ON A COMPILETE SYST EM OF Q INVARIANTS OF TWO TRI.ANGLES BY DAVIDI) o. LEIB A DISSERTATION SUBMrITTED TO THE BOARD OP UTJNIVERSITY STUDI;JS OF THIE JOHNS HOPKI-S 'UINIVERSITY IN CONFORMITY WITTH THE RiEQUIREMENTS FOR T -FE DEGRPEE OF DOC TOR OF PUIFLOSOPRY piESS of TIE N AEW EIA PRINTING COMPANY LANCASTER. PA. 3 UNE, 1909 ON A COMPLETE SYSTEM OF INVARIANTS OF TWO TRIANGLES BY DAVID D. LEIB A DISSERTATION SUBMITTED TO THE BOARD OF UNIVERSITY STUDIES OF THE JOHNS HOPKINS UNIVERSITY IN CONFORMITY WITH THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY PRESS OF THE NEW ERA PRINTING COMPANY LANCASTER. PA. JUNE, 1909 ON A COMPLETE SYSTEM OF INVARIANTS OF TWO TRIANGLES* BY DAVID D. LEIB Introductory. The simpler invariants of two triangles, arising from the connex set up by them when regarded as two cubic curves of the third order and of the third class, respectively, were discussed by Dr. HUN in vol. 5 of these T r a n s a c t io n s, pages 39-55. The present paper exhibits a complete system by means of which all invariant relations can be conveniently expressed, and is largely the result of an attempt to solve a problem proposed by Professor MORLEY: What is the invariant relation between two triangles so that a conic may be drawn on the vertices of one triangle and touching the sides of the other? In this and similar problems, the one triangle was invariably taken as the reference triangle, and the invariant relation consequently expressed in terms of the coefficients of the other, a simplification which we shall use throughout the article. In this way the selection of the fundamental system was largely a matter of experiment in the first place, the object being to select such as would enable one to write down invariant relations in the most simple form, and at the same time to have these fundamental invariants easily interpreted geometrically. In the body of this paper we shall follow the logical order, however, of giving the system first and then proving its completeness. ~ 1. The fundamental system. If we take one triangle as a 3-line given by the symbolic equation (aCx)(/8x)(7x) = 0 and the other as a 3-point (aE)(b0)(ct)= O, where (ax) = amx, + a2x2 + ax3, etc., the constants ai, bi, c~, air,, fry form the domain of rationality for the invariants. The six fundamental invariants A,, D I11, D2, I. and 13, expressed rationally and integrally in this domain are given on page 128 of The Johns Hopkins University Circular, July, 1908. Of these, A\ and D1 are the determinants of the 3-point and 3-line respectively. * Presented to the Society, December 31, 1908. 361 362 D. D. LEIB: INVARIANTS [July That is al 1 7l Al- 2 )2 72 a3 13 73 Following Dr. HUN'S notation for indicating the collective degree in Roman and Greek letters respectively, the invariants are represented as follows: D1(3, 0); A1(0, 3); 1(3, 3); 1,(6, 6); D2(6,6); I3(9, 9). From these can be formed four independent absolute invariants of zero degree, I2 I4 I 1I2 D2' AD1' I, I'3 D2 AID 1 2 I3 and in terms of these four can be expressed rationally every other rational invariant of zero degree in both sets of coefficients. This is a minimum reduced system, since two triangles have projectively four absolute invariants. If now the 3-line be taken as the reference triangle, A1 assumes the value unity and the other fundamental invariants become homogeneous, rational integral functions of the coefficientstof the 3-point, as follows: al b\ ci D1= 2 b2 C2 C3 b3 3 I = a b c3 + a3I bc + 2b3c + a1bc2 + a2b c3 + a b2 c, 2 a3 b2b2 3 C23 D2= a3 6a b3b c c0, a, a2 b1, 2 C, 2 2-= a2a3b3blIlc2 + ala2b2b3c3cl + 3al2c2c3,b2 + a2a3blb2c3c + 13a2b3blbc23 + a3a1b2b3C1C2, 3= a1a2a3b1b2b3c1 0203 It is in this simplified form that the invariants are most available, and it is a matter of very little difficulty to introduce A1 into any invariant form so as to render it homogeneous and of the proper degree in the Greek letters. The invariants D1 and D2 are skew, as is evident from their form. ~ 2. Proof of the completeness of the system. Take the 3-line as reference triangle and call it TY. The 3-point, (a~)(b6) (c) = 0, we shall call T2. If T is taken as (ax) (3)(yx) = 0, we shall define a rational simultaneous invariant of T, and T2 as a rational integral func 1909] OF TWO TRIANGLES 363 tion of the coefficients, which has the invariant property and is unaltered by any permutation of the three lines or of the three points; that is, by interchanges of ai, Ai3, 7i or of ai, bi, ci. When T, is taken as the reference triangle, all rational simultaneous invariants (except A,, which becomes unity) of T, and T2 are homogeneous, rational, integralfunctions of ai, bi, ci, which are unaltered by all permutations of the letters and of the subscripts. The converse is equally true, and any homogeneous, rational, integral function of a,, b,, c,, which is left unaltered by permutation of the letters or of the subscripts, is equivalent to an explicit rational integral function homogeneous in both the Roman and the Greek letters, which has the invariant property under all linear transformations of the plane. Hence the most general form of such an invariant is I = E al aa2 aa3 b{1 bJ2 b3 Ck1 C c2 C73S 1 2 3 1 2 3 1 2 3 where i1 +-2+ i2 + ji3 + k, + = n jl + j2 + j3 =2, i + j2 = k1 + 2 + k3, i3 +j3 + 3 n We next prove the lemma: Every individual term of I, where I is of the nth degree in the coefficients of the 3-point, is the product of n factors of the type aibkcl (i = k l, i= 1,2, 3; k 1, 2,3; 1=1,2, 3). To simplify the proof we write a2 b3 c1 = a2bl C3 -- 1 a3 bl2 =a2 a3b2 C1 -2 These a. and /3 have no relation whatsoever to those used earlier in the article. Clearly as long as every letter and every subscript is present in a term we can continue taking out factors of type aibc c. But some exponents may be zero. In that case suppose i1 say, is not zero. Then the conditions on the exponents give the inequalities: i2 <n, i3 <n, j< n, k < n, j2 + k2> o, j3 + k3> 0, j2 +j 3>, k2 + > 0. Hence, if J2 = 0, k2 > 0 and j3 > 0, and therefore 0o is a factor. If k3 = 0, j3 > 0, k > 0 and /0 is a factor. If j2 = 0 and k3 =- 0 at the same time, f0 is still a factor. If j3 = 0 or k2 = 0 or both are zero simultaneously, a0 is a factor. Hence, if i1 == 0, either ao or /3o is a factor. Taking out the factor a0 (or 60), we 364 D. D. LEIB: INVARIANTS [July have left a term of same type with n - 1 replacing n. This establishes the lemma and we can write any rational invariant in the form I_- L N l a 0k 1 [3- 2 Since I is left unaltered by all permutations of a, b, c and their subscripts, it is left unaltered by all permutations of a and 3 and their subscripts. For if we express the permutations of the Roman letters in cycles, it is easy to express the corresponding permutations of the Greek letters in cyclic form. We have (a b c) is equivalent to ( a2 a ) (/01 2) (1 2 3) " " " (oa 1a2)(a0112), (b c); (a 0o ) (l) ( a2,a2), (2 3)" " " (,aO 8)(aal2)(a2 1) This includes all independent types. Next we introduce the notation: aO + al + r2 = rl,,0 + + 1 2 + 2 =, al a2 + ao2a + 0o a, - r2 112 + - 20 + 02 0 + 1 s82 a0 aI a2 = p,8, I02, (a - (2)( a)( a)(- a,) =r, (I, - 2)(12 - 03)(/3o- 1) =. Since any invariant admits of the permutations (aoaa2a) and (1o 112), it follows that if any invariant I contains the term ao al af,1 32, it contains the 9 terms 3 3 3 Z signifying the sum of the three terms obtained by cyclically permuting i,j, k and i, m, n respectively. If i =j = k or z = m = n, these nine terms reduce to three, or if both sets of equalities exist simultaneously, they reduce to a single term. In any event, we have the product of two alternating functions of ac, al, a2 and f80/3,,32 respectively. By the same argument for all terms of I, we see that (1) -= J-R(r, s, r2 s, 2 p3, s), where R is a rational, integral, isobaric function of its arguments. The subscripts indicate the weight, r and s being of weight 3. As a next step r and s may be assumed to occur to the first degree only, for 0r2 and s2 can both be expressed rationally in terms of the other 5 quantities. In fact* *SALMON, Higher Algebra, Ex. 2, p. 57. 1909] OF TWO TRIANGLES 365 r2 r2 r - 4rp3 + 18rrp3 4r3- 27p_, 2 22 52 = S -8 4S33 + 18,2 3 4 - 4S 27p3. 12 13 2P23- 2 3 Also since the permutation ( al a2) (1 12) changes the sign of r and of s, they can occur only in the combination rs. Hence we can write (2) I= Rl(rl, s, r2, r 2,p3) + rs. R 2(r, S1, r2, Sp]3). Again since ( aip) (a fal) ( a2 2) interchanges r, and s,, I must be unaltered by interchanging ri and so. That is (3) Rl(rl, s, 2, S2, P3) = Rl(s1 rl S2, r2,p3). Hence if I contains a term r s' r2 s2p, it contains the two terms (4) p, (rsa),(r.s2)m( rr + -) = Al, where for convenience we assume k c i, m = i, i - k = a, 1- m = b. Now let (5) p ~(rls, )(r2 2)(rs + r )= A2. Then A(6 + A2 P(rls1) (r2Sm( + )( rb + b) (6)21 = S,(r, + s, r2 + S2 rl r2s2, 3), A - A, p-3(181) (282)m(r- 81)( 2 2) (7) = (r, - sl)(r2 - S2) 82(r1 + S, r2 + -S2, r1s1, r2s 2'P3)' where S, and S2 are rational functions. From (6) and (7) we can solve for A, in terms of r, - s, 5 2- s2 and the arguments of Si. Carrying out the same process for all terms of R,, we get R1 = Ml(rl- 8, r 2+ 82, rls1, r2S2,P3) + (rl- SI)(r2- 2)i/2(rl +s1, r 2+s2, rl1s, r282,)3)9 where M. is a rational, integral, isobaric function of its arguments. R2 can be expressed in the same way. Hence, we have the following theorem. Any rational simultaneous invariant of T, and T2 is a rational, integral isobaric function of r, + 81, 9,2 + s2, r1 s, r2s282 P3 (r, - sl)(r2- 82) and 'rs. Since the above seven arguments are themselves invariants, it follows that they form a rationally complete system. They are not independent, however, for we saw that rs2 could be expressed in terms of the other six. It remains to show that these seven can be expressed in terms of the system proposed in the preceding section. The equalities are easily seen to be rl + S1= I r2 + S2= I, r2 - s 2= D - rl-^ s=l _, r ( S= I ( 2 - D2 ) I I 1 ~~2 S2 - 2' 21 366 D. D. LEIB: INVARIANTS [July The expression for rs2 in terms of the present system is very long. Since rs must be added to our system, for rational completeness we will call it I,. In terms of the coefficients of the 3-point,= (a2,c, - a3 b c,2 )(a3 b, c2 - a, b2c3 )(a b2 c3 - a2b3c ) ( a261 C3- a36b2cl)(a3b2c - al b32)(alb3C2- a2 cbl3) The completeness of the system has thus been finally established. The principle employed in this paragraph, namely, that if an invariant contains a term al ak 1 '/g 3, it contains all terms obtained by permuting the i, j, k and I, m, n, is useful in determining all invariant relations, as we need only concern ourselves about a single term of each type. ~3. The dual forms. We next consider the transformation of the invariants when the role of the triangles is interchanged, that is, when the one originally taken as a 3-line is regarded as a 3-point and vice versa. Obviously in the simplified form of the invariants we need only replace each letter by its minor in D1. We will designate the transforms by primes. To abbreviate we denote the minor of any letter by the corresponding capital letter. A, B1 C, D1 A2 B2 C2 = - DI. A3 B3 This follows from the theory of 3-rowed determinants. I= AB,2C + A2B,3C + ABC2 + AB,3C + A2B,C3 + A,3B2, 3 3 3 = E aC 12 3 - a 3 C2-4 ( al 4 a2 b263 c3l a2b361 C2 C3) This can be expressed easily, for it is of weight 2, and the D's must enter in each term and to an odd degree. This follows from the remark that if the terms of a given type have all the same sign, the D's must enter to an even degree if at all, but if half the terms are positive and half negative, the D's must enter in each term and to an odd degree. Hence in the present case we need consider only 1J, DI, and D2. It turns out that IP = I DI- 6D~. I1 2=D-6D2 1I and D2 stand in a certain formal relation to a square array, A,2A3 B2B3 C2 3 AA BB 3 1 3 1 (3C1' jA1A2 B1B2 C3C2 1909] OF TWO TRIANGLES 367 While D' is the determinant of this array, i2 is the expanded form with all signs positive. If we designate the sum of the positive terms of Di by ' and of negative terms by 0, D'=- -0, I= ~+ +e. The actual calculation is long and the following table of summations is convenient here and elsewhere: 3 3 4 b4 C - al b3 C2 = L' - L", 9 9 a n4 b 3 6 a4 b)2 3 3 ] 21 3 2 cs -l b12 3 C3=- M 9 E a 2 a1 b b3 2 b C 3 E aa2 b2 b3 C C3 = N' - N", Is 18 a3a2 b b 2b C2 el - Z a2 b2 c Cl 2 = D' - D", 21 a2 bIl 2 b 3 C1 2 C a 2 3 C C2 C 3 3 E a 2 2 2 b b2 2 2 2- bC2C = E' - E" (g1 2 2 3 3 1 aCa 2 3 1 23 3 3 Ea' a a a3 1 b b 2 - ab3,b, CC C2 3F' F", Ea232 b3 bc3 Cl E al ab23 b Cc -3 '-G 6 2 63 3 9 ] ala2blb2b3clc2c3 =, 9 a4 =2 b C K2 K. Further let L' + L" = L; etc. In this notation, there is no difficulty beyond long reckoning. Less direct methods offered no advantage. We have then = E' + 3E" - 2D' + 2N' + N"- 5F'+ 6F' - G', 0 = 3E' + E" - 2D" + N' + 2N" + 6F' - 5F" - G". Hence I = 4E — 2D + 3N + F- G, D = - 2(E'E")-2(D'-D")+(N'-N")-1(F'-F")-(G' G"). In I' the D's must enter to an even degree, if at all. It is readily shown that D2= N- 2D+ 2E-F+ G +4H, I D1D2= - N+ 2E+ 5F + G - 4H, D = E+ 2F-2, and from these 2= DJ - 2JD +6D. 2 368 D. D. LEIB: INVARIANTS [July D2 must contain the D's to odd degree in every term. It turns out at once that D'- D1 '2D 3-=A1A2A3BB2B, C C2 C3. The actual expansion and reduction of this form is too long for reproduction here, but we get as a final form Summarizing the results of this section into a compact table, we have: I, -ID -6D2 (I) JI=2 DI) -2JD 1D2 + 6D1, Di D2D 2 = 1 D2 I3 D I3 D3 1 2D1 D2( 1 DZ - 22 D12 ) It is a simple matter to write down the corresponding table when the 3-line is taken in general form.* These formule are very useful, for if we know that the vanishing of an invariant I indicates a certain projective relation between the 3-line and the 3-point, the vanishing of I', obtained from I by these formulae, will indicate the same relation between the joins of the 3-point and the meets of the 3-line. This dual transformation, being obviously of period 2, affords an easy check on the algebra involved in calculating the dual forms. Carrying out the substitutions of (I) a second time we get I'= D(1IDI - 6D2 ) + 6D D2D D 1 D11= D D" D I'=DI, I D2 This2 1isane n 3 1. 3 This is an excellent verification of the dual formulae. ~ 4. Some invariant relations of the three-point and the three-line. We shall calculate directly a few of the simpler invariant relations. (1) The three points are collinear if *FomD,_____ 1=O. * For these forms, see the Johns Hopkins University Circular, July, 1908, p. 125. 1909] OF TWO TRIANGLES 369 (2) The three lines are concurrent if A1 0. The above two theorems require no formal proof. (3) The 3-point is apolar to the 3-line if -,=0. For x1x2x3 0 is the 3-line and (a) (b) (c~) = 0 the 3-point. Regarding the x's as differential operators, and operating on the equation of the 3-point, we get merely the coefficient of ~ 2 23 in the equation, which must be zero for apolarity. This is, a, b2c3 + 2b3cl + a 3bc2 + ab32 ab c3 + 2 + a3b2cl = 0. But that is precisely =0o. (4) The three points of the 3-point and the meets of the 3-line are on a conic if D =0. For the general conic on the meets of the 3-line, i. e., on the vertices of the reference triangle, is alX2X3 + a2 3x1 + ax3 l2 = 0. If the points (at) = 0, (b5) = 0 and (ct) = 0 are on this conic; ala2ca3 + ca2aa3 al3aa2- = 0 alb2b3 abl + a + Oa3blb2 = 0, a c2c3 + a2 c c3 + a3 c C2 = 0. For these 3 equations to be consistent, their determinant must vanish. That is a2 a3 a3a, a a2 b2b3 b3bl blb2 = 0, C C3 C3 1 C C2 which is D2= ~0. For the general case, a study of the degree in the Greek letters shows that A2D, = 0 1 2 is the general condition. The dual of D2, D, D2 = 0, is therefore the condition that the 6 lines be on a conic. (5) We will next take the problem of PASCH * triangles and prove: If the *Mathematische Annalen, Vol. 23 (1884), p. 426ff. See also HUN, these Transactions, Vol. 5 (1904), p. 49. PASCH shows that if there is one such triangle in a collineation, there is a quadruple infinity. 3700 D. D. LEIB: INVARIANTS [J~July two triangles are such that the 3-point is a PASCH triangle in a normal collineation having the 3-line for a fixed triangle, D1 (1 DI- 3D) = 0. To show this, we take again the 3-line as reference triangle. The normal collineation can be written x=- ka. and we want the transform of a to be on the line be. Similarly b is to go into a point on ac, and c into a point on ab. This gives ns three equations like kiaa k2a2 k3a3 bi b2 b3 -0 1 c2 c3 C, C, C WTe can write them k a,1A + ~k2a2 A2A k3a3A 3=0, k1 bi B1~+ k2b2B 2~ klb3 3 = 0, k c, C + k 2 c2QC + k3C3 C3= 0, where the A n, Bi, Ci, are minors as before. For these 3 equations to be consistent, we must have a1AA a2A2 a3A3 b1 B b2B2 bB3= 0. 1 1 c2C2 C3C C1CI1 C2 C1 C3 C3j Expanding this we get 18 18 Ea 13 Z...3 112a3 b23 ~Za2a3 b2 b C' C1C2C- 6a aa a b b b C C C 0. a -1 2 3 1 2C 12 2 231 2 3 This expresses itself readily in the form _ID -3D DD =0 or D(11 DI - 3D2)=0, as we were to prove. (6) If the two triangles are in perspeCtive position, 16 = 0 - It will be fonnd that the 6 factors of 16 correspond to the 6 possible orderings of the vertices of the two triangles. For example, by way of proof, suppose we ask that the joins of 1, 0, 0; 0, 1, 0; 0, 0, 1 to a; 6; e, respectively meet in a point. The 3 joins are a 2X3-a3 x -0, b3XI -61x3O, 0 1 I 2 - 2 1I 0 1909] OF TWO TRIANGLES 371 Hence the condition that they meet in a point is 0 - a3 a2 b6 0 - b = 0, - C2 C 0 or a2b 3c - a3 6bc2 = 0. In the same way each ordering requires a distinct factor of I to vanish. As a corollary we can also state: If there exists a conic with respect to which the 3-line and 3-point are mutually polar, then must I6 = 0. All the problems discussed by Dr. HUN might be treated independently as those of this paragraph have been. But every fundamental system of invariants must be expressible in terms of the present system, since it has been proven complete. Hence by constructing a table of equivalents we can utilize directly Dr. HUN's results. ~ 5. Hun's invariants in terms of the present system. HUN'S three fundamental invariants are defined for the case of a general 3-line and 3-point as follows: 1 — ZBll I2= ( B22 B33 B23B32) Bln Bl2 B13 13 = B21 B22 B23 B31 B32 B33 where BUi-aO( y )i + Si(rya)i + ri( a1 )i, B, = a,(i37) + /3,(y) + y(c)7W, Bk = a.(!3v)k + 3. ( ya)k + 7Y( ax)i, and in turn the symbols (/ry)i, (ya )k, etc., are defined as follows: (/3r) = a(bc//3y) + bi(ca/3y) + ci(ab/f3y); where finally (bc/l/3) = (b) (cy) + (by)(c9), (ca//3y) = (c3))(ay) + (cr)(a,3), etc., the parentheses now indicating row products. If now the 3-line is taken as the reference triangle, we need retain only terms containing the product a/32/ 73 and replace it by unity. This gives 6 Bl =B22 =B33 = E ai 6k Cl, 372 D. D. LEIB: INVARIANTS [July B21 = 2(abc, + ab3 c + a3 bel), B12 = 2(a 2 b2 + a23 c2 + a3b2 C). In general (1) B.i = 2 (a. ck + abkcj + ak bj c.). Then denoting HUN's invariants by bars to avoid confusion, we have 3 6 =- EBii 3E=-3 ai.el=- 31, 3 6 I2= (B-22B33- B23 B3)= 3( ai.Cc)2 - 4[(a3b3Cl + a3b1c3 + alb3c3)(a2b2c l+ ac2blc2 + a b2c2) ~ (a3b3C2 + a3b 2c3 + a2b3c3)(alb, c2 + ab2c, ~ a2b, el) + (a b a C(ab + ac + Cca c + a, + a bc) + (a2bc +23c2 + 3c + 2)(abc3 + 2ab3o1 + 31 )] 9 6 = 38I2- 4(Ea b2b bc3 + 3 Ea2a3b3b, c,1 2) - 3I - (I1 - D + 12]2) = 2I_2 + D- 12I2. The transformation of - I = - ADN is more difficult. Obviously A and D are A, and D1 respectively. We have directly from above equalities (1), 1 2(a2b2c3+ a2b3c2+ a3b2c2) 2(a3,b3c+a,3b2c3+ abc33) -I3= /2(abc3+ a1,b3c +a3bc,) I 2(a3b,3c + a,3blc3+ ab3c,3) 2 (ablc2 + alb2C + a2b1c) 2 (a2b2c + a2,b12+ alb22) 1 Expanding this and using the equality found in evaluating 12, we have 18 12 ' ---=. 1 23231 c2 + Z 2 33C1 2 _7 3 + 16 a a a b b2 C C + 8Ea ab b, 2 + 48a a, b3b b2 b, c c - I ( - D2 + 12z) _ I3 + 12L I- 4D1 D2-I 3 +1 JN2-1 22D1 ( D1 4D2 ). = I + 127 7, - 47)D zD - 13 + IDi -127,7= (zi - 4D). Hence N= 4D, - I D1. Therefore we have as a complete table -D D 1 = 2T2 + DL Al2 - 1212 A =, N= 1- D1 + 4D2,, I =-31,, where the A 's are inserted so as to be correct for the general 3-line. 1909] OF TWO TRIANGLES 373 Using these formulas we shall write down the equivalents of Dr. HUN's results as tabulated on pages 50 and 51 of these Transactions, vol. 5 (1904). The 3-point degenerates, if D1 =0. The 3-point is apolar to the 3-line, if 1 =0. The meets of the 3-line are apolar to the joins of the 3-point, if ID1A - 6D = 0. A conic circumscribed to the 3-point, and apolar to the 3-line exists, if I A1D1 - 2D2 = 0. A point conic apolar to both the 3 -point and the 3-line exists, if D2 D2 =0. A conic circumscribed about both the 3-point and the 3-line exists, if A2 D =0. There exists a point, whose polar conic as to the 3-line is apolar to the 3-point, if D Al(-, DIA-4l- )= 0. The three polar lines as to the 3 -line, of the points of the 3-point, taken two at a time, meet in a point, if A, (, DAl _- 4D2) = o0. There exists a collineation having the 3-point as a fixed triangle, which sends each meet of the 3-line into a point on the opposite line, if Al(ILDAl- 3D2)= 0. The 3-line degenerates, if A1 0. The 3-line is apolar to the 3-point, if 1= o. The joins of the 3-point are apolar to the meets of the 3-line, if I1D1,Z - 6D2 = 0. A conic inscribed in the 3-line and apolar to the 3-point exists, if D1 A - 2D2 =0. A line conic apolar to both the 3 -line and the 3-point exists, if A D= 0. 1 2 A conic inscribed to both the 3-point and the 3-line exists, if D2D =0. 1 2 There exists a line, whose polar conic as to the 3-point is apolar to the 3-line, if DiAl (IDIAl - 4D2) = 0. The three polar points as to the 3 -point, of the lines of the 3-line, taken two at a time, lie on a line, if (I, D1A1-4D2) =0. There exists a collineation having the 3-line as a fixed triangle, which sends each join of the 3-point into a line through the opposite point, if D1(IDI 1 - 3D9)= 0. There exists a collineation having the There exists a collineation having the meets of the 3-line as fixed points, which joins of the 3-point as fixed lines, which sends each point of the 3-point into a sends each line of the 3-line into a line 374 D. D. LEIB: INVARIANTS [July point on the join of the other two, if through the meet of the other two, if ID1 A - 3D2 = 0. I, D A - 3D2 = 0. There exists a point, such that its There exists a line such that its polar conic as to the joins of the 3- polar conic as to the meets of the 3 -point is apolar to the meets of the 3- line is apolar to the joins of the 3 -line, if point, if D2(A2(DI SAl -= 2D0)= D A (I Dz - 2D2)= 0. To this we will add one more, adapted from the same article. There exists a line conic, touching the joins of the 3-point, and apolar to the 3-line, if DI(D1za- 4D) = 0. It will be seen that the essential factor of all these invariants is a member of the pencil, 1 D A1, + XD = 0. It may be noted that if two points of the 3-point are taken as the circular points at infinity, and the third point taken as variable, this pencil is the pencil of circles determined by the Feuerbach circle and the circumcircle. Another pair of dual theorems which can be readily proven are these: If one of the 3 points lies on either of the lines of the 3-line, I-0. If a meet of the 3-line lies on a join of the 3-point, 2D - 2D3 + D1D2 ( D - D) =. The powers of A1 are not inserted in the latter form, and will not be in the future unless for some special reason. ~ 6. A conic inscribed in one triangle and circumscribed about the other. The present investigation has grown by extension from the problem: To find the invariant relation on two triangles so that a conic may be drawn touching the sides of one triangle and on the vertices of the other. This problem together with its applications is especially interesting. Let the three lines be the reference triangle, and take the polar conics of the three points a, b, c as to it. The three polar conics are: (1) al x2X3 + a2x3x + aC3Xx2 = 0, (2) bxx, + b2xx,1 + b,3x, - 0, (3) CI, c23 + c2 x3x + c3xX 2 = 0. 1909"I OF TWO TRIANGLES 375 Let 1 be any line of (t), the polar conic of a, and the poloconic of q will pass through a.* Hence if the three polar conies have q as a common line, then the poloconic of sj as to the 3-line will be on all three points a, b, c. But it will also touch the three reference lines a, i, y, for the Hessian of a triangle is the triangle itself, and the poloconic of any cubic touches the Hessian in three points t; in the present case it touches each of the three lines making up the Hessian, for they enter symmetrically. Consequently the problem is now reduced to finding the condition that the three polar conics have a common line. SAL-MON t gives the condition that three point conics have a common point as 72 = 2 6431. For three conics to have a common line we need only form their line equations and impose the analogous condition. The line equations of (1), (2), (3) are (4) a2 V + a 2 V + a2 - a aa3,3 - 2a, a, 1, - 2a, a,41 = 0 = U, 65 2 ~22+ b2 t2 + h2 e2 - 2b2 b t 5 - 2b 5, - 2b, 5, C, (5) I 2~ l~ I b 1~ V,7 (6) C2 ~2 + C2 t2 2 ~2 - 2c2C323 - 2C3C1 341 - 2c = = IF 1 1 T 2 ~2 3 3 2 2~ 3 1~ 1 C 3 0= Now 31= 0 is the condition that in a net of line conics lU~ mV~ n W there shall be a double point. In that case, the reciprocal of that particular conic of the net vanishes identically. Following SALMON'S notation, ~ we find the equation to be (7) 12> + n2>2Z' + n2 >2" + muck23 + nl 031 + 171412 = 0. The E's are the reciprocals of U, V, W respectively and the f,, are the Clebschians of the same taken two at a time: (8) >3 =4a a, a (a, axx+ ax~x1+ ax xx)=0. >' and >2" are the same in b and c respectively. By the Clebschian of two line conics (a~ )2 and (b )2, We mean the contravariant I abx 12, where I abx I is the determinant. 11 In the present case k23 = (a2b3 - a3b2)2x + (a3b - a b)2 x +(a, b, - a 2x (9) -(a, b, - alba)(a b2 - a2bI)X2X3+( a, b2 - ac2b)( a2b3 - ab) xIx3 + (a2b- a3b) (a3B1 - a,b3)xIx, * SALMON-FIEDLER, - Analytische Geometrie der hoheren ebenen Kuirven, p. 292. The English edition of this book is out of print. SALMON calls the poloconic simply the polar conic of the line. CLEBSCH-LINDEMANN, - Vorlesungen iiber Geometrie, Vol. I, p. 543. KoHN, Encyklopdclie der methematisehen Wissenschaften, Vol. III, part 4, p. 471, gives an account of the poloconic with references. t DURkGE: Ebene uneven dritter Ordnnng, p. 278. T Conic Sections, p. 365-6. Z Conic Sections, p. 366. 11 Ibid., p. 344. CLEBSCH'S form is non-symbolic and 0 equals 012 above. See also CLEBSCHLINDEMANN, Vol. I, p. 277. If ai = bi, the Clebschian is the line equation. Trans. Am. Math. Soc. 25 0 IT Ob 61 6 D. D. LEIB: INVARIANTS [July and 431 and r,, are similar forms in b, c and c, a respectively. The above forms, E and 4j,, are most easily verified by expanding the symbolic forms and substituting the real coefficients. SALMON gives 1 in the form of a 6-rowed determinant. From equations (8) we see that the block 1A B C~ AlI B1 Ct is composed entirely of zeros. Hence 21 reduces at once to the product of two 3-rowed determinants 2a a2a3 2a1a2a3 2 3a1a2a2 (a2b3- a3b2)2 (a3b1-ct b3)2 (a,b, -aba)2 (10) 2b2b~b3 2blb2b3 2 b 2b~b (b~c3- 53c2)2 (bSc1- blc3)2 (bc - )2 2 C2C3 2cc 2 C 2c c2c (c2a3-c3 a)2 (c3a - cla3)2 (cla2- c2a1)2 1 23 12 1 2 3 C2 - 32 (332 The first determinant is clearly 8J D1. The second is seen to be the dual of 2 2 2 C C C 1 2 3 a2 a2 a2 (11) ia1 2 3 b2 b2 b2 1 2 3 Now (11) in terms of our fundamental invariants is D1 -2D2. The dual of this, by the formulhe of (I), ~ 3, is D (, ~LP- 4D2). Therefore (12) M= 83D(3 (,D, - 4D2). To express T similarly we can start with either of SALMON'S forms as given on pages 365 and 367 of his Conic Sections. We shall develop the second form. (13) T= 02 4 +,,8,,+8,,8, 3,,,,) 126. (13) T= 0~~~~13 -4( 0122 0133 + 0211 0233 + 0311 0322 ) +1 Next recall the meaning of the 0's. If we write the discriminant of (14) 1 (~ 2 +M(b~)2 + n (C~)2 we can define them as follows: 0123 is the coefficient of lmnn in the discriminant, 0122 is the coefficient of ln 12, and so on. For the net I U + m V + n W, this gives 6 6 9 (15) 0123 a 2 3l C-2 aa2b2b3c3c1 -2 Ya2 bcc3- D2 -4 1909] OF TWO TRIANGLES 377 (16) 0122 - 4(a1a2bl b2b3+ a 2a3b2b3b2+ a3a- b2 b3), (17) 0133 -4(a1a2c1c23 + a2a3cc2c3+ 3CCc2C3), and similarly for 0211, 033, 0311 0322 Combining these we have 0120133= 16 (a\ a| 1 b2 b3 c 12 c3 + a 2 b2 b2 b3 c 2 6 + a a bl b c2 b3 C C2 C3 + ca aa2a3b1 b2 b3 C C2 3), 0211023 = 16 ( al a2 a2 b2 Cl C2 3c + -2 aa2 a3b b1 c? C2 C2 6 ~ a a 2 C3 a+ E ala2a3 b 2b3 c123 0 - a 2a~~~b~~b aa1c3 3 2 C2 C 2 2 a 3 C), 03110322 = 16 ( al 2 a2 b6 b2 b 21 c2 - a2 C12 a3 b a 2 b3 C2 C2 6 + a, a 3 2 b1 3 2 c2 + 2 2 b2 b 2). Hence 9 6 18) 01220133+ 02110233+03110322=16(Z ala2blb2b3 2C +3 + a~la2a3bb2b3CCc2C) =4 I2 - 4D22+1481 13. As a final step we must evaluate e, which is defined as the coefficient of Imn in the reciprocal system 1 + mZ' + n"= 0. The expressions for, ]', "// were given by (8). The discriminant can be written at once in determinant form, as 0 la3 + m b3+ nc, la2 + mb + nc2 (19) 8ala2ac3b b2b3 C2c3 la3 + mb3 + nc3 0 la, + mbl + inc1 a2+ mb2 + nc2 la, +?-zmb + ncl 0 (20) = 16aa2a3bb2b3cc2c( la + mb, + nc) (la2 + mb2+nc)(a3,mb3+nc3). The coefficient of Inmn in (20) is seen to be 6 16a a, a b b cc63 C C3( C 3) = 16,1 3. Hence (21) ~ =16L1,. Substituting the values (15), (18) and (21) in (13), we have after reduction (22) T= D- 8D2 J + 16D2. Substituting the values of M and T thus found, in the condition T2 = 64M, we have the theorem If the 3-point and the 3-line are such that a conic may be inscribed in the 3 -line and circumscribed about the 3-point, then will 378 D. D. LEIB: INVARIANTS [July (23) (D - 8D1I2 + 16D )2 - 512I3D(1 - 4D2) = 0. Forming the dual of (23) we get a theorem which admits of an important application. The dual is [ D - 8D (D - 2J1 D1 D2 + 6D2 ) +16D4 D2 2 -512D6 (D + - D - +I D7 -)(1 Dro - 2D1 2) = 0, which reduces to (24) D11 [ D (D + 64I2 + 321 D1 D2 -16D 7 - 64D2) -521 3( D - 2D2)1 =0. From this we have the dual theorem, neglecting the trivial case where the three points are on a line: A conic may be drawn on the meets of the 3-line and touching the joins of the 3-point, if (26) D1(D +64I + 32 D D2 —16D7- D 1 64D2)-51213(J D- -2D2)=. Next consider equations (24) and (26) when two points of the 3-point are taken at the circular points at infinity, and the third point as a variable point x. They become equations of curves of the 8th and 5th degrees respectively. We shall discuss these curves briefly, as many properties can be deduced directly. We will designate them by R and Q respectively. The equations of these curves can be written down explicitly by replacing bi and ci by the coordinates of I and J and ai by xi. In the case of conics inscribed in the 3-line and on the 3-point, if two of the points are taken at I and J, all the conics become circles touching the 3-line. In other words; the octavic, R = 0, is the locus of points from which circles can be drawn touching the lines of the 3-line. In other words, R = 0 is simply the equation of the 4 circles which can be drawn touching the 3 lines. In the dual case, the join of I and J, that is, the line at infinity, becomes a tangent to all the conics. Hence they are parabolas. Further the lines joining I and J to x are to be tangents to the conics, so that x is in every case a focus. Hence the theorem: The locus of the foci of parabolas on three points is a quintic curve, and if the points are taken as vertices of the reference triangle, the equation of this quintic is Q=0. It is possible to deduce many properties of R and Q without writing out the * This was further verified by expanding SALMON'S alternative form, Conic Sections, p. 365. 1909] OF TWO TRIANGLES 379 explicit equations. As a first step it is well to note what curves are represented by equating certain simple invariants to zero, when the points are taken as above. — = 0, is the equation of the reference triangle. D1 = 0, is the line at infinity. I1 0, is the polar line of I and J as to the 3-line. D2 = 0, is the circumcircle. ILD1 -2D 2 = 0, is the apolar circle. 1 D1 - 4D2 = 0, is the Feuerbach circle. None of these require proof here except possibly the last. This can be shown directly from HUN's form, for he shows * that N== 0 is the equation of the Feuerbach circle, and we have found NV= I D1 - 4D2. Or it can be proved directly from the easily established fact that I1 D - 4D2 = 0 is the locus of centers of rectangular hyperbolas on the vertices of the 3-line. But the feet of the three perpendiculars of the 3-line are clearly such centers and are also on the Feuerbach circle. Hence since I D1 - 4D2 = 0 is a circle, it must be the Feuerbach circle. (All the conics 1 D1 + XD2 D= 0 are circles, being on I and J, the intersections of D1 0 and D2 = 0). Now take the equation (27) R =(D 4 8D22 16D2 2 51213 D (71 D - 4D,)- 0, which we know to be the equation of the four touching circles. If 3= 0, (DI - 8D12 + 16D )2 = 0. Hence, the reference lines touch the octavic where they cut K-, the quartic, D4 - 8D I2 + 16D2 = 0. If 1 D1 - 4D2 0, K2 = 0. Hence, the Feuerbach circle touches the octavic (or passes through double points) where it cuts K. This is really the well known theorem that the Feuerbach circle touches the inscribed and escribed circles. The quartic K is interesting in itself. From its equation, K= D - 8D2 + 16D =0, we see that if D = 0, D = 0, and if D2 =, either D2 = 0 or DI - 812 = 0. This shows that K is a bicircular quartic, that is, it has double points at I and J. Since the latter are quadruple points of R, all the intersections of K and R are accounted for and easily constructed. There are 12 at the points where the reference triangle touches the R circles, 4 where the Feuerbach circle touches them, and 8 at each of the circular points at infinity. Since R = 0 can be easily constructed, we will now take up Q = 0. We can abbreviate Q = D1[(D - 82 )2 + 321, D1 - 64D2] - 512 (,,1 - 2D2)= 0 *These Transactions, vol. 5 (1904), p. 47. 380 D. D. LEIB: INVARIANTS [July by calling D - 81,= U, 16(ILD - 2D2) V, therefore (28) Q= D1(U2 - DV) - 321V= 0. In this if D1= 0, JI3V= 0. That is, Q cuts the line at infinity where V = 0 cuts it. But I3= 0 is the reference 3-line and V= 0 the apolar circle. Hence The quintic Q =0, passes through the circular points at infinity and has its asymptotes parallel to the sides of the reference 3-line. Again if V- 0, D1U2 = 0. This says that the apolar circle touches the quintic (or cuts it at double points) where U and V intersect. Since Q can be proven rational,* its six double points will not in general lie on a conic. Hence V does not cut Q in double points and we have the theorem: The apolar circle, V= 0, has fourfold contact with the quintic, = 0, touching it at the four points where U= 0 cuts V-= O. This theorem can be proven more directly by taking the dual of V= 0, which is 1 D, - 4D2 = 0, or the Feuerbach circle, F. But, by the wellknown theorem, the Feuerbach circle touches each of the four tangent circles, WR. Since contact is not destroyed in taking duals, V, the dual of F touches Q, the dual of R in four points which was the theorem to be verified. Rewrite Q, thus: (29) Q = D1 U2 2 V D1,D- 1613) = D1 U2 - 2 VW= 0. The curve W = 0 is a cubic on the vertices of the reference triangle, cutting the line at infinity where the sides of the reference triangle meet it. That is, its asymptotes are parallel to sides of reference triangle. When TW= 0, U2 = 0 or D1 = 0, hence IW= 0 touches Q where U cuts W. But since W and Q have parallel asymptotes, it follows that three of the contacts are at infinity. We can also find where Q cuts the circumcircle. In Q, let D2 = 0, and it becomes (30) D (D - 82 )2 - 512, Z3D = 0. So unless the 3-point degenerates (31) (D2 - 8I2)2 -512 h 13= 0. Then let D2 = 0 in R and drop D' as a factor, the result, D2- 8I2 )2- 512 13= 0, *R. F. DAVIs, Educational Times, December 2, 1907, p. 545. 1909] OF TWO TRIANGLES 381 is the same as (31). Hence: Q and R cut the circumcircle in the same points. It should be noted in this connection that the question, whether a locus is on any of the 6 fundamental points - that is, the points of the 3-point and the meets of the 3-line-is equivalent to the question whether the corresponding invariant vanishes, when there is a coincidence among the fundamental points. This is a question of some importance, but will not be discussed in detail here. ~ 7. The Clebschians and their invariants. In the case of two general line cubics ( a )3 = 0 and (b )3 = 0, the Clebsehian is a point cubic defined by (1) X ==abx3=O. Similarly the Clebschian of two point cubics (ax)3 = 0 and (/3x)3 = 0 is defined by (2) X' al 13 I 0. If, as in the case of two triangles, the curves are of both the third class and third order, the two Clebschians are in general different. In forming the Clebschians both triangles or cubics must be taken in points or both in lines. So the work here is slightly different from the preceding sections. If one cubic is taken as the reference 3-point and the other as (a )3 = 0, from (1) we have (3) X = 6(a2ax23x2 + a axa221, x + aax2x- aXx alax2, a2 a- a2 2 - a2ax23 ) If in turn the second cubic is taken as the 3-point (at)(b) (c) = 0 we must replace 3a a2 by (abc, + abc, + alblc), and this gives us X= 2[(a2b3c + g3b2c3 + a c3632)x2 -+- (ab2c + a,2bl2 + a2b2c1)X1X2 (4) + (a3bl1 + a1bc + a, b c) a Xl3)2 - (ab3c3 -+ 3b1c3 - a 3b3C )xx -(l (a b1c + acGlac + a b )2- (a3 + 2b32 + 2b2c3) ] It will aid in clearness to define x geometrically. If we call the reference 3-point d, e, f, then x may be defined as the locus of points y such that the two triads of lines y - d, e, f and y - a, b, c are apolar. %' is the locus of lines V7 having the dual property. From (4), we see that x is on the vertices of the reference 3-point, and, since the two 3-points enter symmetrically, we have the theorem: 382 D. D. LEIB: INVARIANTS [July The Glebsehian of two 3-points is on all six points.* Further since (4) contains no terms in x~x2x3, the reference triangle is apolar to it. Hence, The Ulebschican of two 3-points is a cubic to which both 3-points are apolar.t The dual theorems for two 3-lines can be written down at once. Since the invariants, S and T, of X are mutual invariants of the two 3 -points, they must be expressible in terms of our fundamental invariants. For 5, we will make use of SALMON'S standard form.t In X, the SALMON coefficients have the following equivalents, a=b= c = m = 0, =(I2 a2b3c3 + a 3b2c3 + a3b3bc)3, a3 = - (a3 b2c2 + aAb c02~ a2 b2c), b, = - ( ab3 c3 + a 3bIc3+ ~ 3b c ), b = (a3 bl cI + ~, b 3c + a, b, cc1 = - (a 2b1c1 + a1b2c1 + a1bc2), c = (alb c2 + a2 b1c e+ a2b2ec1). The nonessential, common numerical factor has been dropped. This gives, by substituting in SALMON's form, 5 = - (a, bOc3 + a3bc3 + a 3b3c1)2 (a b2c2 + a 2 blc2 + a 2b2c, )2 - a2 bc + ab + a b c )2(a b1c1 + a1bc, 1 + alb c)2 - (ab2c2 ~ a2 b c2 + -- b2 cc3)2 ('a 3blc + a, Nb c + alb, c3)2 ~ (a2 b3c3 + a3b2 c3 + a 3b3c2 )(a'12 bc1 + a, b2 cI + a, bl c2) (5) (a3 b2c2 + a2 b3c 2+ a 2b2c3 )(a3 b, cl ~ ab3cI + a, b, c3) + (33b2c2 + a 2b3c2 ~ a2b2c3a)(a3bc + a,b3ce + alb~c ) (alb 3c3 + a3b~c3 + a 3b3c1)(alb2c2 + a bAc2 + a bAc ) + (a b3 c3 + a3blc3 + a3b3c,)(a, b2c2 + a2blc2 + a b 2c1 ) (a2 blc, + ab2cl ~a, b, c2) (aAb0c3 + a3b2c3 + a3b3c2). Expanding and collecting the above we get 15 9 9 3 2 2 2b b2 C C2 b2 b2 C2-~ 2 2 (6) 16[ a a.bb2 l 2 33 - E a a b a' 3 or using the symbols of ~ 3, (7) S = 16(N-H-K). * These two theorems lay no claim to novelty, see F. MORLEY, these Transactions, vol. 5 (1904), p. 472, for a discussion. t llihere Ebene Kurven, p. 247. 1909] OF TWO TRIANGLES 383 Employing the fact that the D's must enter in each term to an even power, if at all, we find that the only combinations entering are 12(D 2 + I2) =2N~ 4E~ -4- 10F~+ 2& ~ 8H, JT2 - D0 2 4H, (12 - Df)2 - 32N~ 64H~ 16K, ]1 D1 D= N+ 2E + 5F + G - 4H. It is not difficult to select the proper coefficients; the resnlt is.Q-1QT/ 2 + 12) _ 2 _(1 22 (8) 8G 112(i 1 36 J - 2 I Di.2411 DI D2 = 4 (-1ID, - 3D2)2 (I2 + D 2- 61)22 (9) 2 1 1 ~ Geometrically S = 0 is the condition that the Hessian of the Clebschian of the two 3-points shall be three lines. Since the Clebschian treats the two triangles symmetrically, it is clear that S should be self dual. Applying the formule of ~ 3, we get (10) S =D I. This is a good check on the work. To substitute the coefficients of x directly in SALM\oN's form for T* leads to complicated expressions. The following method presents fewer difficulties. Let (axx)' and (a )3 be two cubics giving the connex (11> (~~)"((aa)(ax)(a~) = 0. The 3 fixed points of the connex are given by (12) ci (ax)(cia)2 = XXi; (i12,3). If the 3 equations of (12) are consistent a1 a1,(aa)2 + X a, a, (a(aa)2 a3C1 (Ca) 2 (13) A(X),=a, ae, (aa)2 a a, (aa)2 + X a, a (ca)2 0. dl a, (aa)2 Cta2a3(aa)2 a3ajaa)2 + X The coefficients of powers of X in A (X) are invariants of the connex. Now let (ax)3 become the reference 3-line and 1 +23 2 3 2a4)a2a+X 2ai aa 2a a (1_4) A(X)=I 2c a~a3 2aa2a3~X 2a a a 2a a2 2c 1 2a a 3a+X * Bhhere Ebene Kurven, p. 248. 384 D. D. LEIB: INVARIANTS [July If the second cubic is taken as a~3 + 3a + c| + 83, + 3a3 + 3b l + 3b32 + 3c 3c1 + 3c2 32 + 6m,1 23 = 0 A (X) takes the form, dropping numerical factors, m + X a3 a (16) A(X)== b3 m+X b, =3+J1X2+J2X+e3, c2 c1 m +X where J,= 3m= 31,, J = 372 - (bc, b c2 + a 3b3), nZ a3 a2 J3 = 3 = )Z b c2 CI m Now the Clebschian of the reference 3-line and the general cubic is X = Cx2 X + b x1 + a3x2 - C 2 - aX2 - X a - 3 = 0, and T for this is T = 6b 1 a3 3 - 8 ( b + 2 + a b) - 27 ( a b + 1 2 2 3 ) + 12(b C2 a 2 c2 2 a+ 3 b3 +2 a3 b3 + C2 2 b, cl + a2 b2 c, + 2 2 a% ). Or we can write (17) T= 6x - 8y - 27z + 12t. From the expressions derived above we have 3 6 (18) (3_-)3 = E b C + 3 b2cca2C2 + -6bcc a2a3b —6- + y + 3t (19) (3 + 21 3 J,)2 =c2 b\2 C+ 23 2 c + 2b, 1 C, 3 b 2x+ x, \V "1 1 22/ 2 3 1 3 1 2 2 32 3 3x y 3 (20) S(32 - ) = -b3c + 3,363= 3 + a a Elimination of x, y, z, t from these four equations gives (21) 7= 4(312 -J2 )3 - ) 7 (J3 + 23 - J2)2 + 2S(S - ). But I, is identical with the JI of the present system, J2 and J3 are I, and - 13 of HuN's system, and S is given by (9). 1909] OF TWO TRIANGLES 385 Substituting these values in (21) we find 1= - 8[( I 2 - D2 )3 - 21613 - 182 I( D4 4 ) (22) + 108(II2 + DRD2) - 54(D1I- -I D)2 + 361, DI (I - 2)- 64812D 2 ]. Applying the dual formulas we have T' -- DT, again a check on the work. As a further example, the invariant condition on the two triangles, that the two Clebschians may be apolar, will be calculated. The outline, only, of the work shall be given. The form of x is given in full by equation (4). To get the equation of X, the coefficients of the corresponding terms can be derived from X by replacing each letter by its minor in D1. Call the minors A,, B1, etc., as in ~ 3. Then '= (AB, C 3 AB2, A C)2 + (AB2 + BA2B C' A 2B2CA1)tI, + (AB, + C A + ABCt + ABC) 3- (ABC3 +A3B,+A3B3C,), - (A2B1C1 + AB2 C1+ AlBl 2) - (AB, C2 A 2,3 + A2,B C3)0 3. To find the apolarity condition, regard, as usual, the %'s as differential operators and equate the result to zero. This gives (ag2b3C3 + aC3b2C3 + 3 b3c2)(A2B3C 3 + A3B2C3 + A3B3C2) + (alb2c2 + a2b, c2 + a2bc2el)(A1B2 C2 + A2BiC2 + AB2C01) + (a3b1lc + a1l b3c + albc3)(A,3BlC + AlB,3 C + ABC3) (23) + (al + 3 + A361C3 + C b33Cl))(A B3C3 + A3BCi3 + A3B3C1) + (a2,1c, + cbc, + abb2 )(ABC + A(AB,2B + A, BlC2) + (a3b2c2 + ac3c, + a2,2c3)(A3BC2 + A2B3C, + A2B,3)- =0. By replacing the capitals by minors in the small letters and expanding, the whole can be summed thus: 9 9 6 6 2(-,a 2 2 3 2( 3 2|61 3 1 231 2 3) + 3(Za1a2b2 3e332- IEa2b3 C2 c2) ~~~~~~~~~~(24) 2(o9 9 +Expressed in t( -f te f am al iara, ts i)0 Expressed in terms of the fundamental invariants, this is (25) D_ - _ID 3- 31D~ + 9JD = 0. 386 D. D. LEIB: INVARIANTS [July Hence: The two Clebsehians of two triangles are apolar if D3 - I D - 31, D + 9J2 D1 = 0. Of necessity, this form is self-dual. The invariants of X' are seen to be the dual of those of X. But S' = S and ' = T. Hence: If neither triangle is degenerate, S = 0 is the condition that the Hessian of one Clebschian break down into 3 lines and at the same time that the Hessian of the other break down into 3 points. There are a number of other interesting cubic covariants and contravariants of two triangles. But the limits of this paper will not permit of a detailed treatment, the above discussion of the Clebschian being itself meager and incomplete. ~ 8. Self-dual invariant forms. A number of self-dual forms have appeared in the course of this article, and the question naturally arises as to the number of such self-dual forms which are independent. Such forms are of special importance, because the vanishing of one of them indicates a mutual relation between the two triangles. The most general self-dual invariant of given degree can be found by carrying out the dual transformation on the most general invariant of that degree, and asking that the invariant be left unaltered save for multiplicative powers of D1. For example, suppose the most general self-dual invariant of degree 3 is required. We ask that D3(aI3 + bD3 +cI 2DI + dLD2 + eII2+fID2+gD1D2+ h2D, + 2k3) -a(I3D3 18D2D2+ 108TD)D2 —216D3)+bD6 +cD(I2D2 -12DID2D+ 36D2) + d)(IDj-6D.tD2) (1) +e(lI+2D3 - 2ID2 D2 + 18IDLD2 —672D2-D-36D3) +f (- ID3D2+ 6D D) 2)-g(D )D2) +h(DJI2-2JIDD2+ 6D2D2) + k(2D31- 2D3 + LDD - -2D2D) Equating coefficients, we get the equalities c =- 9a, k = 54a, 6c — - (f+ h), g = - 3d. Introducing these in the left of (1), we have as the most general self-dual form of degree 3, (2) a(f3-9J-+1083,)+ d(D-33DD2) +c(I2D1-6,D2) +h(I2D —IJD). 1909] OF TWO TRIANGLES 387 Since a, c, d and h are arbitrary constants it follows that the expressions in parentheses are all self-dual forms, as can be readily verified. In the further study of self-dual forms, we get the following system of five independent ones, which form a complete system: (1) D1, (2) 1 D -3D2, (3) 1- 61, (4) 2D1 - I1D, (5) \ - 91 + 108I3. To prove this system complete let us designate forms (1), (2), (3), (4), (5) by A, B, C, E, and F respectively. We then have the following equations: D =A, 2- -3 (B — 1A)= (A7l- B), I = ( -C), I3= f (F - 1C 312) 1 (F- 3 C1 + 13. That is, D1, D2, I, and I3 can each be expressed in terms of I1 and self-dual forms. From (4) or E we get the further equation A I E= (A - C) - (AIZ - B) or 6E= 2B - AC - AI, and hence (6) 2 6E+ A - 2B a formula which enables us to reduce higher powers of I,. Hence from this series of equations we see that any invariant can be expressed as a rational function of the five self-dual forms A, B, C, E, F, and of 1. to the first degree. That is any invariant I can be written as (7) I= S1 + I S2, where S1 and S2 are rational functions of A, B, C, E, and F. Now suppose Iis self-dual and written in form (7). Since S1 and 82 have been proven self-dual, it follows that (7) can be true only for kI S.2 = 0, as it can obviously not be self-dual, 1_ not being self-dual. Therefore if self-dual I= SI, that is, a rational function of A, B, C<, E, F. This proves the theorem. Any form which can be expressed in terms of these five is self-dual, and it is frequently easier to show the self-duality of a form by expressing it in terms of 388 D. D. LEIB: INVARIANTS [July A, B, C, E, F, than by applying the dual transformation formule. Only the first two of the five have a simple geometrical interpretation known to the writer. ~ 9. The fundamental invariants under a special Cremona transformation. It is the purpose here to show the effect of a special Cremona transformation on the fundamental invariants, and to apply the result to a single example. Take the simple, quadratic involutory transformation k y --. The effect of this transformation on the invariants is to replace each symbol by its reciprocal. Since all invariant forms are homogeneous, k will appear to the same power in each term, hence may be taken as unity without loss of generality. The fundamental invariants, under this transformation, become D' - 1 -T2 (1) D I1 1' A I II I,31 J, D1, 1, 1 -T3 3 -33 As a typical illustration we take the rational quintic, t -a. x) (i- 1, 2, 3). This quintic has 3 cusps, 3 double points, 3 double tangents, 3 flexes, and is of class 5. It follows from the simple counting of constants that any two of the singular triangles can not be taken arbitrarily, but have a single relation connecting them. We shall give in detail but a single one of these relations. Taking the triangle of double points as the reference 3-line, we ask: What is the condition on the triangle of cusps? From the theory of Cremona transformations,* it is well known that if the above involutory transformation is carried out on the given quintic, with the double points as fundamental points, the quintic goes into a rational quartic with 3 cusps, and with simple points at the 3 fundamental points. If we next carry out the same transformation regarding the cusps as fundamental points (which change of fundamental points is evidently equivalent to taking the dual), the rational quartic goes into a conic which touches the original fundamental lines, and has the three cusp points as ordinary points. But we already know * CLEBSCH-LINDEMBANN, vol. I, pp. 478, ff., where a number of references are given. 1909] OF TWO TRIANGLES 389 from an earlier paragraph the condition that a conic may be inscribed in the 3 -line and circumscribed about the 3-point. It is (2) (D4 - 8D2/2 + 16D2 ) 2- 5123D(/ I 1- 4D ) - 0. Hence to find the condition on the cusp triangle, we must take (2) and carry out the above transformations in reverse order. Being of period 2 the inverses are the same as the direct transformations. That is, transformation (1) must be carried out first, then the dual transformation, and finally (1) a second time. This gives an expression of the 24th degree in the coefficients of the original triangles. If on the other hand the cusp triangle is taken as the reference 3-line the condition on the 3-point of double points turns out to be of the 18th degree by a similar argument. It is DID5 16 (I12D2 -4ID1DD + 4D -D4D2,-16 3DI) 1 2T IV \ 1112 43 32 2D 2 1 (3) x (2D3'I-2D|+i~A1D -2D ID2D )2 -8D2(2f2D6D2- 3JgD + D2+ 6DlD )(2DI3a- 2D + D1DD2-DD2D)= 0. This is sufficient to illustrate our point. For by direct attack this problem is practically impossible, yet by this transformation it becomes comparatively simple. Other examples can readily be found to further illustrate this idea. Incidentally this gives a convenient geometrical interpretation for I = 0, the only one of the fundamental invariants not previously interpreted. For if the transformation y, = 1/x is carried out on the three points of the 3-point, they are carried into three new points, whose equations are a2a3 t- a3 a3, 2 c+ a, =23 0, b2b3 l1 +~ b3 b12 + bb,, = 0, c2C,3 + C3C1,2 + C1 C2 = 0. Clearly the condition that these shall be apolar to the reference 3 line is = 0. Or as a definite theorem: If two triangles are such that when one is taken as a 3-line and the other as a 3-point, and the transformation yi = l/x 'is carried out, the 3-point is apolar to the 3-line, then will 2 =0. 390 D. D. LEIB: INVARIANTS OF TWO TRIANGLES This gives, however, no direct information as to the projective relation between the 3-line and 3-point in their original position. Other interpretations of 12 = 0 arose, but none so direct and simple as for the vanishing of the other fundamental invariants. JOHNS HOPKINS UNIVERSITY, Ftbruary, 1909. VITA DAVID D. LEIB was born October 28, 1879, near Allen, Pa. His parents were Samuel and Mary (Deitch) Leib. After passing through the public schools, he entered Dickinson Preparatory School, now Conway Hall, Carlisle, Pa., in 1897. In the fall of 1899, he entered Dickinson College, from which he was graduated with the degree A.B. in 1903. For the next three years he was Instructor in Mathematics and Physics at Pennington Seminary, Pennington, N. J. In October, 1906, he entered upon graduate studies at The Johns Hopkins University choosing Mathematics, Applied Electricity and Geological Physics as his subjects. The courses pursued were under Professors MORLEY, COHEN, WHITEHEAD and REID and Dr. COBLE. To each of these he desires to express his thanks for the great interest shown throughout the courses. Especial thanks are due Dr. COBLE for frequent and valuable suggestions, and above all Professor MORLEY for his uniform kindness and inspiration throughout the three years, and for his guidance during the preparation of this dissertation.