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X(2), X(3), X(6), X(55), X(956), X(999), X(1001), X(1012)

X(47037) → X(47043)

infinite points of K002

vertices of the Thomson triangle Q1Q2Q3

points of nK0(X6, X5275) on (O)

see points below

Geometric properties :

K1263 meets the Thomson cubic K002 at three points on the line at infinity and six points on the Jerabek-Thomson hyperbola, namely X(2), X(3), X(6) and the vertices of the Thomson triangle.

K1263 meets the circumcircle again at thtree points T1, T2, T3 that lie on nK0(X6, X5275). These are the antipodes of the points (apart A, B, C) of pK(X6, X7580) on (O). X(7580) is the reflection of X(1012) about O.

The tangential of G is P2 (see below) and K1263 is a pK with respect to triangle T1T2T3 with pivot G and isopivot P2. It follows that K1263 passes through the vertices S1, S2, S3 of the cevian triangle of G in T1T2T3.

The Thomson isogonal transform of K1263 is a central cubic with center O, passing through X(2), X(3), X(376), X(573), X(991), X(5657), X(5731). It is a member of the pencil mentioned in page K758 which also contains K764, K1260, K1261.

Points on K1263

P1 = -3 a^4+a^2 b^2+2 a^2 b c+2 a b^2 c+a^2 c^2+2 a b c^2+4 b^2 c^2: : , SEARCH = 5.402320120132397, on the lines {2,6}, {55,536}

P2 = a^2 (a^4+a^2 b^2-2 b^4-2 a^2 b c-2 a b^2 c+a^2 c^2-2 a b c^2-2 b^2 c^2-2 c^4): : , SEARCH = 2.604821552828271, on the lines {3,6}, {55,2810}

P3 = a^2 (a^4 b+a^2 b^3-2 a b^4+a^4 c+a^3 b c-2 b^4 c-2 a b^2 c^2-2 b^3 c^2+a^2 c^3-2 b^2 c^3-2 a c^4-2 b c^4): : , SEARCH = 5.249320297336108, on the line {6,31}

P4 = a^2 (a^4-2 a^3 b-2 a^2 b^2+2 a b^3+b^4-2 a^3 c+8 a^2 b c-2 a b^2 c-4 b^3 c-2 a^2 c^2-2 a b c^2+2 b^2 c^2+2 a c^3-4 b c^3+c^4): : , SEARCH = 2.832758877051794, on the line {63,517}

P5 = 3 a^4+2 a^3 b-4 a^2 b^2-a b^3+2 a^3 c-3 a b^2 c-b^3 c-4 a^2 c^2-3 a b c^2+2 b^2 c^2-a c^3-b c^3: : , SEARCH = 4.81339205863944, on the line {527,551}

P6 = 3 a^4-2 a^3 b-4 a^2 b^2+a b^3-2 a^3 c-4 a^2 b c-a b^2 c+b^3 c-4 a^2 c^2-a b c^2+2 b^2 c^2+a c^3+b c^3: : , SEARCH = -0.01503367970514763, on the line {55,519}

Remarks :

• the polar conic (C) of P2 passes through G, T1, T2, T3 and obviouly P2.

• by Poncelet porism, ABC and T1T2T3 circumscribe a same conic which is the Kiepert parabola (P).

Recall that ABC and Q1Q2Q3 circumscribe the Steiner inellipse.

• K1263 meets the sidelines of Q1Q2Q3 again at R1, R2, R3 which are the orthogonal projections of P2 on these sidelines. The lines {P2, Ri} are parallel to the asymptotes of K002 and K1263.