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4(y+z)(z+x)(x+y) - (x+y+z)^3 = 0 |
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infinite points of the sidelines of ABC midpoints of ABC other details below |
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Geometric properties : |
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We meet K1309 in Mathematical problems, problem #1830, by Wolstenholme, Joseph, 1878, Macmillan and co, London. K1309 is the complement of the Tucker cubic T(-1), the locus of M whose cevian triangle has an algebraic area opposite to that of ABC. Recall that these areas are equal when M lies on T(1) = K016, the only Tucker cubic with three concurring asymptotes. The hessian of K1309 is also a Tucker cubic, namely T(2/3), the locus of M whose cevian triangle has an algebraic area equal to 2/3 of that of ABC. The infinite points of the sidelines of ABC are points of inflexion and the inflexional asymptotes are the sidelines of the antimedial triangle GaGbGc. K1309 passes through the midpoints Ma, Mb, Mc of ABC and the tangents at these points are the sidelines of ABC. K1309 meets the median AG again at A1 = -2 + 2 √5 : 3 - √5 : 3 - √5 and A2 = -2 - 2 √5 : 3 + √5 : 3 + √5. The points B1, B2 and C1, C2 are defined cyclically. Note that the tangents at A1, A2 (and Ma) to K1309 are parallel to BC since the polar conic of the infinite point of the sideline BC splits into the median AG and the line GbGc. K1309 is invariant under isotomic conjugation with respect to MaMbMc, which is also G-Ceva conjugation with respect to ABC. K1309 is invariant under the three oblique symmetries whose axis is one median and direction the corresponding sideline of ABC. K1309 is obviously a self-permutting cubic. If P = p:q:r is a point on the curve, then the six points p:q:r, p:r:q, q:r:p, q:p:r, r:p:q, r:q:p lie on the curve and on a same ellipse E(P) with center G, homothetic at G to the Steiner ellipse. These six points are not necessarily all distinct, in particular when P lies on a median of ABC. In this case, E(P) is tritangent to the cubic. This is the case of E(A1), E(A2) and E(Ma) which is the Steiner inellipse. *** There is only one other cubic with analogous properties. Its equation is : 4 (3 x+3 y-4 z) (3 x-4 y+3 z) (-4 x+3 y+3 z) - 3 (x+y+z)^3 = 0. Its anticomplement is T(1/3) and its hessian is T(6/25). |