WHOXLE SERIES No. 34. NEW SERIES NO. I' IPublications OF THE Tlnivertit~ of lennllvania. Mathematics, No. i. Contributions to the Geometry of the Triangle. ROBERT JUDSON ALEY, A. M. Properties of the Locus r- Constant, in Space of n dimensions. PAUL RENNO HEYL, B. S. PUBLISHED FOR THE UNIVERSITY OF PENNSYLVANIA WITH THE AUTHORIZATION OF THE COMMITTEE ON PUBLICATION. PHILADELPHIA. 1897. CONTRIBUTIONS TO THE GEOMETRY OF THE TRIANGLE, BY ROBERT JUDSON ALEY, A. M., Professor of Mathematics in the University of Indiana, Bloomington, Ind. Fellow in Mathematics in the University of Pennsylvania, 1896-97, on the George Leib Harrison foundation. Accepted by the Department of Mathematics of the University of Pennsylvania, in partial fulfillment of the requirements for the degree of Doctor of Philosophy. ISOGONAL CONJUGATES. Definition.-Two lines drawn from a vertex of a triangle making equal angles with the bisector of the angle at that vertex are called isogonal conjugate lines. It is known that when three lines drawn through the vertices of a triangle are concurrent, their isogonal conjugates with respect to the angles at these vertices are also concurrent.* The two points of concurrency thus determined are called isogonal conjugate points with respect to the triangle. If I and I' are two isogonal conjugate points with respect to the triangle ABC, and if the ratios t of the distances of I from the sides a, b, and c are 1: m: n, then the ratios of the distances of I' from the sides are -: I t n * Lachlan's " Modern Pure Geometry," Art. 101. t Hereafter the ratios of the distances of a point from the sides of the triangle will be spoken of as the Ratios of the Point. t Casey's " Analytic Geometry," first edition, page 320. (3) 4 GEOMETRY OF THE TRIANGLE. Let I and I' (Fig. 1) be two isogonal conjugate points with respect to the triangle ABC. * Let us assume that II,:IIb: IIc —=l: m: n. We are to prove that! 'I I`-_ 1- a b c I m n AD and A G, BE and BH, and CF and CK are pairs of isogonal conjugate lines. The triangles FFaC and II"C are similar, as are also the triangles FFb C and IIC. f. II' FF —IC: FC, ) and IIb:FFb=-IC:FC; (2) (whence II. FF,- IIb:FFb, o 'or I: IIb =FF: FF (3) But I II=l:m. FF,: FF,=- lm, (4) (or FFb: FFa,, - Now since the lines FC and KC are isogonal conjugates, the triangles FF C and KKb C are similar, as are also the triangles KK, C and FF C. 5 f.. PFFa: KKb-FC: KC, (5 ) tand FF: KKa=FC:KC; (f whence FFa KK=FFb:KK, ( or FFb: FFa=KK:KK,. But from the similarity of triangles (7) KKa KKb —IIa: I'Ib; 1 m W8) ** -L~ia'll I 9 m' * Notation. If we are considering any point P, the feet of the perpendiculars from it to the sides of the triangle will be designated by Pa, Pb, Pc, respectively. GEOMETRY OF THE TRIANGLE. 5 In exactly the same way it may be shown that llb ^lC?m'n 3n".. ' 1 1 1 (9).'. rj I'I::. a II. ISOTOMIC CONJUGATES. Definition.-If from a vertex of a triangle two lines are drawn cutting the base of the triangle in X and X', and if XX' has the same middle point as the base of the triangle, the two lines thus drawn are called isotomic conjugate lines. If any three lines AD, BE, and CF (Fig. 2), drawn from the vertices of a triangle, are concurrent, their isotomic conjugates AG, BH, and CK are also concurrent.* The two points thus determined are called isotomic conjugate points. If T and T' are two isotomic conjugate points, and if the ratios 1 1 1 of Tarep:q:r, then the ratios of T' are:: pa2 qb2 re2 The triangles FFaC and TTaC are similar, as are also the triangles FFb C and TT, C. 1. TT: FFa- TC: FC, (1) [land TT,: FFb-TC: FC; (2) whence TT: TTb=FF.: FF. (3) But TTa: TT,=b: q, (4).F. FF:FF,=p: q. The triangles FFaB and KKaB are similar, as are also the triangles KKbA and FFbA. 5 f. FFa KK= FB: KB, (5 ) land KK.b:FFb= KA: FA. But FB-KA and KB=FA, (6).. FFa: KKa= KK: FF,. * The proof of this may be found by a direct application of Ceva's " Proposition." 6 GEOMETRY OF THE TRIANGLE. (7) Hence FF,. FFb=KK,. KKb. (8) Now KK.. a + KK. b-=2 (where J =area of triangle ABC); FF.. FF (9) also from (7) KKb FF. FFb (KK ' Substituting this in (9), we have (10) KKa. a + FF. Fb b 24. KKa (11) Hence a. KK'- 2 J KK+FF,. FF. bb=0. (12) KK 2DJ+/4 z2 - 4a. b. FFa. FF (12).. KK= 2 — a (13) or KK =4t JV2-abFF.FF, a a From (4) (14).-. We have also (15) (16). FFa FFbp: q, q.FF-p.FFb. FF,. a+ FFb. b=2, FF 2 J-FF.b a ba Substituting this in (14), we have (2A-FFb. b) (17) q =p. FFb, a (18).-. (19) Similarly FF- 2J qp bq-ap 2ap bq+ ap Substituting these values in (13) d=~ 4A2 4a:'abpq (20) KKa (ap + bq)2 a GEOMETRY OF THE TRIANGLE. 7 A a p - bQ Q' a { (ap+bq) a (ap + bq): i: (ap -- bq) a app+bq J ' 2 bq 2Jp a(ap + bq) rap+ bq (21) Similarly KKb= 2, Jap or - -. b(ap + bq) ap + bq It is readily seen that the second values are not relevant to this problem. We now have from (20) and (21) (22 KKT: KK - 2 2bq K 2 ap (22) KKa KK - a(ap+ bq) b(ap+ bq) bq ap a b _ 1 1 The triangles KKa C and T' T' C are similar, as are also the triangles KKb C and T' T' C; 23).. KIa T' Ta'=KC: T' C, ( 2) land KKb: T'Tb'KC' T'C; (24) whence T' Ta: T' Tb'KK: KK,. (25) But KK: KKb= a2p b2q (26). T' T'Tb'- a2p b~ In exactly the same way it may be shown that I I 1 1 (27) T'T: T'T -b-:, 2 1( ).. (28) T'T': T'Tb': T'T '=-.... a~p b q ' 8 GEOMETRY OF THE TRIANGLE. These relations between two isogonal conjugate or two isotomic conjugate points, enable us with ruler and compass to construct the series of points whose ratios are a: bn6: cn where n is any positive or negative integer. If we start with the centre of the inscribed circle, whose ratios are 1:1: 1, and take its isotomic conjugate, we have the point whose ratios are -: -:. The isogonal conjugate of this last point is a point whose ratios are a2: b2: 2. By repeating this process, taking alternately the isotomic and isogonal conjugates, we form the series of points whose ratios are an: b": C, where n is any even positive or negative integer. If we start with the symmedian point (Grebe's point), whose ratios are a: b: c*, and take its isogonal conjugate point, we have 1 I 1 the point whose ratios are: -:. The isotomic cona b c jugate of the second point is itself, as it should be, for the point is the centre of gravity of the triangle. If we take the isotomic conjugate of Grebe's point, we have the point whose ratios are a' ~S -3-. The isogonal conjugate of this last point is the point whose ratios are a'3: b3: 3. By repeating this process, taking alternately the isotomic and isogonal conjugates, we form the series of points whose ratios are an: b: cn, where n is any odd positive or negative integer. III. COLLINEARITY OF NAGEL'S POINT, ITS ISOTOMIC CONJUGATE, AND THE ISOTOMIC CONJUGATE OF THE CENTRE OF THE INSCRIBED CIRCLE. In the triangle ABC (Fig. 3) 0 is the centre of the inscribed circle, and therefore its ratios are 1:1: 1. Let Z be the point isotomic conjugate to 0. According to ~ II, the ratios of this point are: * Schwatt's " Geometrical Treatment of Curves," Art. 6, page 3. GEOMETRY OF THE TRIANGLE. 9 (1) 1 1 1 a2 b2 C2 If the vertices of the triangle are joined to the points of tangency of the inscribed circle, the three lines AP,, BP' and CP ' will be concurrent at P.* Let us find the ratios of the distances of this point P from the sides of the triangle. The triangles P,'xC and PPaC are similar, as are also the triangles Pc'y C and PP, C, and (2) whence P 'x PPa P' C: PC, Pcy: PP,=PC'C: PC; PP,: PPb =P x: PC'y. Let MC be the middle point of AB. The triangles P'xB and MzB are similar, as are also the triangles Pc'yA and M, 'wA, (3).*. P x.: Mz =BP,': BM,, 7 1' -s —o: ' c. Px = 2M z(s-b) c el / Whence But (4).,. Also a p/x=2 Ji (s-b) c ac P'y M, w-PA: M A, -s-a: 1 c. PcY — C Mcw — b,, 2 w (s-a) be (5) Whence But (6).'. * This is a quite well-known proposition. It may be easily derived by a direct application of Ceva's Proposition. 10 GEOMETRY OF THE TRIANGLE. We now have from (4) and (6) (7) p., p.,y2 4 (s-b) 2 a (s-a) (7) PCtz; P,'y: ac be ac be s-b s-a a b 1 1 a(s-a) b(s-b) But PC x: Pc y -PP,: PPb, 1 1 (8).. PP: PPb -a): ' b — a(s-a) b(s-b) In exactly the same way we find (9) PP P,= -- -7 PP PPc b(s-b) c(s-c) 1 1 (10).. PP PPb PP ' - - a(s-a) b(s-b) c(s-c) be ac ab s-a s -b s-c = bc(s-b) (-c): ac(s-a) (s-c),: ab(s-a) (s-b). Let Q be the point isotomic conjugate to P. By ~ II, the ratios of Q are ^(11) s-a s-b s-c (11) -- ' - I -- * a b c The point Q is known as Nagel's point and has many remarkable properties.* We will now prove that the three points P, Z, and Q are collinear.t For the ratios of the point P we have from (10) be ac ab PP: PP:PP= -- --: —. PP Ps Pc - s- -b s-c * Schwatt's " Geometric Treatment of Curves," Art. 66, page 35. t This might be proven by trilinear co-ordinates, but the purely geometric method is preferred. GEOMETRY OF THE TRIANGLE. 11 p be pp pba, s-ab (12) Then 1 PP=-s-b' pp pab I — C But pab + I+ =2 1 pb s-a s-b + S-C 2 J (s-a) (s-b) (8-c) whence p -- = _abc [(s- b) ( -) (s —) + ((-a) (s- b)]' 2 J (s-a) (s-b) (s-c) (13) or p -- abe (s-a) (s-b) Substituting the value of p in (12), we have 2 (s- b) (s-c) PPa = a (s —a) (s- b) (14) PPb2 zJ (s-a) (s-c) '14) Pp 2- ) P- b z (:-a) - b)' 2 J (s-a) (s- b) c T (s-a) (s- b) In the same way, using the result of (1), we find ZZ, 2 J b ZZ-a -2ab' 2J ae (15) ZZ b T ab' 2 J ab ZZ - 2 ab' and using the results of (11), we find QQa 2a (s —a) (16) 4 QQ - 2J (s-b) (16) b s2 =; Q=2 J(s-c) L & Cs 12 GEOMETRY OF THE TRIANGLE. From (14) and (16) we have (17) PP.- QQ,{ (s —(b) (s-c) s —a} 2 J (s(s-b) (s-c)-(s-a) Z(s-a) (s-b) a s 7(s-a) (s-) 5' 2J a[2ab —s(b+c)] a' sz (s-a) (s- b) From (15) and (16) 2) J be s —a <18) ZZa Q Qa- a ab 8 _2D f sbc-(s-a)2ab-) a sl.ab J' 2aJ $ a ab-s(ab+ac) a slab ' _2 f a[2ab-s(b+c)] a s2'ab ) Now from (17) and (18) (PP- Q Q): (ZZa- Q Qa) 24 a[ab-s(bc)]2 a[ab-(b+c)]2 a[a +)] a s s2(s-a)(s-b) a s v ab 1 1 (19) or PQ ' Z'= (s-a)(s-b) ' ab =-z a- z (s-a) (s-b), =-ab: 2'ab-s2. In exactly the same manner we find 20 PPb- Qb ZZ,- q, ab: ' ab-s2. () PPC- Q Q: ZZc- QQ,= ab: 2ab -s2. Any two of these proportions are sufficient to show the collinearity of the points. The triangles P' Q, and ZQ" Q are similar, PQ': ZQ"=PQ: ZQ. GEOMETRY OF THE TRIANGLE. 13 But PQ: ZQ"'= ab: ab-s2; PQ: ZQ =.ab: ab-s2, PQ-Z: ZQ=Z' ab-Z ab+s2.: ab-s2, (21) or PZ: ZQ —s2: z ab-s2. IV. COLLINEARITY OF THE ISOGONAL CONJUGATE OF Z, THE ISOGONAL CONJUGATE OF P, AND THE SYMMEDIAN POINT. If in any triangle (Fig. 3) Z1 is the point isogonal conjugate to Z, P1 the point isogonal conjugate to P, and K the symmedian (Grebe's) point; then are the three points Z1, K, and P1 collinear. Since Z1 is the point isogonal conjugate to Z, we know by ~ I, and ~ III (1) that its ratios are a2: b2: c2. (1).'. (2) That is But (3).. (4) Whence Z1Za' ZZ,,: ZZ,,-= a2. b: c2. ZlZla p a2 Z,Z,=-p b2, ZZl=C p C2. p (a3 + b3+ )=2 J; 2J zz 2 J a2 ZlZlah — 3 Za3 2 b2 z zlb a3 L 1 1c It is known* that * Schwatt's " Geometrical Treatment of Curves," Art. 14, page 7. It may also be computed as we computed the distances of Z. 14 GEOMETRY OF THE TRIANGLE. KK. 2 J a 2' a2 ' KK~_ 2 J b KK = KKc 2 a2 a2 As P, is the point isogonal conjugate to P, we know by ~ I, and ~ III (10) that s a s —b s — (6) PPica: PPlb: PPC be = ae a ab From this we can readily find r pp 2Ja (s-a) 1 la s a2 —Za3 2 J (s-b) s 2,a2 —2' a3' 2Jc( s-c) P1 A sla 2 la3 From (4) and (7) we have a s —a (8) Z^- Pi =i2J a a { ^ - -^ra_3 (8) ZZa PlPla la3 s.a2-, a3 =2 as Za2-a Z a3-s Za3+a Za3 =2 a a3(s a-S a3) 2 J as (a 2a2- a3) - a3 (s 27a2 —_'a3) And from (5) and (7) we have { 1 s-a (9) KIKa-PPl2 A 2a a2 s a2 — a 3 -2 f s a2 a3-S 2a2+a a2 ) -a2 1-r Y a 2 (s a2-2 a') ' 2,t (s 2' a —'al) ' GEOMETRY OF THE TRIANGLE. 15 From (8) and (9) (10) (ZZi —PPJ):(KKa-PP-.) 2 i (s aa2 —a ) ( Za2 ( a-a3 ) {..~a3 (s1Aa2.- -Ya3))X (sa a2 ---2a3) s 1 - 2'a3: a2 In exactly the same way we find, 1 ZZlb —PplPb: KKb-P1Plb a3 ' a23 (11) and ZZ,c-PP,1Pc KKc-P,1P lc: a Any two of these results are sufficient to show the collinearity of the three points in question. Since the three points are collinear, the triangles Z1Z,'P and KKI'P are similar. ZP, KP =ZZ,: KK', = ZaZ —pPi: KK —PiPa, s 1 - V3 a' 2 - 8 a2: a3. ZP-KP1: KP1= s8 a2 —a3: a3, (12) or Z,1K: KP1=s S a2 — a3: a3. V. COLLINEARITY OF S', K, AND D. If S' is the point isogonal conjugate to the middle point of the line connecting the two Brocard points, K the symmedian (Grebe's) point, and D* the point isotomic conjugate to K, then the points K, S', and D are collinear. * D is also the point of concurrency of the lines connecting the corresponding vertices of the fundamental triangle and Brocard's first triangle. Schwatt's "Geometric Treatment of Curves," Art. 7, page 4. 16 GEOMETRY OF THE TRIANGLE. S and Q' (Fig. 4) are Brocard's points, and S is the middle point of Q2'. It is known t that J a (b2+ c2) | aa b2 (1) J b (a2+c 2) ss a c (a2+ b ) SS- a2b2 ' (2). SS: SSb: SS= a(b2++2): b(a2+c2): c(a2+ b'). Since S' is the point isogonal conjugate to S, we have by ~ I (3) S'S' S': S'S-= + b( '+ o): c(+b); P | al a(b2 + C2) (4).. S'Sb' b( P b(a 2 (+C2) S'S/=- + — l S SC c(a2+ 62) (5) But P{b2+c+a2 + 2+ b 2J, a'a4+3 2ra2b2+ -2 (a2+ b2) (a2+ c2) (b2+c2) 2 1,,(6). P 2 J (a2+b2) (a2 +c2) (b2+c2) ~~~~(6) 0=~. a4+ 3 I a2b2 Hence from (4) f 2 J (a2+ b2) (a2+c2) S S-a ' Sa4+3 2Cab2 (7) s,^ 2 1 (2+c2) (a'+ b2) S b ' a4+32 a2b2 2 4 (a'+c2) (b2'+c) c ' za4+c3 Ia2b t Schwatt's " Geometric Treatment of Curves," Art. 16, page 11. GEOMETRY OF THE TRIANGLE. 17 The distances of K from the sides have already been used in ~ IV (5), and are KKa 2 J a 24b (8) KK- -2 j-a KK- 2ac l K SI a2 D is the point isotoiic conjugate to K, and so by ~ II, I I 1 (9) DDa DDb:DDC= a b 3 C DD =-, D a3 (10).. DD, DDC- C (11) But 12 + + )=2J, a2b2+ b2C2 +c 2a2?o 2 2 2 A2, p ti a2b2c2 d,2 (12).. P- a2b2 27J a2 62 24 b2c2 DD= 2 --- 24 a2 c2 (13) Hence from (10) DDb- 2 a2b2 b ' Za2b2' 24 a2b2 DDc - c ' a2b2. From'(8) and (13), we have (14) KKa —.DDb —2 ( a2 ab a~ vL a~> b~, 18 GEOMETRY OF THE TRIANGLE. = 2 2 ( a Za2b b2c2 'a2 2aJ 'a 2 & a2b2 / 2 J a4(b+ C2)-b2C2(b2+c2) a * Ja" 2 a2b 2 J (a4 -b22) (b2+c2) -a ' a22 a2 b And from (7) and (8), we have (15) KK —S'S'= 2 D (a- a(,a+~2)3 2(ab2) 2 2_aJ /2a4+ 3a4 2 a2b2 -a4 a2_- a2_a2b"2 l~ ~a2( \a4+3 a 2b2)' 24J (a4-/b'c2) (b2+c2) a Za2(a44+3 a2bb2)' Hence from (14) and (15) (16) KKa DDa: KK -S'Sa 2 J (a4-b2C2) (b2C+e) 2 (a4-b2') (b'2+2) a * Z 'a27a2b2 a ' Sa2('a4+3 1a2b)' 1I ~ I421 - Va4+3,a2b: ~a2b2. - ab2 ~ a4+.3 a b2b32 - In an exactly similar manner, we find Kb —DD b: KKb,-S'Sb'= 2a4+3 2a2b2: Za2b2, KK,-DD,: KKC-S'S '=Sa4+83 Za2b2: Sa2b2. Any two of these proportions prove the collinearity of the points in question. Since the points are collinear, the triangles (Fig. 5) KK"D and KK'S' are similar. KD: KS' =KK": KK', =KKa — DD' KK S' — S'S _=_va4+3 a2b2: a2ab2, KD-KS': KS' ---2a4+3 a2ba2 —Za2b2: a 'a2b2 S'D: KS'" ab+2 Sa2b2: ~a2b2. (18) Or KS':S'D = aSb (Za2)2 GEOMETRY OF THE TRIANGLE. 19 VI. COLLINEARITY OF H, M' AND D. If H is the orthocentre, I' the point isotomic conjugate to the centre of the circumscribed circle, and D the same point as in the previous proposition, then H, M', and D are collinear. We must find the ratios of the point II. The triangles (Fig. 5) HcxC and HHaC are similar, as are also the triangles Hfy C and HH^ C..Hex: HIH. —c C: ffC, and IHy: HHb,=H, C: HC. (1) Whence Hcx: Hfy=HH. H IH. The triangles AHaB and HxB are similar, as are also the triangles BHA and HcyA. (2).-. Hx: AHf = HB: AB. By substituting values for the known terms, we have 2 1/ a2C2-4 z2 Hx: -- =:c, a c 2`4 (3) or Hcx=- 2J /a2C2 4 J2. ac2 Also Hy: B, =AH: AB, ~and H iy 24 I/b2C2-4 42 and ICr^y: -b c: C, 2,4 (4).'. I2,y-2 1/b=2 42. From (3) and (4), we have 2z1 2 2C (5) Hfx: -Efy=J /a2c2-4J: Jb2c24J2, /Ca2C2 -4 12 b2c2- 42z a b 20 GEOMETRY OF THE TRIANGLE. (6) Hence IHcx: Hy /a2c2- (2 2zab2- a4) l/b2c-_ (2 2a2b2-_a4) a b a2-b2 + 2 -a + b2+ c2 a b 1 1 -a(-a2+ b2+ C2) b (a2- b2- C2) (7).~. HH~ ' HHb, __ —^^ ^ ^ _ (7) H -a ( —a2+ b2-+{c2) b(a2 — b2+2)' In a similar way, we find (8) IHb~ -—.= b (a2-b 2+c2) c (a2+b2 — 2) Finally we have (9) HH' HHb: HH'E 1_ 1 1 a( —a2+b2+e2) b (a2 —b2+2) c (a2+b2-c2) But M, the centre of the circumcircle, is isogonal conjugate to H, and by ~ I its ratios are (10) a (-a2 —+ b2+c2) b (a2-b2-+c2): c(a2+b2 —c2). M' is the point isotomic conjugate to M and by ~ II its ratios are 1 1 1 (11) a3 (-a2+ b2+c2) b3 (a2-b2+c2) c3 (a2+b2-_C2). The ratios of D are from ~ V (9) 1) 1 (12) a3: b3-: c From (9) we have ^ HH, --- a(-a2+b2+c2)' (13) -H -C- c H- -H l c (a —+ b + C2)'?o H~c= e a~ "b b — ~)' GEOMETRY OF THE TRIANGLE. 21 But 1 - I -1 - 1 =2zI But ( a2+b2+c2) + (a2- +c2 + a2 + c) b =2 Now for brevity let us put a 2+ b2+c2 2 2 k2 a2 + + 2 — b k —2_-a2 2 as__ b" + c (14) and.. a2-b2+c 2 a2+ b-c2 k 2 - -- k;__c~ Making these substitutions, we have (15) p +)k2_a2 + k2 b2 k2_C2)-4 ((k2 —b2) (k2 —2) + (k2-a2) (k2-C2) + (k2-a2) (k2-b2) 4, (k12-a2) (k2-b2) (k22) //3k4-2(a2+ b2+ c2)k2 + a2b2) ) P (k2_2) (k2_-b2) (k-2_c2) (k2_ a2) (k2_ b2) (k2_) -4 (16) or p 4 D (k2-a2) (k2-b2) (k -2C2) (16) or 4( T — a2b2-k4.'. From 13 HIa= 2 A (k2-b2) (k-C2) (1-7) 4 -2 zJ (k2-a2)(2_k2) (17) H 2 J (k-a2) (k 2 —c) 17b 2Ta2b2-k4 H_ 2D (k2-a2)(k2_b2) [ "c * Vab2 — k4 (18) From (11) M'Ma': M'Mb': M'M, 1 1 1 a3(k2-a2) b3(k2-b2) c3(k-c2) 22 GEOMETRY OF THE TRIANGLE. M a(k2- a)' M'M' - ) "- b3(k 2_b ) (19)... M'Mj= [ Mm c-ca32_r2) ' (20) But p { (a) + b(k-b) + (k-) =2, (a2b(k2-a2) b2(k2- b2) (k2c2) _ 2) '" (k4 az~b —k2a4b2+ la4b') '? 2 -b2C2 2 -' 2) 2,~ — 2, 2k ---) 2,J; (21) or 2 J a2b2C2 (k2-a2) (k2-b2) (kC2) P' k4 d'a2b' — k'2 a4b'2+2a4b4 Then from (19) (22) M'MJ 2 a2e2(k2 —a2) (k2 —c2) a 'l k4a 2b2 — k2 a4 b2+:a4b4 MIM-=2J a2b2(k2_a2) (k2_b2) c k4 Za2b —12 a4b2+- a4b4 2 kl a2 b'(k2 —a2)(k2 —b2) From ~ V (13), we have DD - 2,. b2' (23) - DD I- b a 24J a2b2 DDC *a=b From (22) and (23) we have (24) M'M'- DDa 2 J b2C2(k2-b2)(k2 e2) b2c2 a k_4 Ia2b2- k2a4b2 +la4b4 7a2 b2 ) 2 J b2c2 If a2b 2(k2a2+ b2a2- _a2b2) + k2a2b2c2 a L 2a2b2 (k4'a2b2 —k2f a4b2+-a4b 4), 2Ja2b2c-2 _ 2 _ ab+22 2 ab2+2 b22C2 -- a k a2b —t a (2 4 'C2b2 - 2 a(b X2a4)b ' GEOMETRY OF THE TRIANGLE. 23 and from (17) and (23) (25) HHET-DDa a 2 a2b2- -k 4 a2ba2 ' _{(a la-b)(- - ~)a b} 2 JF k4Z a2b2- k2(b 2+c2) a2b2+k4bc2 } Ca a,2b2 (2a2b2 — k4) ' 2z2k2 { +k2 -a2b — (b2+C2) a2b2+k2b2c22 a ~Y 2a2 (2 a2b2-k4) J, 22 k - k~ 'a2b2+a2Sa2 b2+k2b2c2 a. a, a2b(2a(a2b2- k4) }.. From (24) and (25) (26) HH,- DDa,. M'Ml'- DD 2 212 { - Ja22aa222222 c-t2b2+ 2 _b 2 a b 2a2b2(~a2b2 —k4) 2 2Ja2b2c2 f -k2 Va2b2+a22' a2b2+ k2b22 c a 2 27 (2 4 2a' 2 —,2 2 a4b2+ _ a4b4 j k2 a2 b2c2 - a b2 -k k4 -a2b2- k2a4ab2+ t44 = 2 (k4 Vaa2b2- 2a4b2+ a 4b2 a4b4) a: a2b22 (aa2b2- k). In an exactly similar way, we find (27) HIH,-DDb: llI'Mb'-DDb - k2 (k' 2a2b2_k2 2a4b-b2-+ a4b4) a2b2c2 (a2b2_ k4) and HHc —DDc: M'Mc' —DD = k2 (;k4 a2b2-2 j2a4b2+ 4V a4b4): a2b2C2 ( a2b2 — ). Any two of these results prove the collinearity of the points in question. Since the points are collinear, the triangles HH~D and l'Mi1D are similar. Hence H:!1I'M-HD: T 'lD. 24 GEOMETRY OF THE TRIANGLE. But HH,: 1'M = I —I -t DD, I: M' —DDa - k(k4 V'a2b2k2 2'a4b2+tva4b4): ab2C2 (vCa2b2-_4),.'. HD J:1'D -k2 (k4 a2 ba_-k2 ~a4b4+ va44) a b2e (a22 —k). (28) and IID —M'D: M'D =k2 (k.4 7a2 b2 _2 2a4b2 + a4b4) - a2b2c2 (Va2b2-_4): ac2bC2J ( Va2b2-k4); or HM': ~M'D= -k2(k' 'a2b2-2 a4b2+ a 2 4b4) - a2b22 (v a2b2 k):a2b2c2 (2a2b2-k4). VII. COLLINEARITY OF Y', F, AND D. If Y' (Fig. 9) is the point isogonal conjugate to the centre of Brocard's circle, F the centre of the Nine Points circle, and D, the same point as in ~ V, then are Y', F, and D collinear. Y the centre of the Brocard circle * is the middle point of the line joining M, the centre of the circumcircle and K, the symmedian point. The ratios of M, ~ VI (10), and K ~ IV (5), are already known, and from them the ratios of Y can readily be found. The ratios of Y are a [a2 (b2+c 2 —a2)- 2b2c2]: b [b2(a2+ c2- b2) 2a2c2]: C[c2(a2+ b2_) +2a2b2]..'. by ~ I, the ratios of Y' are 1 1 (1) [a2 (b2+ c —a2) + 2b2c] b[b' (a'4+ c2- b-' 2a'22] [C2 b2 + 2a' b'2] c [c2 (a2+ b2- c2) + 2a2- b2]' * "Modern Plane Geometry," Richardson & Ramsey, page 44. GEOMETRY OF THE TRIANGLE. 25 From these ratios we readily find (2)........... " 172 i [b2(a2+c2-b2)+2a2c2] [e2(a2+b2c2)+2a2b2] ) a a (3 cz'a262(2 2a2b2 -2a4) 5 2J E [c2(a2+ b2_ 2) +2a22b] [a2(b2+c2 —a2) +2b2c2] b b 3 53a2b2(2 2'a 2b2-27'a4) Y'-,-2 J D [a2(b2+ C2 -a2)+2b2c2] [b2(aC2+c2 b2)-+2a2c2] ) c c 3 3 2a2b2(2 a2'b2 —2a4) F the centre of the Nine Points circle * is the middle point of the line joining HT, the orthocentre, and M the centre of the circumcircle. From the known ratios of these two points ~ VI (9) and (10), we readily find the ratios of F. They are b2(a+C2 -b2)+c2(a2+b2 c2) a2(b2+c2-a2)+c2(a2+ b-c2) a b a2(b2 + 2-a2) + b 2(a2 + c2 -b2) c From these ratios we easily find fFF 2. b2(a+c2 — b)+C2(a2+b2- 2) a 2(22a2b2-2_a4) (3) FF - ^(3) FF- 2 J a2(b2+c2-a2)+c2(a2+ b2-c2) - 2 (2 2 a2b2_ —a4) 2 a a( b2+c2-a2)+ b2(a2+c2 — b2) FF= ce 2(22.a2b2 —2a4) We already know ~ V (13) that 2 A b2e2 DD 2 -. 2 [DD 2 a a2-b2 (4) 2J a2c2 (4) 1 DD 2 d a262 2J c a2b2 DD -- b S,a2 b. * Lachlan's " Modern Pure Geometry," page 70. 26 GEOMETRY OF THE TRIANGLE. Hence from (2) and (4) (5) Y' Ya'-DDa 2 A [b2(a2+c2-b2)-2a2c2] [c2(a2+b2-c2)+2a2b2] bc M - al3 5 a2b2(2 Z a2b2-Za4) "a22J ' 2_A f [b(a2+2-b2)+2a2c2 [c2(a2+b2-c2)+2a2b2]-6b2c2 s a2b2+3b2c2 Z a' a t3 Z a2b2(2 Z a2b2-2 a4) 2 A 1 2 [4a4b2c2-2a 2b4c2-2ac2b2c4-2b4c4+a4b4-a4c4+b2c6+b6c2-a2b6-a2c6] a T 3 a2b2(2 Z a2b2 — a4) and from (3) and (4) (6) FFa —DDa 2A b2(a2+c2-b2)+c2(a2+b2-c2) b2c2 -a l 2(2 2 a2b2 —2 a4) Z a2b2 2A b2 (a2+c2-b2) Z a2b2+c2 (a2+b2-c2) Z (a2b2-4b2c2 2a2b2+2b2c2 2a4 a- 2 (2 2; a2b2-Ea4) Z a2b2 ' 2 4 a42c2-2 a2bc — 2a2b2c4-2b4c4+ ~a4b4+ ca4 +b2c6 + b6c2-a2c6-a2b6 "a{ t2 2a2b2 (2 Z a2%2 —2a4) From these two results we at once have (7) Y Ya- DDa: FFa- DD - 3 2 =4:3. In a similar manner we find Y' '17- DDb: FFb — DDb= 4: 3, Y' Y '- DD,: FFC - DD,= 4: 3. Any two of these results show the collinearity of the three points under consideration. Since the points are collinear, the triangles Y' Y,'D and FFID (Fig. 9) are similar. Y' Y': FF- Y'D: FD, 4: 3- Y'D: FD, 4-3: 3 - Y'D-FD: FD. (9) Hence 1: 3 - Y'F: FD. GEOMETRY OF THE TRIANGLE. 27 VIII. PROPOSITION. The bisector of the angle between two similarly situated lines in two similar triangles, which are in perspective, is parallel to the bisector of the angle between the sides of the triangles opposite the vertices from which the similarly situated lines are drawn. Let (Fig. 6) ABC and A'B'C' be two similar triangles in perspective, AE and A'F are two similarly situated lines intersecting in L. ML the bisector of A'LP is parallel to KD the bisector of C'DB. From the similarity of the triangles, ang. A'HB'- ang. AFB, and ang. A'FC = ang. AFB. We have then ang. A'LA=180~-(ang. A'HB'+ang. AGC'). From this we have at once ang. A'LM-I= (ang. A'HB'+ang. A G C'). But ang. AGC'-180 —(180~0- ang. A'HB'+ ag. HDF), = ang. A'B' — ang. HFD. This comes at once from a consideration of the angles of the triangle GED. We now have ang. A'LM= ang. A'HB' — ang. HDF. In the triangle HKD, we readily find ang. A'KD= 180~-(180~-ang. A'HB'+ i ang. HDF), =ang. A'HB' — ang. HDF. Hence ang. A'KD = ang. A'LM, KD is parallel to LM. .28 GEOMETRY OF THE TRIANGLE. IX. PROPOSITION. The angles between the corresponding sides of Nagel's triangle and the fundamental triangle are A OO, BOM, and COIM; where 0 is the centre of the incircle and Ml the centre of the circumcircle.* 180~ — ang. A OM — ang. A' OM. Since the lines OM and A 0 are respectively parallel to the lines HQ and A"Q ang. A' 01M- ang. HQA", and since HB" is parallel to C'A' ang. (HB",B C) ang. ( C'A', B C). But ang. ( C'A',B C) -= (A + C) (A,B, C are the angles of the triangle ABC.) ang. ( C'A',B C) - ang. B', = ang. B". Ang. (C"B",B C) = ang. (HB",B C) - ang. HB" C", =ang. B"- ang. JIB" C", = ang. A" Q C"- ang. IB" C', ang. A" QH, ang. A' OM. ang. (B" C",B C) - ang. A OM. In an exactly similar manner we find ang. ( C"A", CA) = ang. B OM, ang. (A"B",AB) = ang. COM. * Schwatt's "Geometric Treatment of Curves," Articles 72, 73, 74, and 75. It is -there proven that (Fig. 7) H (the ortho-centre), Q (Nagel's point) and the three vertices A"B"C' are concyclic; that OM is parallel to HQ; and A 0 parallel to A" Q; BO parallel to B"Q; CO parallel to C"Q; and HB" parallel to C'A'. Triangles A'B'C/ and A"B"C" are similar. The triangle is found by taking M A"I=M aA', MIB"-=I=bB and M C"-=M C'. There is an interesting analogy between this theorem and a similar one for Bro-,card's triangle. See Schwatt's " Geometric Treatment of Curves," Article 24. GEOMETRY OF THE TRIANGLE. 29 X. A GEOMETRIC METHOD OF FINDING THE CENTRE OF ANYELLIPSE ISOGONAL CONJUGATE TO THE POLAR OF ANY POINT WITHIN THE TRIANGLE. Let ABC (Fig. 8) be any triangle, and P any point within it. XY is the polar of P. Draw AP, BP, and CP and produce them until they meet the circumcircle in A', B', and C',, respectively. It is a well-known property that BC' and CB' intersect on the polar,* as do also C'A and A'C, B'A and A'B. Let these points of intersection be L1, L,, and L3, respectively. Now let M1, M,, M3 be the points isogonal conjugate to L, L,, L,. Then it will be found that AM,, B1V,, CM, mutually bisect each other at the point S which is the centre of the ellipse isogonal conjugate to the polar of P with respect to the triangle ABC. Ang. BA.1i is subtended by arc B C+ arc B C', since ang. CAM -= ang. BAC'. Ang. ABMx is subtended by arc A C + arc A C', since ang. CBM1 = ang. ABC'. Hence the two angles BAM2 and AB-M4 are subtended by a whole circumference and are therefore equal to two right angles, and therefore the lines AM2 and BMl are parallel. In exactly the same way it may be shown that AM, and CMI and also BM, and CM2 are parallel. Therefore in the hexagon AM2CM1BM3 the opposite sides are parallel. We will now drop perpendiculars from L, L,, L3 upon the three sides of the triangle, and we will let the feet of these perpendiculars be indicated by Lla, L17,, LL; L2a, L2b, etc. We will also drop perpendiculars from A', B', C' on the sides of the triangles and we will indicate their feet in a similar manner. The triangles L2L2bA and C' Cb'A are similar, as are also the triangles L,L2,A and C' CC'A. * Lachlan's " Modern Pure Geometry," Article 246. 30 GEOMETRY OF THE TRIANGLE. 'C'C': LL,, C'A: AL,, and C' C': L2L, - C'A: AL. Hence C' C': C' C =L2L2b: L2L2,. But the triangles BC'C' and CC'C,' are similar, since they are both right angled, and have the acute angles C'CCb' and C'B C' equal. C,': C'C,'c- CC' BC'. Hence LL2b: L2L2 = CC': B C'. The triangles L2L2 C and A'A',C are similar, as are also the triangles L2L2b C and A'A' C. A'Aa': L2L2 A C L2 C, and A'A':,L2b = A' C: L2 C. Hence A'Aa': A'A'= L2La: LLb. But the triangles A'BAa' and A'AA' are similar, and so A'Aa': A'Ab' BA': AA'. Hence L2L2a: LL2 = BA': AA' -BA'. CC': AA'. CC'. Also L2L2,: L2L2 = CC': B C' -AA'. CC': B C'. AA'. L2L2: L2L2: L2L2c = BA'. CC': AA'. CC': AA'. BC'. In exactly the same way, we find LiLa: LL: Lb C = BB'. CC': AB'. CC': A C'. BB', and LLa:L3L LL3b: L3L3-BB'. CA': CB'. AA': BB'. AA'. Now, by ~ I, the ratios of Ml= BB'. CC" AB'. CC' A C'. BB 1 1 1 M2 BA'. CC AA'. CC' AA'. BC' M — 1. 1. 1 M CA'. BB'' CB'. AA' * AA'. BB GEOMETRY OF THE TRIANGLE. 31 Let p be the multiplier which transforms the ratios into the actual perpendiculars. Then remembering that M2 is outside the triangle ABC, we have for M2 ( a b + \ 2 J BA'. CC'- AA'CC' + AA'BC' a. AA'. BC'- bBC'. BA'+ C. CC'BA' r p BA. BC'. AA'. CC' - 2 D BA'. BC'. AA'. CC' or P - aAA'. BC'- bBC'. BA'+ cCC'. BA' 2e AA'. B C' Hrence M a — aAA'. BC'- bBC'. BA + cCC. BA' In exactly the same way, we find JI[MM=1_ 2 JAA'.CB' aAA'. CB'+ bBB'. CA'- cCA'. CB' Now, remembering that in an inscribed quadrilateral the product of the diagonals is equal to the sum of the products of the opposite sides, we may write b. BB'a. AB'+ c. CB', and c. CC'a. AC'+ b. BC'. Making these substitutions, we have 2 J AA'. BC' M2M2= a. AA'. BC'- bBC'. BA'+ aAC'. BA'+ bBC'. BA' 2JAA'.BC' aAA'. BC'+ aAC'. BA" 2J AA'. CB' M33a- aAA'. CB'+ aAB'. CA'+cCB'. CA'-cCB'. CA' 2__ AA'. CB' caAA'. CB'-aAB'. CA" 2MM AA' BC'. CB' A a AA'. BC'. CB'+A C'. BA'. CB 2 J AA' BC'. CB' 3 3a a AA'.BC'. CB'+ AB'. CA'.BC'' 32 GEOMETRY OF THE TRIANGLE. From the similarity of the triangles, readily seen in the figure, (Fig. 8), we have the following: AC': CA' AP:PC, BA':AB' ---BP:AP, CB':B C'- PC:BP. From which we have at once AC'.BA'. CB'- CA'. AB'.BC'. Using this relation, we have 2,J AA' BC'. CB' i2 2a2- a AA'. BC'. CB'+CB'.BC'.CC' Hence MM211a- M —I33a. Now BM3 is parallel to CM12, and therefore the triangles BMM11i13a and CI42M2,, are equal, and BM13- C12. From this it immediately follows that AM2-=BM11, and AM3 - CM, Hence the figures BM3fa2 C, AM3i1L C and AM2M, B are parallelograms, and their diagonals AMllf, BM and CM3 mutually bisect each other. The point of intersection S of these lines is the centre of the ellipse in question; for the six points A,B,C, JM,M M, are on the ellipse, and the three lines AMl1, B1~J2, C I3 are chords of the ellipse. But three chords of an ellipse can mutually bisect each other only when they are diameters. Hence S is the centre. PROPERTIES OF THE LOCUS r = CONSTANT, IN SPACE OF n DIMENSIONS. BY PAUL RENNO HEYL, B. S. It is to be observed, first, that in a space of n dimensions, a straight line drawn perpendicular to two given intersecting straight lines will be perpendicular to every straight line in their plane; for three such lines if drawn in n-fold space would determine an ordinary solid space contained in the higher space. Every line in the plane of the two given intersecting lines will lie in this solid space, and the proposition first stated is known to be true in 3-fold space. The Theorem of Pythagoras can be extended to n-fold space. This theorem, as known in two and three dimensions, asserts that if there be a series of two or three mutually perpendicular straight lines, AB, BC, CD, for instance, then the square of the straight line joining the ends of the series is equal to the sum of the squares of the separate lines of the series; that is to say: A C2= AB2+ BC2 in 2-fold space, and AD2= AB2+ BC2+ CD2 in 3-fold space. Now let AB, B C, CD, DE, be a series of four mutually perpendicular straight lines. DE is perpendicular to AB and to BC and hence to AC (since AC lies in the plane of AB and BC). Also DE is perpendicular to A C and to CD and hence to AD. AD and DE being perpendicular, AE2== AD2+ DE2 But AD2- AB2+ BC2+ CD2. Hence AE2 AB2+ BC2+ CD2 DE2. 3 (33) 34 PROPERTIES OF THE LOCUS r = CONSTANT, The theorem is thus capable of indefinite extension, and may be formulated as follows: If there be a series of n mutually perpendicular straight lines, AB, BC, CD,....... L M, MN, then AN_= AB2 + BC2+....MN2. Let us now consider the locus r = a in n-fold space. In 2-fold space the locus is the circle x2+ y2= a2. In 3-fold space the locus is the sphere x2+ y2+ z2= a2. In 4-fold space, by the generalized Theorem of Pythagoras, the locus will be represented by the equation x-+ y2+ z2+ t2= a2, and for spaces of more than four dimensions the equations are of similar form. We can find the volume of a sphere by integrating its circular plane section along a third dimension from - a to + a. If r be the radius of the circular section, a Volume = rfr2dz, where r2_a2-z2. -a a a Hence Volume = r f(a2- z) dz=2 a 2(a2- 2) dz. -a o Let z = a sin 0, then we have 2 2 4 Volume — 2 7 a3 cos3 d o = 2r a3. 2- a3. 1. —3 0 In like manner let us find the content of the 4-fold analogue by integrating its solid section along a fourth dimension from - a to a. For any value of t, the solid section will be the;sphere 2+ y2+ z2 a2 — t2-=r2, and the content of the 4-fold locus will be 4 a 4 a 9 a 3 - r- rdt -- (a2- 2) t= (a2- t)t. -a — a o IN SPACE OF n DIMENSIONS. 35 Let t a sin o, and we have 7r 8ar4 2 87ra4 1. 3 " 2a4 3 Jos do- 3 2.4 2 2= o For the 5-fold locus 2 a Content =- rdur, where r2=a2- u2 -a 2 a a = -i-Ja2- u2)2du= 7 (a2 U2)2du. -a o Let iu a sin o, then we have 7r C2 24 82 2 2a5 x 2 Content == a os5 d 5a5 == 4 - _ 8 a 1.3.5 15 0 It will be seen that we must integrate each time an expression involving one higher power of 1/a2- u2, and hence an expression involving one higher power of acos 0. In general, the content of the n-fold locus will be T A,= 2 An_ a cosn 0 d 0, 0 7;T 7T 2 2 = 22An,2a tosn d 0 oscos- d 0, o o 7; 7 ' T- 2 3An3a cosn 0 d T7rcOSn-1 d 0coSn-2 Od 9, 0 0 0 and so on. Carrying this down to A2 in the right-hand member, and noting that A,= content of circle = = a2 36 PROPERTIES OF THE LOCUS r= CONSTANT, 7r f7 7tr 7 2 2 2 2 An-2n-2ra Jcosn 0d 0 COSn-1 / do.......... cos3 do. 0 o o o If n be even A_ _ _2-2a) (1*.3.5.... (n-1) 7r 2)(.4.6 (n-2) Al, 2.4. 6..n 2 1.3.5..... (n-) /2 \..e..o and if n be odd A2n-2ra(2.4. 6 (n-1)1 35 (n-2), 1.3.5.2I3.5:..... / (n —) 2 **.. -(a)....... (1.3) In each of these continued products there are (n-2) fractional factors. When n is even, the 1st, 3d, 5th........ of these contain -, giving in the result ( -). When n is odd, the 2d, 4th, 6th....... contain —, giving in the result ( 2 )-. Hence, when n is even n-2 n 2n-2 7r an |.2 2 A - Ap- 2.4.6.... and when n is odd n-3 n+l- n-1 A %a ~ 2- 2 2 2 — 2 a /r \ 2 2 a2 1.3.55.n 2 1.3.5.....n For purposes of computation these formulas may be written 77n a~2 7a2 A2n — = -- A2n-2.......... (i) n! n 2n7 n —1 a2n-1 27ra2 2- 1 3.5.. (2n- 1) A2- 3 2* (1 1) IN SPACE OF n DIMENSIONS. 37 where n is any integer. Let us see if these formulas will hold when n —. A1 would be the content of the locus x=-=a. The content of this in line units - 2a. From (ii) we see that 2n-1 A2n-3 2n-1 A2n-1; and if n-2, 2 7T a' 3 3 4 3 A —2a2 A3 -2a2 3 as it should. Our locus cannot exist in 0-fold space, as at least one dimension is required for the radius vector. We find A0 from (i) equal to unity; but we can assign no meaning to this. By means of (i) and (ii), the following table was computed. It gives the numerical measure of the content of the locus for a =1 up to 20 dimensions. Dimensions. Content. Dimensions, Content. 641 2 885 1 2 11 -1.885 10395 - 2 = — 3.142 4 3 3 4 r =- 4.189 2 4 - = 4.936 87r' 5 - 5.264 15 - 6 - 5.170 16 3 7 10 4.725 105 7r4 8 - 4.061 24 32w4 9 45 3.299 72.552 10 -- 2.552 120 - 12 720-1.336 13 128 -0.9110 135135 14 5040 0.5997 5040 2567E' 15 2 -- 0.3816 2027025 16 -- 0.2351 40320 17 512wT 0 1410 34459425 W9 18 - 3 0.08207 362880 19 — 0.04663 654729075 10 20 - 0.02579. 3628800 38 PROPERTIES OF THE LOCUS r = CONSTANT, It will be noticed that there is a maximum value for the content at 5 dimensions. The reason for this we can see by throwing (i) and (ii) into continued products. Thus we have for (i) rna2n (rT a2)n r a2 ra2 ra2 *ra' A2n- n! n! 3. As long as n is less than Ca2, the successive factors are greater than unity, and A2n increases with n; but as soon as n is greater than ra2, A 2nbegins to decrease. If a - 1, the maximum value for A2. will happen when n =3, or 2n = 6. This agrees with the values in the table. Treating (ii) in the same way, we have 2n7. n-la2n-1 A2n-1 1.3.5....(2n-1) 1 r27ra2 2ra2 27ra2 2.a aTr L 3 5.*..2n-1J' In this case the maximum happens when 2n - 1 is the greatest integer less than 27ra2. If a 1, the maximum occurs when 2n - 1 = 5, which agrees with the table. The two functions A2n and A2f _ fit into one another; they are special cases of a more general single expression in Gamma functions, which can be obtained by integrating the general expression for An without regard to n being odd or even. In both A2n and A2_n-, we see that when a has any finite value, a maximum must occur for some value of n; and that as n increases indefinitely, the last term of the continued product approaches zero; that is to say, in a space of infinite dimension our locus can have no content at all. We might be pardoned for supposing that in a space of infinite dimension we should find the Absolute and Unconditioned if anywhere, but we have reached an opposite conclusion. This is the most curious thing that I know of in the Wonderland of Higher Space. I might call attention here to the analogous theorem for the boundary of the locus r a in n-fold space. The area of a triangle is one-half its base times its altitude, hence the circumference of a circle is twice its area divided by its radius. The IN SPACE OF n DIMENSIONS. 39 volume of a pyramid is one-third its base times its altitude, hence the surface of a sphere is three times its volume divided by its radius. It can be shown by developing some of the elementary geometry of n-fold space that the law appearing here holds throughout higher space, and that the volume of an n-fold pyramid is equal to - (altitude) [(n-1)-fold base], hence the n boundary of the n-fold locus r = a is equal to n times the content divided by the radius. That is 2n 27 7na2n-1 _ 2n - ____ _27_ 2 n — (n-1)! 2n-1 2n7:n-la2n-2 and B A 1 — A and B2n — a A2n- 1.35..... (2n-3) 1~.3.5~. (2n(2n — 3 ) Throwing these into continued products, we have a 2 7ra2 ra2 7a_2 _1 B =2-=-2ra 'J B2n2ra1 2 3 n -1 and B2n1= 2[27a * 2,ra2 2ra' 2-23] an3 5. 2n -3 An passes its maximum when n becomes greater than 7a2. B 2 passes its maximum when (n - 1) becomes greater than 7ra2. Hence An reaches its maximum before B2,. A2ni passes its maximum when 2n-1 becomes greater than 2i7a2. B2n-_ passes its maximum when 2n-3 becomes greater than 2ra2. Hence A2n- reaches its maximum before B2ni Both B2n and B2n-_ are special cases of a more general single expression in Gamma functions. The curve in Plate II shows how the content and boundary vary with n for a unit radius vector. All this work forces upon our consideration the idea of a fractional number of dimensions. We are put in somewhat the same position as were mathematicians when they were confronted with the necessity of extending their ideas of exponents so as to include and interpret fractional and negative exponents. 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