X845

X165X_166∩X173X_503

X845 = X63-of-excentral-triangle
< A B C >: Reference Triangle △ABC [ A B ]: Sideline AB (Base c = 1) [ B C ]: Sideline BC (Side a) [ C A ]: Sideline CA (Side b) { 165, 166 }: Central Line L(165, 166) { 164, 362 }: Central Line L(164, 362) A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6122, 0.5999) X_173 = CONGRUENT ISOSCELIZERS POINT (0.5924, 0.2741) X_165 = CENTROID OF THE EXCENTRAL TRIANGLE (0.4762, 0.0583) X_166 = GERGONNE POINT OF EXCENTRAL TRIANGLE (0.5156, 0.1554) X_164 = INCENTER OF EXCENTRAL TRIANGLE (0.4930, 0.1115) X_362 = CONGRUENT CIRCUMCIRCLES ISOSCELIZER POINT (0.5299, 0.2081) X_845 = ∩X(173)X(503) (0.4168, -0.0880) A B C X173 X165 X166 X164 X362 X845 (a, b, c) = (5, 6, 7) | AB = 1

X845 (X165X_166∩X173X_503) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

Trilinears: \(- x + y + z : :\), where \(x = x(A,B,C) = \sin^{2}(A/2) \sec(A/2)\)
Search Value (6, 9, 13): \(15.329029407710\)
▸ Incident Central Lines4 lines

{1, 7370}, {164, 362}, {165, 166}, {167, 7991}

▸ Functional Relations2 relations

Notes

This point solves a problem posed by Antreas Hatzipolakis in Hyacinthos #2768, May 3, 2001.

Let Ia, Ib, Ic be the excenters of a triangle ABC. Let A' be the Clawson point of IaBC, and define B' and C' cyclically. The lines AA', BB', CC' concur in X845. (Angel Montesdeoca, August 12, 2018)