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X(2), X(4), X(6), X(115), X(125), X(338), X(523), X(1640), X(14086), X(14223), X(36189), X(41254), X(51480), X(60500) X(65608) → X(65624) vertices of the cevian triangle A1B1C1 of X(41254) vertices of the anticevian triangle A'B'C' of X(523) = Schroeter triangle, see ETC X(8286). A', B', C' and X(14086) are the square roots of X(115), X(14086) being that interior to ABC. |
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Geometric properties : |
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K1367 is another example of pK with pole X(115). See also K237, K238, K239, K672, K877, K878, K1144, K1378 and the table below computed by Peter Moses. It is the only pK with pole X(115) through the points X(2), X(4), X(6) and then also through their X(115)-isoconjugates X(115), X(125), X(338). The barycentric prooduct of K1367 by X(99) is K1368. *** A selection of pK(X115, P) through at least 9 centers. Recall that every cubic passes through the eight fixed points A, B, C, X(523), X(14086), A', B', C'.
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Remarks : • pK(X115, P) meets the line at infinity at X(523) and two other points which lie on the circum-conic with perspector the barycentric product of X(523) and the complement cP of P. When P lies on the Euler line, this conic is a rectangular hyperbola. • pK(X115, P) passes through a given point M ≠ X523 (and obviouly through its X115-isoconjugate M') when P lies on the line MM'. For instance, pK(X115, P) passes through X(2) and X(115) when P lies on the line {2, 99, 111, 115, 126, 148, 543, 574, 620, 671, 2482, 2549, etc}.
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