and tan ( may be considered to have the same sign;. V4c2 + 4ac.e sec a = - -; b b(2ae- hd) eb - 2cd K 8 8a /c (a + c)- ~ 4)/V c( + c)+ 4,/c(a +c) () 16. (1) When the axes are oblique, let the origin be transferred to a point a1, 31 as before, and the equation then reduced to polar co-ordinates;.. asin2 0+ b sin 0 sin (w - 0) + c sin2 (o - 0)} p + sin wo d' sin 0 + e' sin (wo- 0) } =; or { (a- bcost + ccos2w)sin20+sinto(b-2ccosw) sin cosO + csin2wcos2O}p + sin w (d' - e cos w) sin 0 + e' sin w cos0 = 0; hence {2 c sin w cos 0 + (b - 2c cos w) sin 02 p + 4,c sin a {(d' e cos,) sin 0 + e' sin wo cos 0; APPENDIX II. 203 d' - e' cos w 2 c sin w let = =tan;....... (1) e sin w b - 2c cos w 4_C2sin2 tw sin' to si. - sin2 (0 - ) p + 4ce' os(-) = e' sin" c and sin2 (O - S) o + e cos (0 - o) =; c cos which is the equation to a parabola whose axis makes an angle S with the axis of x, and latus rectum e' sin2 S 2L -- - -- c cos * (2) To find ai, /3 and L. d' 2c sin2 w b cos w - 2c We have -- = cos ow - = e b - 2 ccos b - 2c cos wt o2a/3 + bae + d b cos ow- 2c or b/3 + 2cat + e b - 2c cos wo 2a 2ae - b b cosw - 2c hence b b (b3, + 2cai + e) b- 2ccosw H 2a bcos w - 2c and -=be' b b - 2ccos w 2ab - 2 b cosw + 2bc 2 (a - b cos o + c) 2a' b (b- 2c cos o) b - 2ccos w and =...... (2) aud e'=- or b31+ 2cal+e +.e is the equation to the axis; e H sin. Hsin w V/c sin w also L = -- tan sin -- sin - = 2c 2ba' 2ba' /a since sin 3 may always be considered positive; H sin2 w. /c ~2 b a' 204 GEOM1ETRICAL PROBLEMS. He sin2 o (2 a e - bd) sinin2 (be - cd) sin2 w hence L- 2b \/ca' 2b V/ca' 4 /c. a - KIsin2 ctw or L............ (3). 4 Vc. a'd As in the last case e'2 - 2 K3, + N, N e'2 N H2 b' 2K 2~K 2K 8Kba'2 ' 2 b b2 and —;.. = K2, K 2 Ic i 4c N Kb'? N KI or = - - = - (a - b COS + c cos w) 2 K S 2 c2 a' 2 K 8 a' c' N K 'T N K Kfsin2 w (a' - c sin' w) ) — - 2K 8 a -c 2K 8ac c 8 a' M H Hsin2 c similarly, al 2f- 8a'a + 8a'2 17. Let b2<4ac, and the co-ordinates rectangular, to find the polar equation from the centre. Change the origin to the centre by putting x = ' + h, y = y + k;.'. ay'! + b xy' + c'2 + (h, k) 0. Again let this equation be transformed to polar co ordinates by putting,e' = p cos 0, y'= p sin 0;.'. (asin2 0 + b sin 0 cos 0 + c cos 0) p2 + (p (h, k) = 0;... (1) or [a - {(a - c) cos 0 - b sin 0 cos 0 ] p -+- p (h, k) = 0; b and if tan 2 = -—, we have a - c a- (a - c) sec 2 cos 0.cos (0 - 2c)} j2 + (h, k) - 0;...(2) but the polar equation to the ellipse, referred to the centre, whose axis-major makes an z ~ with the axis of x is { - E' sin' - E' cos 0 cos (0 - ) 2 )} = B2; APPENDIX It. hence equation (1) may be made to coincide with this equation by putting 1 - E2sin2 - a, (a - c) sec2 - ~ = ~ and = -;... (3) B2 (h, k) 9 (pOh, k) 2- 2E sin'2 2a 2 - E a+ c - -- ~;or. E2 - 2Esin a — c Er cos2 ~ a - c 2 - E2 a + c a + c a E2 (a-c)sec23 v/(a +c)2 +m ' f+m ' the negative sign being inadmissible since 2 a a a - = 1 4 —, and,2 > 1; E2 v/a2 + mi \/a" + m a' 2.. 1 - is negative and cannot = E. V/a'2 + m, 2 a' hence = 1 + ^v/,~ + rn 2 E /aP2 m (a' - v/a r + m)2 or E2= = - +;.. (4) a' + x/a'2 + m m also B 2= E-q) (h, k) E2p (h, ks) _ 2(h, k) (a - c) sec 2 x x/af' + (a' + /a' + m) hence B = 2 (h ) ( /a + )..................... (5) m - 2-1 (mh k) (a +/, +,).......... (6) Bhe (i E4ac, (- a'-va, ( + m)... (7). When b2 < 4ac, a sin0 + b sin 0 cos 0 + c cos' 0 =- (2a sin 0 + b cos0)2 + (4ac - b2) cos 0} 4a is positive; or from equation (1), ) (h, k) is negative. 206 GEOMETRICAL PROBLEMS. From Art. 9, if e' = bf3 + 2ca + e; and p (a, 3) = 4); whatever be a, /3 we have 2c e'2 - 4c4(= m {( -k)' + - (hk- g), and when a = h, 3 = k, e' becomes = 0; and E =q5 (h, k);., - (h, k) {hk-g. 4. -4cp(h,k) 2e -(hk - g), or 2 ') = ( -g. b m b Hence 2 = - (a' + a + ).............. (8) Bc - (a' - \/a'2 + m) (;............ (9) (a' - \/a"'2 + m)3 hk -g. L m............. (10) In the ellipse p (h, k) is negative, and since B is positive, equation (3) shews that in the ellipse (a - c) sec 28 is positive;.. (a -c) sec 2 = + v/a2 + m. In the hyperbola equation (3) becomes E2 (a - c) sec 2 E2 (a - c) sec 2 __ - __ or - B2 - (h, k) o BQ= (h, k) and since in this case m is positive, m /hk -g\ M - mh2 Mm - H2 0 (h- k) =- ---, P(hv, c) 2 b ) 4a 4am 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. I - Esin2 a Also B2 - 2 (h, k); 2 ca + c E' ' E (a - c) sec 2' APPENDIX II. 207 and if d( (h, k) be positive, (a - c) sec 2 is positive = /a' +; 2 a' 2 V/a'2 + m.'. 14 - -, or E =, +-, + -;.. (11) 2 a orr =2 (a - c) sec 2 m = 1+,/aa2 + mr' a'\ +Va2+ - t2V 2 +m-a>(h Ic); (13) E* -T ) or B =-(v/a2+m a') h-;........ (14) "k - A2 = - (/a2(+m + a )) (h_; 2..... 2 ((15) If P (h, k) be negative, (a -c) sec 2 2 is negative - /a'2 + m; 2 a' 2 /a' + m 1. = -, and E = Es vV/ /a'+m2m-' 2 2 E20(hk) _ 2p (h, k) (a - c) sec 2 ' /a'2 + m - a 2 ( + \/a' + m) + 6) A2J ~(h, k); (16) B2 2 (Vi/ac' + m - a') A2 - E _ 1 =- - -(h, ); (17) or B2 (a' + /a'2 + m) (h;......... (18) A2 =(a2 + m - a) (......... (19) 18. Let b2< 4ac, and the co-ordinates inclined at an angle o9, 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 - b cos c + c cos'wo) sin 0 + b'sin w sinOcos 0+ csin2 cos20 p2 + sin2 w p(h, k) =; or t (a -bcosw + ccos2w) +b'sinwsin0cosO - (a- bcosw + ccos20)cos'O p + sin w. ( (h, k) =; hence (a'- c sin2%) + b'sin o. sincos0 - (a'- 2csin2 j)cos20O p2 + sin2 t. ( (h, k) = 0;.. (a'- c sin2 w) - (a' - sin2w) sec 2 os cos ( - 2 c)} p' + sinw ( (h, k) =O;.................. (1) -- b' sin w by putting tan 2 - --.............. (2) a - 2c sin2'w...... and the polar equation to an ellipse from the centre, whose axis-major makes an angle (8) with the axis of x, is (1 2in - EK sin ) - E cos (0 - 2 =) p2 = B2; and equation (1) will coincide with this equation if 1 - E' sin2 (a' - c sin2 w) B2 sin2 w ( (h, ck) E2 (a' - 2c sin2 w) sec 23 and = - si2 (h............ Be sinz o (p (h, k) 2 - 2 E sin2 2a' - 2c sin2 w or E- _ _~_i. or - 2E2 sin'2 a' - 2 c sin2 2 - E6 af hence =; E' cos 2< a' - 2c sin w 2 a or ~= +2 T;=~ (a 2csin2J)sec 2Y APPENDIX II. 209 E2 Now in the ellipse B- is positive, and 4P (h, k) is negative;.. (a'- 2c sin2 w) sec2 is positive from equation (3); hence equation (2) gives (a'- c sin2w) sec 2~ = </(a- 2c sin2)2 sin+ /a'2 +m'; 2 Ca' or E~ 2a 1 2 + t/ m'; E. - 2_;................ (4) a' + va/1 + i' - E 2 sin2 w. Q (h, Fk) 2sin2, B2 (a! - 3~ a2M) (hi k) (a' - 2csino) sec2 = m m (a' - a' ) + ') (h k) (5) B2 (1 - E -2(a / + T)3(h, )() L; = B. E; (7) m' m (h, ) (hk - g) and as before, - -; 7m b A2 (a + a + )............ (8) 2 (a'', (/a + r' 3(hk-g).... (io) L an-d............. (9o) m b In the hyperbola, equation (3) becomes 1 -E2 sin2 a' - c sin2 w E2 (a' - 2c sinw) sec2, and B" sin2w( (h, k) B sin2 w5 (h, k) 0 21 0 GEOMETRICAL PROBLEMS. Hence, (a) if ( (h, k) be positive (a' - 2c sin2w) sec 2c is positive; ~~2 a'__ __ a 2 and =,1 + (- l in) csec = 1+ / + j32 si n_ p hk)( B72 = n S1 ( -I ) 2 (\/a'2 + m - a')(. (11) (a' - 2c sinw2) sec 2 m A B2 Bp ( m' q,(h,,k) A = (^+ m' + a')............ (12) E2 r - 1 or A8 - (a + a"+in) (h zc~g..)....(13) B o= (a - va2 + m') (.........................(14) (/3) If (p (h, k) be negative, (a' c sin'o) sec 2 - /a' + me', 2 2 aD - _,2 m 2 \/a'2 + m and = - /a + and E Va="~m' E2 Va" + n a nd E v a1'2 + m' - a'; A2 =.2 / =-2 (,//2+' (h,) =(/a2 +m'-a') hkg. (16) E;'- 1 a - m - ' b' sin w Also since tan 2 = - sin ( 9 a' - c sinw we have, (1) If b' =0, tan 2 = 0;.. =0, or 90. b' sin to v /(a' - 2c sintow (2) If aIc, tan 2i - a' 2c sin (b - c cos ) sin w t. = tanw, or c-. 2 c cos2 - b cosw 2 APPENDIX II. 211 (3) If a' - 2csin2w = 0, or a - b cosw c cos 2w = 0, tan 2 = co, and S = 45. (4) If a' - c sin2w =, or a - b cosw + c cos2w = 0, 1- E sin2 ( = 0;.'. E = cosec (. B2 (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 5msin(w - ) A ^. sin.. a, = h - 3,-; S/3 = k - _ sin' co sin2o hk-g H2 - mM also - b 2am2 K - mN 2cm2 and when m' becomes very small, (a'- v/a'2 + m)3 m -- -- - -yS, nearly; 2 K2 - m N (K2 - mN) sin4 w 8 a'3 cm2 16ca'3 henceL= K sin2 ( hence L = t - _ 1 MN2 2).....ee (A) - 2K );............... (A) KsinSw and when m vanishes L = -- _a. 4,/ca'3............... (B) a;- b cosw + c cos 2wc - V/a' + m' Also tan b' - si b sin o a - 2csine - a'+ 2< a'j or - = b' sin w 2csino / m 2 + ); 0 2 212 GEOMETRICAL PROBLEMS. a_, q \ a/'2 + a ( 2a cosec 2 = - =; b' sin o b' sin tan s c sin o - b'sinw sin2$= x -; 2cosec2 b \ 4ac, m a' +2 a c sin2 w m m a a or sin2 in- +o - or -ld- 7 ' 1 + a' a 4 \ 2c 2a c sin2w VI K /K2~f i'mN m' mS ( m V 2 I K2 +4'a' 4a'2 J =m - t - - i -.1 + - - (a - b cos w + c cos 2 w) hKe K KN mN or 3) - - -,- (a - b coso w)... (C) 2K~ 8 aac Similarly, a, = - 8- 2- (c - bcos + a cos )).... (D) When the two axes are tangents to the curve, M. = 0, hk - g mh2 _ M mk,2 _ N N= 0; and = - b 2am 2cm! hk-g _ h2 k2 ~ 2a 2c - either of which expressions may be put for - in the values of ~, 2B,~ L.b values of A, B, L. 19. To determine the position of the axis, the co-ordi APPENDIX LI.21 213 nates of the vertex and the latus rectum of the parabola whose equation is a y2 - bcvy + cx2 - dy - exr + f = qp (x~, y) = 0'; where b = 4av; d-2= 4af, and e 2= 4cf; the co-ordinate axes being inclined to each other at a given angle v Change the origin to a point a, f3 in the curve; and transform the equation to polar co-ordinates; ~~~asin20-bsi~~~~~~~~~~~~~~~~~~n 0sin (w -0) c sin'2(w~)} ~ sin {.) d' sin 0 + e' sin (w - 0) = 0; where d'= 2af- ba — d; el'=- bf3~2ca -e; and (P (a,3) = 0; or i a+ b cos w + c cos'n) sin2 0 - sin w (b + 2 c cos w) sinO0 cos90 + c sin~w coslo p + sin w I (d -'ecos w) sin 0 +-e'sin wcos0 0; and since b2 = 4 ac, { (b + 2 c cos w) sinO0 - 2 c sin w cos 02 p ~ 4cesin w { (d' - e' cosw() sin 0~+ e' sin w cos 0}. Lt d' - e' cos w = 2sin w e'sin w b + 2c cos wj 4c2sisin2i(0 )p 4ce'sin.2 which is the equation to a parabola whose axis makes an angle ~ with the axis of x, and latus rectum e'si n' 2 e' sinw 2L =- Sins d' 2aJ3 -ba -d 2c sin& bcos w + 2c Also -:Cos W e' 2ea -l)3 - e b +2cecos W b +2c cosw therefore by reduction 2 (a.b cos w+ c) (bf - 92e a + e) = db + 2 ae + (2 cd ~ he) cosw; 214 GEOMETRICAL PROBLEMS. but db = 2ae, and 2cd= be;. - 2 (a + cosw + c) e' = 2d (b + 2ccosw); 2 d sin w sin ~ hence 2L =, a + b cos wA + c v/. sin w and sin= -; /a + b cos W + d V/c sin2 W (a + b cos w + c) (1 The axes of x and y are both tangents to the parabola; and if a1, b, be the distances from the origin at which the curve meets the axes, al =/2 = /; c 2a a d / b, b 2b, or -=2615 =- -= -; a a a1 a al d /e b, - C sin2 a 2b -l sin2w a a a1 b 2 b, b 1 - cos + - 1+ — cos)+ - a a/ a, a, j 2 al b2 sin2, (a2 + 2 ab, cos w + b' (2) 2) Hence if AB, AC (fig. 165) be two tangents to a parabola meeting in A, and CB be bisected in D, since (2 AD)2 = a12 + 2 a b, cos a, + bi2, 2AB. 4AC. sin2 o (A AABC)2 8AD3 JD' A" ( The equation to the curve may be put under the form l /c /a 2 a? V + yv -1) 4 x y; f ft — ~l f APPENDIX IL. 215 B, y i ^xy or ( + ( -- 1 - hence V/- A ) = 1 (/3) 4 a, b' And if g = 90, the latus rectum = ~ + ~' (7) (alr + bl)2 20. To find a, 3 the co-ordinates of the vertex. d12 - 4aq (a, 0/) (2bd + 4e) a; or d '= 4bda; e" - 4cp (a, 3) (2be + 4cd) 3;.-. e12 = be[3; and - (a + b cos co + c) e' = (b + 2c cos c) d; - (a + b cos w + c) d' = (b cos 4 + c) d; d 2c b 2 - +- - cos w d (b cos w + 2C)2 a \a a a =9 4 b 1( + -- cos + o ao a a a,2 (b, + al cos t) (aolp + 2 a bl cos Cw + bj2)2 b, a, (,1 + b, cos ) )2 Similarly, 3 = (a,2 - 2a,b, osco + b 2)2 I a 0 / +l4 COS W \ 2 (k) If ~ _ =t, then -; hence if m be v b a 1 + m cos / constant, - is constant; and if parabolas be drawn touching a two given lines AB, 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 A. 21. To find al, /13 the co-ordinates of the focus. L sin (w - S) d c/ac sin2 w a\ - a 4 -+- a + 2 2 sin w 2 (a + b cos w + c)" 216 GEOMETRICAL PROBLEMS. 4cd (a + b cos w + c) cd or al 4b (a + b cos w + c)8 b (a + b cos w + c) b26 al, 26 b 2bl bl2 1 + -- cos +aC a alj b12 aL hence a, = - --..I.......... (1) a, + 2 a, b, cos (w + bl a12 b1 Similarly, 3= a + 2ab, os +b (2) ale + 2 a c b cos + b.............. /3 a1 Ca, /1 and or if m; -;.........(3) ai bo b, al 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 A. 22. If S (fig. 165) be the focus, and AS - pi, bisect BC in D, then pj a2 +- 22a 31 cos w + 31 = (bi2 + 2alhl (cosw + al2)- + (a~ + 2aCTbl cos w + b6/)" al2 b1,2 2or p:^- a a1 - 3bl;...... (4) aor Ca+,aibl cos o + b6' (AlB. A.4C)2 hence 4AS (B A.C) or A.AD = AB.AC;...(5) AD2 and if SE, SF be drawn parallel to AC, AB respectively; AS' = AE. AB = AF. AC and a circle may be described round the points B, E, F, C.............................. (6) 23. When a parabola slides between two straight lines inclined at a given angle w, to find the locus of the focus. 2 alb" sin2w 2Pl3 sin2, 2a1, 3 sin 2 (a, + 2 albI costw + b62)i alb1 p APPENDIX II. 217 p2 4 sin4 at232 LZ 1 2 coos w 1 4 sin4 w al, a 1/1 / 2 L2 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, a, b, (b, + al cos tw) b1 a1 (a, + bl cos w)s then m 4; y= -; 2 a2 b2" sin" w L -; rl3 where r2 a12 + 2a1b1 cos w + b,; and if - m, we have' al L? (m + cos w)2 L 2sin2w (1 +2mcos m2)' y (1 + m cos o)2 L 2 sin2S m /1 + 2 m cos w + m" 2w (m + cos t))2 or - a 1 + 2m cos + m2 ~I2y 2 - + COS f 2 Y ) m (i+cosw )i L where a - sin2 sin to hence m + cos = / (,/2 2 / q,2 2 - + - V2+ sin2 w; \a a a -- + cos / = in m a' a a 218 GEOSMETRICAL PROBLEMS. { V ' - - / + si' - COS} a \a a" X { / y / V ~sinlw) -to COS} l ra a \a a2 25. To find the equation to a conic section passing through five given points A, B, C, D, E. Produce BA, CD (fig. 166) to meet in 0; and let OAB, ODC be the co-ordinate axes; 0A = a, OD b, OB = a, OC = b'; then the equation to the conic section is Xoc y + - + -+ -1 = 0; a b aI b I and in order that this may pass through the point E whose co-ordinates are a", b", we have (a b \a b \ Xa"b" + b- ) j- -I)=o, sr ( ---f ~ a b ) b W —)-, or (+_4 -_) I + -1 xy-a"b"( -+ —1 (+1,- I =o, \a b a \a b \a b j 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 -- - 1 - — 1I + A=y0;$ \a b a' b or Ay2+ Bvy+- C +D y+Ew+ 1 0; and if X be assumed so that B2 = 4AC, we have M" 2xy y2 /I 1 I 1 whih, =s 2 + + Y (-+a + = O09 aa Vaa'bb' f b aab b which is the equation required. APPENDIX II. 219 Also the equation to a line through the origin parallel to 2C the axis of the parabola is y +- = 0, (Art. 11, Equation 6). Therefore the line y = = // -a x 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 - y + -x y:\ a b \a b ) and if h, k be the co-ordinates of the centre, the equation from the centre becomes {a (a b \ b' aa b at 1 h k 1/ h k \ or \h + - y - + j - 1 - 1) 0; b a a ba b I/A k \ l {h k \ Xk+- + -1 +,- -+ - 1 =0; a b' a \a b; /k h\ {h k k h h k \b ai \a \b a' \a b 2 k2 2h2 /l 1\ - / 1\ or — - + - +7 0 (a) bb b' aa' + \a a which is the equation to a conic section, the co-ordinates of a + a' b + b' whose centre are - -- a - 4 4 220 GEOMETRICAL PROBLEMS. b b' If - = - the equation becomes a straight line a a l/h k I h k.- +- 1 - + -+ - - I =0; or b a b I a b 2h 2/k b + b' 2h 2k +~ +07 ~7~.~; - - 7+ ' ' 0 b a' bb bb' a + a b + b 27. If u - 0, v o, w = o be the equations to three straight lines, then /u + -\/yav + \/ow = 0 will be the equation to a conic section which touches the three straight lines, where a and 3 are arbitrary constants. Since V/u + v/av + /,3w = 0, u = av + /w - 2 V af /v o;... (u - av -- 3w)2= 4avw, or u2 + a2v2 + 3W2 - 2auv - a 2uw - 2 w 2af vw = 0; and when u = 0, a2v2 + /32W)2 - 2avw = o; or av - 3 = 0; therefore the straight line u = 0 will meet the conic section in the straight line whose equation is av - t3w = 0; or in one point only; hence u = o touches the conic section. Similarly the straight lines v = o, w = o will touch the conic section. 28. If t = o, v = o, w =0 be the equations to the three lines BC, AC, AB (fig. 196), which touch a conic section in the points a, b, c respectively; since av - 3w = 0 is the equation to a straight line passing through A the intersection of v = o, w = 0, and also through the point of contact with u = 0, it is the equation to Aa. Similarly the equations to Bb, Cc are u - /3w = 0, u - av = 0; and at the point of intersection of Bb, Cc, (u - 3Co) - (ut - av) = 0, or av - 3w 0o, which is a point in Aa; therefore A a, Bb, Cc meet in the same point. APPENDIX II. 221 29. To find the condition that the straight line whose equation is u + Av + Bw = 0, may touch the conic section whose equation is V/u + \/av + /3ow = o. We have u - av + (3w + 2 \/x-t /v w; hence at the points of intersection with the line u + Av + Bw = 0, (A + a)v + (B + f3)w + 2 t/al3 /vw 0; (2) this will in general, when combined with u + Av + Bw = 0, give two values of x, and intersect the curve in two points; but when equation (2) is a complete square, \/(A + a)v + V/(B + ) w== o, or ( + a)v = (B +3)w; therefore x is determined by a simple equation and has only one value; in which case equation (1) becomes that of a tangent. Hence (A + a) (B + /3) = a; or a + + 1 =0, which is the condition required*. 30. To find the condition that - 1 may be the a / equation to a tangent to the curve ay2 + bxy + cx2 + dy + em + f = 0 referred to two tangents as axes. a3+ + bar Y + + ea +f= 0; or (a3'-ba.+ca2) () -(2a2-ba+d/-ea)-+a2dl3+f=O; (1) \-/ a and in order that - + - - 1 = 0 may be a tangent, - must ap Oa 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." 222 GEOMETRICAL PROBLEMIS..'. (2a /3 - baf+d3 - ea)'= 4(a/3 - ba/3+ ca2) (a/y+ d3 +f); or (b' - 4 ac) a2f - 2(2 a e - bd) a -2 (2cd - be) a3 +,2 (2bf- de)af3 + (d2- 4af)/32 + (e2 - 4cf)a2 = 0; and when the axes of x and y are tangents d' - 4af= 0, e2 - 4cf = 0;.. af 2 - 2kD a -J 2g = 0,............ (2) which is the condition required. When b" - 4ac= 0, we have H3 + Ka - G = 0.... (3) But 113 + KTa - 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 E, F; EF will be a tangent to the parabola. 31. If a conic section touches four given straight lines, viz. the axes of a, y, and the two straight lines - + = 0 a y - + - - 1 =0; to find the locus of the centre. a / From equation (2) we have a3 - 2h 3 - 2k a + 2g = 0; a'/'- 2h/' - 2a' +2 2g = 0; (a'- /3) - 2(3' - 3)h - 2(a - a)k 0; (4) which gives a relation between h and k, and is evidently that a a i p of a straight line passing through the points - 2, 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,/3, 7, 3, e, let any four of them, as a,/3, 7, form a quadrilateral figure q,; the centre will be in the line 8, joining the middle points of the diagonals of q; similarly, let e and any three of the others, as a, f3, 7 form a quadrilateral figure q2; the centre will be in APPENDIX II. 223 the line E2 joining the middle points of the diagonals of q.; therefore the intersection of,9 2 will be the centre of the conic section. 32. From Art. 11. If x, y be the co-ordinates of the focus of a parabola, Kx + Hy - G = 2Ky cos; and Hx = Ky; and Ka+ H/3 - G= 0, (Equation 3, Art. 30); K. (x(- a) + H(y - /3) = - 2Kycosw, or - (x - a) + y - 2X = cos w; Y.. x + 2 cosw xy + y- - ax - 3y = 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 AB, 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 AB, AC. From equation (3) it appears that if the curve be a parabola, the locus of D will be a straight line, viz. the line joining the points of contact of the two tangents AB, AC. af (/3) Since -, 3 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 I y x y I y - + - - -+ — 1 09 -, + -- 1 - o a hb a b' 224 GEOMETRICAL PROBLEMS. or u, ' = O, = u" = o be the equations to the three remaining straight lines; then the equation to the conic section is X/u + v/au-' + /3tU = 0; where a and 3 are constants to be determined. A A B B let - — +, +, + = X0 o, 1 + A + B =; a a a b b' b" b 1.. + A ' + Bu" = Ax, and A=; 1 1 b" b' 1 1 b' b I 1' b" b' Similarly, if 1 1 a a" On' 1 1', a a a a B' -, 1 1 a" a' u +A'u' + B'u" = Xy; and when the axes of x and y are tangents, (Art. 29) a f3 a / -+-+ 1 = o A + B+ 1 = 0o A' 4+ a or + I1 a a" + 1 -+ = 0, 31 _ _ _I I - I I a a a a a (3 1 ---- + - + = 0, an f1 1a a1 I [b b"J \b' bJ {b" b'J and if a = a' (a -a 7) (b b) 9 (a a )) (b -b ' )3 = fa bab b 1 1\ /I 1 I a a') b" b'= p ab ( ");(b b- ) (b+P =0', a -a '(a ) ( -) or (a' - P) b-b + (3 -+ P-) =0, APPENDIX II. 225 (a' - P) (!=. - +(' -P) (2 ) =0; a a a a).'. a -P=o, 3' -P=, or a'= P, t' =P, (a a,) b b") (b b') (a a a.. a=, '3= a a (b' b') \a' a) bi b') hence / -+) (t 7-7) (+- 1) // 1 \/I 1 \ a y +/ (1 a+) (1 b) ( y ) a a/2 ) (b b)a b" = 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- o y y ordinate axes; and let -+ - - 1 =0 + - 1 = 0, be the a b a b 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 /V y + - 1 +Va + y = 0............ (1) and - + -1 =( + - 1 + + By; a b' a b 1 1 1 1...-=- +A,,-=+ B, a a b b or = (- B = - B; \a a + b+ b hence - + - 1 = 0; (Art. 29). P 226 GEOMETRICAL PROBLEMS, and Vaa + /3b" + / -- = 0; a b from which equations a and /3 are determined, and these values when substituted in equation (1) will give the equation required. There will in general be two conic sections corresponding to the two values of a and/3; but when - + — 1 0, or a b the point a", b" is in one of the given straight lines, a, 3 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 / - + + la + \//y = O (1) a b be the equation to the conic section; and a', b'; a", b" the co-ordinates of the given points; then /a +,- 1 + \/.ta + a"13b' =0; a b - /+ -1+ / aa" + V/3 = b a b 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 coordinate axes, then if - + - 1 = 0, be the equation to the a 3 chord of contact of the two tangents, the equation to the conic section is a+ Y section i +1 + Xxy = o, where a, 3;, X are to be determined. If a, b; a', b'; a", b"; be the given points, APPENDIX II. 227 + b i)2 +\ab 0, (a- - ) +X a =b'0; he a b /ab fa ' \b hence - + - V 1 T -+ - 1 /na' '/-bb' /J'b' ab(1 o a - \/ ab' -V /a b V * /aa" N/b b V/a" " Lab Similarly, -.., av (2) a v v/ab" -Vab )a. ( from which equations -, and may be determined, and the a p equation to the conic section is ( y (a 2 b 2 ab _ + +- I = y 0o. From equation (1) it appears that when a conic section passes through two given points, and touches two given lines AB, 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 ab= a'b'; in which case 3 = 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. 37. 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 x; and let the equations to the four lines joining the given points be I y v - ~-+" = =f-0,, + -1= = 09 &c.; a b a b then the equation to the conic section is u' + Xu' = = y2 + Bxy + C2 + Dy + Ex + F; and since the axis of x is a tangent, E2 = 4 CF) 1+ a'X; (a 1 ( 1 butC= I/i+,a -E'a-""-I+ l + F,) = X; aa a a \a a \a a 228 GEOMETRICAL PROBLEMS.. {( + ) +x + )}2=+ ), + 4{aa"a}; (; ---,) X2+2{&+ 1) (1, + 4) -2 (,I,,, + 47)}x+ (5) =o; [\aa a a a"a a a" aa a a /I 1 \X 1I 1- 1 \2.. +. (-,,)- 2+2(+ + -,. + = 20; - -,, +,,,,, Ya a a/a aa a' a a a a a a 1a 2 \ a a a"a aI a a (a aI a f1 3 /1 1 (^ 1^/1 12 /1 3 2* X+77 1 ---T, +2 ---- " —, ---lX+ -— [ 0; a a \) a a a _ a a a l I l _ 1/1 1\ / 1 1 Lor the r ln j n th gv p t m a a \a a I \a a1 \a which determines two values of X. Since XX' p of there are evidently two curves whose equations are ut tu + Xu "= o 0 and uu"k + --. u U'" o=0.;\a a x 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 in P~, Pg, P,, P4 respectively (fig. 188); and let A be the required point of contact; then (Art. 54.) if A be supposed for convenience to the left of P, APPENDIX II. 229 1 1 1 1 AP, + ~AP = A- P + -P1 1 1 1 or AP - o A Ar PI P2 = 2P4 AP3 hence PP2 P3P4 hence AP1. AP2 AP3. AP4 or if AP1 = x, P1PS = a2, P P3 = a3, P1P4 = a4; a, ag3 - a4 x (? + al) (x + a3) (xI + a4) - a2a4 Va2a4 (a -a3) (a4 - a3) a2 + a4 - a3 (a2 + a4 - a3) 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 x be negative, A will be to the right of PI. 37. (i3) 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 QN parallel to PM, and make z SPT= / TPM, an z TQS = Z TQN; then PS, QS will intersect in S the focus of the parabola. Draw SY, SZ perpendicular to TP, TQ respectively; then Y and T are points in the tangent at the vertex; join YZ and draw SV perpendicular to YZ; V will be the vertex required. 37. (y) To describe the two parabolas which pass through four given points. Let A, B, C, D (fig. 211) be the four given points, produce 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 D intersect in FE; (Art. 62); and Art. 25. Cor. 1. y = A /-; x a-a 230 GEOMETRICAL PROBLEMS. is the equation to the two lines parallel to the diameters of the two parabolas; hence if FK/= FK' be a mean proportional between FA and FB, and FL a mean proportional between FD and FC, the axes of the two parabolas will be parallel to KL, 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 lHD, HA at the points D, A, and KI'L the direction 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. 37. (S) 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 E; and AD, BC to meet in F; 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 respectively; then YZ will be a tangent at the vertex; and if SV be drawn perpendicular to YZ, V will be the vertex required. 38. Let y + ml + c1 = O, y + mx + C2 = 0, y+ m,3x+c=o y+m4+ c=o be the equations to four sides of a quadrilateral figure inscribed in a conic section whose equation is ay2 + bxy + cx2 + dy + ex +f = 0, then the equation to the conic section may be put under the form (Ym c(y +m+ C,m) +c) (y+mS + c2) (y+m 4 4) = 0; (A) or (1 + X) y2 + {(n1 + 3) X (m2 + M4)} y + (m,m + \Xm"m,) X2 + {cl + C, + X (c, + c4) y + {m13+ 4n mc1 + { (m2c4 4++ 4 m2)} +4- 13 4+ \ Xc:C = 0o; APPENDIX II. 231 and in order that this may coincide with the given equation to the curve, we must have the five following relations: a {(mn + ma) + X (mm + m4)} = b (1 + X); (1) a (mm,^ + Xmm4) = C ( + X); (2) a {1 + C3 + X (c2 + 4) = d(1 + X); (3) a {clm, + c3'm + X (cm4 + c4m2)} = e (1 + X); (4) a (cc3c + Xc2c) =f(1 + X). (5) Equation (A) equally represents the equation to the conic section when y + m2x + c,= O y 4- m40 + c4 = O, are the equations to the diagonals of the quadrilateral. 39. If any three of the quantities nm, m2, m3, m4 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 AB 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'A' parallel to DA, and A'B', B'C' parallel to AB, BC; the remaining C'D' 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 AB, 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 232 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 AB, BC; then ac will be parallel to CD; for abc may be considered as a quadrilateral figure whose evanescent side at a 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 AD, 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 respectively; 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,, c2, c3, c4 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 b 44. If mi + m3 = -; equation (1) gives m2 + m4=-; a a and when b = 0, the axis of v is parallel to one of the axes of the conic section; hence if m3 = - m,, m4= - 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 x and y respectively, then b = 0, and if APPENDIX II. 233 pq be any chord parallel to CD, and y = mrnx, y = m3n be the equations to Cp, Cq respectively; since pq is bisected by CP, n3 = - mi; hence m4 = - 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 inscribed 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 quadrilateral 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 A, 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 AB, (fig. 172) the second and fourth sides will be in the directions of the tangents at A and B, and if z1 = mn = -; m2 + m4 will be 2a constant for all chords parallel to AB; but if PP' be drawn through C parallel to AB, since the tangents at P and P' are parallel, m, = mn, and mr + m4 cannot = - unless the tangent a at P is perpendicular to the axis of x; and CT is parallel to the tangent at P; hence m, + m4 = cot ATC - cot BTC is constant for all chords parallel to AB. 48. If the origin be changed to any point in the conic section, a, b, c remain unchanged, and f= 0; hence if ml, m., 234 GEOMETRICAL PROBLEMS. mW3 be given; m4 and X will be known from equations (1), (2); and -- - X from the equation (5); but if pI, pI, p3, p4 be C2 *4 the perpendiculars from the origin upon the four sides Ct Cs P,==-; P == — &c; pV1 + m; 1 & + PIPP PiP3 V/(1 + m22) (1 + -42) P2P4 (1 + i12) (1 + m32) 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, P1,p will bear an invariable ratio to P2P41 (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 + 3 = -; then c2 + 4=-; and when d = o, the a a origin is the middle point of the chord which is the axis of y; and if c3 = - Ce, 4 = - c2. 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 A; the two remaining sides RQ, PS will meet the chord in two points C, c equidistant from A. In the same manner it may be shewn that PR, QS meet the chord in two points equidistant from A. (2) If PQ, SR be any two chords passing through A the middle point of BD; (fig. 174) RQ, SP will meet the chord BD in two points C, c equidistant from A. Similarly QS, PR will meet BD in two points C', c' equidistant from A. 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 A; and if PQ, SR meet the tangent in two points E, e equidistant from A; PS, RQ will meet the tangent in two points C, 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 Q; and if PQ, QS meet the chord or tangent CAc in two points E, e equidistant from A, SP and the tangent at Q will meet the chord or tangent in two points C, c equidistant from A. 5.()f 50. (1) If c1, =, then c,, =-; and if A be the a a origin, and ABC (fig. 177) the axis of y, AB. AC = f; hence 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 points G, H such that AG. AH = AB. AC. Similarly if the diagonals SQ, PR meet the line ABC in G', H'; A G'. AH' = AB. AC. (2) If the points E, F coincide in 0 (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 AG.AH= AO2. (.) 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 A 02 is constant for all chords drawn through 0. 236 GEOMETRICAL PROBLEMS. (4) If c1 = 0, then c3 = oo; the first side QP (fig. 180) passes through the origin A, and the third side RS is parallel to ABC; hence if PS, QR meet ABC in G and H, 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. AH is constant for all positions of PR. 51. If d = 0, the origin is the middle point of a chord; f and if clc = —, then c, + 3c + X(c2 + C4) = 0; C0e3 + XCc4 = - ceC(1 + X); or cl(c2 + C3) + X2 (c1 +c4) = 0;. (. (C -c + C) - C (Ci + 0) (- ==0; hence c, (c, + C3) (c, + C4) = 2 (c +C3) (c, + 4), or clc, (c - Cl) = C3e4(C2 - Cl); '. 63e4 = C12 - - hence if A be the middle of the chord BD (fig. 182), and PQ, QR meet AB in C, C' such that AC. AC' = AB2; then RS, PS will meet AB in c, c' such that A c. A c' = AB. (2) If c, = 0, or the first chord PQ passes through A, (fig. 183) ce = oo, and QR is parallel to AB; hence if RS, SP meet AB in c, c'; Ac. Ac' = AB for all positions of SP, 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 Ac. Ac' = AB2 and is invariable for every position of S. APPENDIX II. 237 52. From equations (1) and (2). l m+ +m b m+ +m4 b If = -; then = -; but if ml = O, m, = o, m1 m c m2n 4 c or the first and third lines are parallel to the axis of x, then 1 1 b -+- -; hence if PQ, RS be two parallel chords; (fig. m, m4 c 185) the sum of 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. ~C JR C3 d c2 + C4 d 53. If -; then = -; and since l C3 f C2c4 f d (l_ 1 1 I +; if-+1=- (+; AB ACJ el cC \AB ACC I 1. then - + - is constant; and will be the same whether we C2 C4 take the second and fourth sides or the diagonals of the quadrilateral. (2) If cl = 0, then C3 = 0, and the first and third chords PQ, RS (fig. 186) pass through the origin A; therefore PS, I 1 QR meet ABC in two points E, D such that A — + is 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 fixed point A will meet a given line ABC in 1 1 two points T, T', such that A-T + A- is invariable. 54. If d = 0, and f= 0; the axis of y becomes a tangent 1 1 1 1: to the curve, and from equations (3) and (5) -- + - = - + -; e 3 c2 C4 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 P1, P2, P3, P4 respectively, and the direction AP1 be considered positive, 1 1 1 1 P AP+ = -- + - and if the diagonals meet the tanI P A, AP, AP-4' gent in Q2, Q4, 1 1 1 1 1 1 API AP3 AP4 I P2 A Q2 A Q4 55. If c and f= 0, the equation to the conic section becomes ay2 + bxy + dy + ex = 0; which is that of a hyperoola having one of its asymptotes parallel to the axis of x; and C1 C3 C2 C4 1. 3 =, -; or if A be the origin in the curve; and the 1 m3 ms M2 nm four sides of a quadrilateral figure meet a line passing through A and parallel to one of the asymptotes in P,, P2, P3, P4; then AP,. AP = AP2. AP4; and if the diagonals of the quadrilateral meet the same line in Q2, Q4; AQ,. AQ4 = AP. AP4. 56. If a = 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, mm, - mmno = 0; in this case the axes are parallel to the two asymptotes. Hence if a quadrilateral figure be inscribed in a rectangular hyperbola, the product of the tangents of the angles which the first and third sides make with either of its asymptotes, 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 = o, the axis of y becomes one of the asymptotes, and c, + c3 = c, + c4; hence if four sides of a quadrilateral figure inscribed in a hyperbola meet one of its asymptotes in the points PI, P2, P3, P4; P, + AP A= AP, + AP;.'. PP,=PP,; also PIP4= P2P,; 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 Q2,Q4; then AP, + AP, = 4AQ, + AQ,: or P1Q2= P3Q4; similarly P QG = P4 Q. 58. If a and f= 0, c1c3 = c2c4. 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 P1, P2, P3, P4, and Q2, Q, respectively; then AP,. 1A3 AP.. AP4 = AQ2. AQ; and when P1, P3 are fixed points in the axis of y, AP2. AP4, and AQ,. AQ4 are invariable. 59. If b and c = o, the curve becomes a parabola with the axis of v parallel to the axis of the parabola; and from 1 1 1 1 equations (1), (2) we have - + - = - + -. Hence the Ml m3 M2 M4 sums of the cotangents of the angles which each pair of opposite sides, and the diagonals of a quadrilateral figure inscribed in a parabola make with its axis in the same direction are respectively equal. 60. Since the form of equations (1), (2) which connect Wm, m2, m3, m4 is similar to the form of equations (3), (5) which connect ce, c2, c3, c4; and the quantities are similarly involved in equation (4); if any relation be found between mi, m25, n3, m4; a corresponding relation may be determined between c,, c0, e3, C4; 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 mn, m, are such that mnm3, or 1 1 in, + m3, or - + - is constant, the locus of the intersection 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 cl, c3 from the centre such that cc3, 240 GEOMETRICAL PROBLEMS. I 1 or C1 + C3, or - + - is constant, the locus of the intersection of Cl 03 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 A to a conic section, the straight lines joining the points of intersection of MP, NQ, and of MQ, NP 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 + m,1 = 0; y+ 2 m2 +c- C=0; y + mv = O; y+ m4V +4- = 0; hence the equation to the conic section is (y 4 mI) (y + m3) + X\(y + M2a+ - C2)(y - + 4a - = 0; (1) and in order that this may be identical with the given equation to the conic section ay2 + bxy + cx2 ~ dy exf =; we must have aX(c2 + e4) = d(1 + X); aX(c2m4 + c4m2)= e( + X); aXcc4 =f(1 + X); 1 1 d m2 m4 e +- - _; - -- +- --; C2 Ca f 02 1C e but for the intersection of y + M2at + C2 = 0; y + rn1x + C4 = 0, we have (\ 1 1 \ fm2 mA4 d e (+)y+(h+ - X ) 2 =0, or -y + v = 0; \C2 C4 \C2 C f +F hence the second and fourth sides intersect in the straight line dy + e x + 2f = o, which is independent of mn, m,; and since equation (1) will equally represent the equation to the conic section when y + m2x + c2 = 0, y + m4V + C4 = 0 are the equations to NP, MQ; the locus of S the intersection of MQ, NP is the straight line dy + ex + 2f = 0; and the locus R of the intersection of MP, NQ is the same straight line; therefore APPENDIX II. 241 dy + ex + 2f= 0 is the equation to the straight line joining R and S. Since RS is independent of the position of AMAN, and APQ, let P and Q move up to T, and M and N to T', so that AT, AT' are tangents drawn from A; then aP, NQ become coincident with the chord of contact TT'; and since they always meet in the same straight line, KS 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 = o. 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. W~hen tangents are drawn from A, tie chord of contact is RS passing through S; similarly when tangents are drawn from R the chord of contact is AS 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 2t2 GEOMETRICAL PROBLEMS. 66. To draw a pair of tangents to a conic section from a given point A without it; and also to find the chord of contact. Through A draw any two chords AMN, APQ; join MP, NQ intersecting in R; join A2Q, 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 T, T'; 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 T; join TA, 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 tangents 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 T, the tangents at A and B pass through T; hence TA is a tangent at the point A. 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 A, 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 K; and let AC, BD meet in L; join HK, HL, KL; then the tangents at A and C meet in HK. (Art. 64.) Similarly if EC, AD meet in K', and CD, EA meet in H'; the tangents at A and C meet in H'K'. Hence if HK, H'K' intersect in T, TA, TC will be tangents at the points A and C. Again, the tangents at A and D meet in HL; and the tangents at A and B meet in KL; hence if TA meet in HL, APPENDIX II. 243 KL in T', T"; T'A, T'D will be tangents at A and D; and T"A, T"B tangents at A and B: and four tangents TA, T"B, TC, T'D have been drawn at the points A, B, C, D of the conic section which passes through the five points A, B, C, D, E. 68. If RST' be the chord of contact of a pair of tangents drawn from a point A without a conic section (fig. 189); AMlN any chord passing through A, 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 AM = MB; draw BSTR perpendicular to MB; then BR is the chord of contact of two tangents drawn from A; and since AM does not meet the curve again, SP must be drawn parallel to AM; and the locus of the intersection 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 AB; hence the point F is fixed. (2) If a 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 A. 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 F. In this case AF = 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 ai, N, and P, Q respectively; the intersection R of MP, ANQ will be in the plane of contact of the enveloping cone whose vertex is A; and the locus of R 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 A; and this will be a straight line on the plane of contact of the enveloping cone whose vertex is A; and the same will be true when the lines AMN, APQ are drawn through A 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 S the point of intersection of MQ, NP is the plane of contact of the enveloping cone whose vertex is A. 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, /, y,, e; and let their equations be respectively Ua = O~ u 0,u= O, 23 = O~, Ue = O ~ U = 0; APPENDIX II. 245 also let u1 = 0 be the equation to the diagonal BE; and U= 0 the equation to the conic section; hence uGu. + XI Ua = mI U;, Uu + X2uu3B = m2 U; (** (nm?, - maui) U = X1UoaUuu, - X\UpUaUS. (1) If uz and u3 = 0, the values of w and y will be the coordinates of the point of intersection of the sides a and ~; and from equation (1), (mlu - mun) U = 0; but U cannot = o, since the intersection of a and S is not a point in the conic section;.'. nuy - -mu = 0; or a and S intersect in the straight line whose equation is mu -m,2u = 0. Similarly, if in equation (1) we put t3 and u = o, the lines 3 and e intersect in the straight line muz, - mnu-z= o. Now mu - m,ug-= 0 is a straight line passing through y, '; hence the intersections of a, S; /3, e; y, e lie in the same straight line whose equation is mn,~u, - m2,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 AD, BE intersect in eb; and AD, CF in ad. In like manner by varying the order of the points A, 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, c, d, e, f the points of contact; then (Art. 63) AD is the chord of contact of tangents drawn from a' the intersection of dc, 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); therefore AD, BE, CF meet in the same point. 76. Let ABCD (fig. 195) be a quadrilateral figure inscribed 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 A and C; then since the opposite sides of the hexagon intersect in a straight line, if AB, DC be produced 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 ABDCA, 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 A and D will meet in a point; or the tangents at A and D will meet in EG. Similarly, the tangents at B, C meet in EG. These properties have been already proved (Art. 63). 77. Let ABCD (fig. 195) be a quadrilateral figure circumscribing 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, c, 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 ac, bd pass through the intersection of AC, BD. (3) Let A, a, B, b, C, D be considered the angular points; then Ab, Ca, BD meet in a point. Similarly, A c, 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, c, since ABC may be considered as a hexagon whose angular points are A, c, B, a, C, b; the diagonals Aa, Bb, Cc 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 in c; c 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 AB, DC to meet in E; AD, BC to meet in F; also let AC, BD meet EF in H and K respectively; and let AC, BD intersect in G. (1) ab, dc, (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, dc intersect in AC. (2) ab, de intersect in the line EF joining the points E, F of the intersection of the pairs of tangents at the extremities of the chords ac, db (Art. 65); hence ab, do intersect in H. Similarly ad, be 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 ac, bd will meet in G, the point of intersection of AC, BD. 80. A conic section is to be inscribed in a quadrilateral figure 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 sides. Join AC, BD meeting in G; let AB, DC meet in E; and AD, BC in F; produce AC to meet EF in H and aG to meet DC in c; join Ha, He meeting BC, AD respectively in b, 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 Aa, aB, BC, CD, DE, EA may be considered as the six sides of a circumscribing hexagon; and the three diagonals D a, 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 AB. In like manner the points of contact with the remaining sides may be determined. 82. Having given five points A, 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 determined; 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 in a direction parallel to BC; bisect DF, BC: the line joining the middle points will pass through the centre. Similarly, find a chord DG parallel to EA; 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 A, B, C, D, E 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 A, B, C, D, E. 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 AKI'; B, C, D the three given points; BE the direction of a chord passing through B; E 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 A in the direction of the tangent AK', AB, BE, EC, CD, DA; and since the opposite sides meet in a straight line, if AB, 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 AB, 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 E. 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 AB, 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 cd, be the direction of a chord of the conic section 250 GEOMETRICAL PROBLEMS. which passes through c; produce ba, cd to meet in E; join EB and let be meet EB in F; 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 c in the direction cB; 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 EB in F; join Fa, which will intersect cd 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 OA, OB in given points A, B, (fig. 201). Let AD be the direction of a chord of the conic section through A; D the extremity of the chord which is to be determined; then ADCB may be considered a hexagon inscribed in the conic section whose sides are the tangent at A, AD, DC, CB, the tangent at B, and BA; and the opposite sides meet in the same straight line. Let CB meet OA 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 A, 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 A; K, K', K" (the points of intersection of AB, DE; BC, EA, and of DC and the tangent at A) lie in the same straight line; hence join KK', and let DC meet KK' in K", K"A will he the tangent required. The construction here given is more simple than that in Art. 67. 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 A 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 AT at the point A (Art. 89); and draw OG meeting AT in T; then OGT is the direction of the diameter to AF; and if OD'= Od' be taken a mean proportional between OG and T, 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 AF; and from E one of the given points, draw D'E, d'E meeting OK in H, K; take OL = 01 a mean proportional between OH and OK; then (Art. 51, 3) L and I will be the extremities of the diameter. 92. Find the extremities of a chord GH drawn in any given direction. Through the given point A, (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 D'E, d'E 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 semiconjugate diameters; draw PT parallel to CD which is therefore a tangent at P; let CP = a', CD = b', ACP= 0, Z CPT - L PCD' = a; then b'cos (a - 0) = a' sin 0; b 252 GEOMETRICAL PROBLEMS. b b' sin (a - 0) = - a' cos 0; a.'. b'2 sin 2 (a - 0) = a'1 sin 2 0; b/2 draw PE making z EPT' = z CPT; and make PE=,; a join CE; then zCPE = - 2a; and if Z PCE=; L PEC= 2a -ap; P.E b' sin siin 20 PC a'; sin (2a-p) sin (2a-20) " or CA bisects angle PCE. CP2 Hence make Z T'PE= CPT; and PE = —; join CE, and bisect Z PCE, this will give the direction of the axis-major. The axis-major CA always falls within the acute angle PCD'; for tan 0. tan (a - 0) = 2; a.'. tan (a -0) 2 tan ( 0) <tan (- 0; hence a < -; since 0 may be taken <-. 2 2 Draw CB, PM perpendicular to CA, and PN perpendicular to CB; then if the tangent at P meets CA, CB in T', T; CA, CB are mean proportionals between CM, CT'; and CN, CT. To find the foci, take BS, BH = CA; and S, 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. 253 95. Having given five points, draw a tangent to the conic section which passes through them, from a given point without it. 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 A, B, C. If from any point P in the conic section Pal, Pbl (fig. 207) be drawn through A, B; then the straight line a,b will always touch a conic section. (Appendix I. Art. 5). Let alb1, a b2, ab,3, a4b4, a5b5, 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; D'AF', 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 A, B, C. Let P be one of the given points; through A, B, draw 254 GEOMETRICAL PROBLEMS. the chords Pa,, Pb, (Art. 82); join ab,. Similarly, we can determine a2b,, ab3,, a4b4, ab, 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 alh,, ab,, ab,, a4b4, ab,; find D,, E, 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 F'; 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. 84-88, although the conic section itself is not traced. 100. To inscribe in a given conic section a polygon whose n sides taken in order shall pass through n fixed points, A1 A42... A If (n - 1) sides of a polygon taken in order pass through (n - 1) fixed points, the nth side will always touch a conic section. (Appendix I. Art. 6). From any point P in the conic section let a polygon be formed whose sides successively pass through A,, 2A...A,,__; and let al b, be the position of the nth side; in like manner let ab,, a3b,, ab4, ab, be four more positions of the nth side, and from An draw as in Art. 97, two tangents BB,_,, B',B'_,, to the conic section which touches the five lines ab,, a,b,, a,3b, aLb4, a,b5; these will be the nth sides of the two polygons which can be inscribed in the conic section so that its sides may pass through the n given points; draw B,-_ B_ through A4,-,, B_,-2 B, through A,,, &c., BaB through A3, B2B, through A,; then BB,_l will pass through A,, and B, B,...Bn; B',B,'o..B,' will be the two polygons required. 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 n sides taken in order shall pass through n 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 nth side which passes through A, is a tangent to the conic section which touches ab,, a2b2... ab,; if A, be without this conic section, we can draw two tangents, which will give two polygons. If,, is a point in the conic section, there will only be one tangent, and only one polygon can be described. If A, is within the conic section, it is impossible to draw a tangent from A,, and the problem is impossible. Hence if A be the point of contact with a, b,, and AA, be drawn, find B the extremity of the chord AA,; 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 Problems, No. 15, Thursday, Jan. 7, 1847, 1...4.) Let AD be the axis of v; then if y - mx = 0; y + m x= 0; y - m2, = 0, and y - mn4 - c = 0, be the equations to AB, AC, the tangent at A and BC respectively, the equation to the conic section is (y - mla) (y + mnx) + X(y - 2) ((y - I - c4) = 0; or (1 +X) y-X (Mn2+ml4) xy+ (Xm2m 4-m12) t- XC4Y+Xm2c4x=0; and if the equation to the conic section be ay2 + bxy + c22 + dy + ex = 0; we have XC4 d d Xm, c e X( ) =; or e, = (mn + m); and X - = C X(m -+-m b;oe~ b -X = d; 256 GEOMETRICAL PROBLEMS. e d e m. - -; hence c4 = -m —; d b b therefore the equation to CB becomes y + = m4 x +; which is the equation to a straight line passing through a fixed d e point whose co-ordinates are-; - b 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. (/3) If AD be a chord of a surface of the second order; let any plane pass through AD, and let two chords AB, AC be drawn in the plane section (which will be a curve of the second order) equally inclined to AD; 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 AB, AC; AB', AC' be drawn in two different 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 AB, 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 AD. 104. Let AD be a fixed line, AC, AB two lines equally inclined to AD; 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 1I, 257 105. If AB, AC be two chords of a conic section, to find the condition that BC may pass through a fixed point. Let y + mx =o; y +m,x =; y + m2 + c2=; and y + m4x = 0; be the equations to AB, AC, BC and the tangent at A respectively; then the equation to the conic section is (y + nms) (y + m, x) + X (y + mn2x + c2) (y + m.4) = 0; and making this coincide with the equation to the conic section ay2 + bxy cx2 + dy + ex = 0, we have a ( (m1 + m3) + X (n, + 9n4)} = b (1 + X); a (mnm, + Xm2,m) = c (1 + \); aXc, = d (1 + X); aXcmn = e (1 + X); e a1 lac,.. 4 =, and -=- ( 1; d X \d dI fac \ em, f 1 cc2 or m^ - ( 1) + ---- =_ 1 +4 = A) (1) (mi + 4nm) ad -'1 + m4 + d- d ' (2) Hence if m,m,, or ml + m3 be constant; we hIave an equation of the form 3 + m2a + C0 = 0, where /3, a are constant; and the line y + mx + c2 = 0 passes through a fixed point a, 3. Let 3 + mna + c2 = 0, be the general equation connecting 212 and c,; ac e C. + \ c=-' (rn+m,)(ar 2d1) -- c +-3)=R d - d a d R 258 GEOMETRICAL PROJBLEMoS, and eliminating 2,, an equation of the form 'm3 -- A (in1 + m3) -F B = 0, is determined where A, B are constant. If this relation holds, the line BC will pass through a fixed point;. (mi - ) (i, - A) is constant, or - - - - -. is constant. 1 1 If B=, - +- is constant. 106. If the angle BAC be a right angle, ImM = -; 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 AB 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' respectively; join BC, B'C' meeting each other in 0; then AO will be a normal at the point A. (Art. 106.) APPENDIX II 1 1. IF C'L' (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 0 be the centre of the circle; draw OH perpendicular to CL'; with centre O and radius OH describe a circle; from G draw CGgCIL touching this circle; then since CL 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 FE, FQ respectively; then C'L' will be constant for all positions of '. For z C''L' = z QFR and is constant; hence C'L' is constant. 3. 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 AB, 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 B, F, A and R, z ARP z ABF - z ABK z LACK; hence KC is parallel to FR. 260 GEOMETRICAL PROBLEMS. Again, since a circle may be described about the points C, G, F, B, z BCG -- z BFG 2 right angles == z IFQ + z BFG;.'. z IKFQ =z BCG = z BCL = z BKL; 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. I. Art. 3.) 5. To inscribe geometrically a triangle, in a hyperbola, whose sides shall pass through three fixed points. From Appendix i, Art. 4. we have, (a + a3) tit - -b3 (t1 + t2) + a3 - a = 0. L a b3 Let a3 = -; P3 = - a3;b a3 then (a 4.- a) t, t - /3 (t + to) + (a - a3) = 0. (1) a2 b, a2 b Similarly, if af =-; /3:= ba2; am=; /~= we have (a + a,2) tt, - I2 (it 4+ / ) + (C - 02) I 0; (2) (a + al) t2t3 - 3 (2 -1+ ) + (a -1a) =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 a2 b3 a2 a2 b2 a2 a2 b, a' (3 b a3 Ca2 b a 1i b al, hence if CA (fig. 146) be the semi-transverse axis; P an angular point of the triangle inscribed in the circle APD so that its sides may pass through the three points above determined; then L ACP 01; and CR = a sec 0 = the abscissa of the corresponding angle of the triangle inscribed in the hyperbola so that its three sides may pass through the three points a3, b3; a, b2; a,, b,. 6. To inscribe geometrically in a parabola, a triangle whose three sides shall pass through three given points. Substitute for yl, y2, /3 in equations (1), (2), (3), Art. 6, Append. i. the values 2at,, 2at2, 2at3 respectively; and let a (a - a3) a b O3 3 --- +- 93 a +- as Ct 4J 3 a (a - a,) a bo,= --- - a + a -a- + 2 a(a - a,) abl C -= — 7 ] ---- a + a, a + a a.. (a + a3) tt - /33 (tl + t9) + (a - 3) = 0, &c. hence t1, t,, t 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, 93; Ca, 13; al, 31, respectively. Hence if E (fig. 15) be the vertex of a parabola; B the focus; with centre E, and radius EB describe the circle BDA, and let C be one of the angular points of the triangle GEOMETRICAL PROB)LEMS~ inscribed in the circle B DA, so that its three sides nay pa ss ti rough th ree points whose co-ordinates are IQ3, a3.I2' J2 ac, P~ draw BF perpendicular to AB meeting AC produced in F; then LCEB =03 L CAB=03, and BF =2at,= y,; C) 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 a3, b, 03, &2; a bl, Hence in each of tihe 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.Frm the equations in App. i. Art. iO, it wvill easily be seen (1) thbat when a polygon is to be inscribed in ain ellipse so that its n sides may pass through n given points a,, &, 03,5 b2;,a, b,; the problem w ill be equivalent to that of inscribing in the circle described upon tbe axis-major,a polygon whose it sides talken in order shall pass through the n fixed pounts a, a a (2) When the polygon is to be inscribed in a hy perhola; let a circle be described upon the tr'ansver e axis; and in ti is circle let a polygon ble inscrllbed, wvhose ni sides sh 11l pass through -the qifxed point. 0 ia1 02 02a a,' b a3 then if P be anyv angular point of this polygon, as in Art..s, CR (flig. 146) will be equalI to tI e abscissa of the corresponding, angular poin-t of the polygon which is to be inscribed in the hyiperbola. (30) W hen the polygon is to ble inscribed in a parabola; let E (fig. 1 5) 1)0 the vertex;B th e focus; withi centre E APPENDIX III. 263 and radius BE describe a circle; and in this circle let a polygon be inscribed whose n sides shall pass through the n fixed points a, (a -- a) ab, a (a -- a) a b a - - a + a + a2 a + a2 a+ a',, a,2 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 corresponding 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 n sides of the subsidiary polygon are to pass; fromr whence the abscissa of the corresponding angular points of the polygon inscribed in the conic section; and the angular points themselves may be determined. 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An Appendix to GOODWIN'S Collection of Problems and Examples, containing the Answers, and the Solutions of some of the more difficult, of the Questions A Collection of Problems in Illustration of the Principles of Theoretical Hydrostatics. By W. WALTON, M.A., Trinity College. MW Plate I. - ~. r L 4 A 5 6 A' C 7 6 A ~ c B C 15 13 i JrA Ix 20 E A4 C B P C 21 D A B 26 c A c B 26 K D E A G B J _ __ _ _ I ~2^^\n 27 12 C F P B D B D Mc7 ed6k & TJnZoe,.lbo iO _ ___ I Plate 2. I L 32.D _c A i / ~ c 33 / AR r RQ 1'-^ ----^ A B E 37 40 C 43 / 49 / C/ C / - C X f,, —,,.1 X^B A' B 2V M 44 c X B I 52.Meaccale & F,lvwr., Jit:LO -j Plate 3. 660 1 65 — -Dew c 61 64 D__A BLI 7 66 c a c Ac B 6c9 '2 P 6f, c~~~~~ c 74! A B -9 c —= --- —~ C 76' a -B --- —— _ ---B C A A 60\ B D C 82 B 83, B c m A AMecicaZfe & 7Palmer, ZyJ-i,. 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