X1655

2nd HATZIPOLAKIS PARALLELIAN POINT

Continuing from the description of X1654, let h(B,A) be the distance from the point BA to the line CA, and define five other distances cyclically. If P = X1655, then
< A B C >: Reference Triangle △ABC < K_a K_b K_c >: Anticomplementary Triangle △K_a K_b K_c [ A B ]: Sideline AB (Base c = 1) [ B C ]: Sideline BC (Side a) [ C A ]: Sideline CA (Side b) { 2, 274 }: Central Line L(2, 274) { 2, 39 }: Central Line L(2, 39) { 8, 192 }: Central Line L(8, 192) { 21, 385 }: Central Line L(21, 385) { 193, 452 }: Central Line L(193, 452) A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6122, 0.5999) K_a = Anticomplementary Vertex K_a (1.6122, 0.5999) K_b = Anticomplementary Vertex K_b (-0.3878, 0.5999) K_c = Anticomplementary Vertex K_c (0.3878, -0.5999) X_2 = CENTROID (0.5374, 0.2000) X_274 = ISOGONAL CONJUGATE OF X(213) (0.5168, 0.1843) X_39 = BROCARD MIDPOINT (0.5778, 0.2305) X_8 = NAGEL POINT (0.4694, 0.1333) X_192 = -CEVA CONJUGATE OF X(2) (CONGRUENT PARALLELIANS POINT) (0.6147, 0.2635) X_21 = SCHIFFLER POINT (0.5287, 0.1771) X_385 = -HIRST INVERSE OF X(6) (2.2863, 2.0795) X_193 = -CEVA CONJUGATE OF X(2) (0.7878, 0.4690) X_452 = -HIRST INVERSE OF X(449) (0.5268, 0.1722) X_1655 = 2nd HATZIPOLAKIS PARALLELIAN POINT (0.5787, 0.2312) A B C Ka Kb Kc X2 X274 X39 X8 X192 X21 X385 X193 X452 X1655 (a, b, c) = (5, 6, 7) | AB = 1

X1655 (2nd HATZIPOLAKIS PARALLELIAN POINT) 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 \(f(a,b,c) = bc[-1/a^{2} + 1/b^{2} + 1/c^{2} + 1/(bc) + 1/(ca) + 1/(ab)]\)
Barycentrics: \(g(a,b,c) : g(b,c,a) : g(c,a,b)\), where \(g(a,b,c) = -1/a^{2} + 1/b^{2} + 1/c^{2} + 1/(bc) + 1/(ca) + 1/(ab)\)
Search Value (6, 9, 13): \(1.445819564171\)
▸ Incident Central Lines6 lines

{2, 39}, {8, 192}, {21, 385}, {193, 452}, {350, 1107}, {668, 1500}

▸ Functional Relations4 relations

Notes

h(B,A) + \(h(C,A) = h(C\),B) + \(h(A,B) = h(A\),C) + h(B,C) (Antreas Hatzipolakis, Anopolis #20, 1/20/02)