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X(4), X(265), X(476), X(523), X(15328), X(34150), X(41512) P1 = barycentric quotient X(1989) ÷ X(34150), on the lines {4,476}, {265,523}, now X(51349) in ETC P6 = {15328,41512} /\ {34150,P1} other points below |
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Geometric properties : |
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K1280 is a focal nK cubic with singular focus the orthocenter H of ABC. See Special isocubics, §4.2.2, and also CL028. The real asymptote is the perpendicular at O to the Euler line which meets the cubic again at X(15328). X(15328) is the tangential of H and its polar conic meets the cubic again at H, X(265), X(476), X(523). In other words, the tangents to K1280 at these points pass through X(15328). Every circle with center H meets K1280 at two points collinear with X(15328). K1280 meets the Euler line at H and two other points P2, P3, symmetric about X(5) and antipodes on the circle passing through X(110), X(265), X(476), also on the circum-conic passing through X(265), X(925). The orthic line of K1280 is the perpendicular at X(5) to the Euler line i.e. the perpendicular bisector of OH. This line meets the cubic again at two imaginary points P4, P5 which are actually the X(1989)-isoconjugates of the circular points at infinity.
• their midpoint is X(34209) and their barycentric product is the barycentric square of X(2166). • they lie on the polar conic of H in K1280 which is the circle passing through H, whose center is the reflection of X(110) in X(403) and the reflection of X(2071) in X(125). This point lies on the lines {4,52}, {30,74}. • they lie on the circum-conic (C1989) with perspector X(1989). This is a hyperbola with eccentricity 2, passing through X(476), whose center is X(14993) and whose axes are parallel and perpendicular to the Euler line. Recall that X(14993) is the G-Ceva conjugate of X(1989). Remark : every circle passing through X(265), X(476) meets (C1989) again at three points which are the vertices of an equilateral triangle. The sidelines of this triangle are tangent to the parabola with focus X(265) and directrix the orthic line above. In other words, the parabola is inscribed in every equilateral triangle. See A Generalized Morley Configuration, § 4.2.
• Let Ω be a point on the orthic line. The line HΩ meets the circle with center Ω passing through X(265) and X(476) at two points on K1280. • Let Ω' be a point on the line {30,74}. The circle with center Ω' orthogonal to the circle passing through X(110), X(265), X(476) meets the perpendicular at X(15238) to the line HΩ' at two points on K1280. Note that this circle passes through P2, P3 and belongs to the pencil of circles with Poncelet points X(265), X(476). • Let (C') be the circum-conic with perspector X(3003) and center X(113), the rectangular circum-hyperbola passing through X(110). A variable line (L) passing through X(110) meets (C') again at Z. The isogonal transform (H) of the parallel at O to (L) meets the parallel at X(265) to the line HZ at two points on K1280. Note that (L) and (H) meet on the stelloid K613 = nK(X112, X14590, X4). |