X502

ISOGONAL CONJUGATE OF X501

X502 is the isogonal conjugate of X501.
< 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) { 10, 191 }: Central Line L(10, 191) { 594, 3678 }: Central Line L(594, 3678) { 1089, 3969 }: Central Line L(1089, 3969) A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6122, 0.5999) X_1 = INCENTER (0.5714, 0.2333) X_10 = SPIEKER CENTER (0.5204, 0.1833) X_191 = -CEVA CONJUGATE OF X(1) (0.4859, 0.1209) X_594 = ISOGONAL CONJUGATE OF X(593) (0.5025, 0.1672) X_3678 = com(EXTOUCH TRIANGLE) (0.4786, 0.1283) X_1089 = TRILINEAR SQUARE OF X(10) (0.4606, 0.1381) X_3969 = com[INVERSE(n(INCENTRAL TRIANGLE))] (0.4350, 0.1040) X_502 = ISOGONAL CONJUGATE OF X(501) (0.5927, 0.3141) A B C X1 X10 X191 X594 X3678 X1089 X3969 X502 (a, b, c) = (5, 6, 7) | AB = 1

X502 (ISOGONAL CONJUGATE OF X501) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

Trilinears: \(f(a,b,c) : f(b,c,a) : f(c,a,b)\), where \(f(a,b,c) = bc(b + c)/[a^{3} - b^{3} - c^{3} - bc(a + b + c) + ab(a - b) + ac(a - c)]\)
Barycentrics: \(af(a,b,c) : bf(b,c,a) : cf(c,a,b)\)
Search Value (6, 9, 13): \(-2.150446441377\)
▸ Incident Central Lines7 lines

{10, 191}, {12, 14873}, {42, 14874}, {58, 21043}, {313, 17791}, {594, 3678}, {1089, 3969}

▸ Functional Relations2 relations

Notes

Let IaIbIc be the incentral triangle. Let BCA″ be the triangle similar to IaIbIc such that the segment AA″ crosses the line BC. Define B″ and C″ cyclically. The lines AA″, BB″, CC″ concur in X502. (Randy Hutson, 9/23/2011)