X251

ISOGONAL CONJUGATE OF X141

< A B C >: Reference Triangle △ABC < M_a M_b M_c >: Medial Triangle △M_a M_b M_c [ A B ]: Sideline AB (Base c = 1) [ B C ]: Sideline BC (Side a) [ C A ]: Sideline CA (Side b) { 2, 1369 }: Central Line L(2, 1369) { 6, 22 }: Central Line L(6, 22) { 2, 32 }: Central Line L(2, 32) { 37, 82 }: Central Line L(37, 82) { 184, 263 }: Central Line L(184, 263) A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6122, 0.5999) M_a = Midpoint M_a of BC (0.8061, 0.2999) M_b = Midpoint M_b of CA (0.3061, 0.2999) M_c = Midpoint M_c of AB (0.5000, 0.0000) X_2 = CENTROID (0.5374, 0.2000) X_1369 = ANTICOMPLEMENTARY CONJUGATE OF X(6) (0.3769, -0.0086) X_6 = SYMMEDIAN POINT (LEMOINE POINT, GREBE POINT) (0.6000, 0.2672) X_22 = EXETER POINT (0.4143, -0.1221) X_32 = 3rd POWER POINT (0.6400, 0.3332) X_37 = CROSSPOINT OF INCENTER AND CENTROID (0.5567, 0.2158) X_82 = ISOGONAL CONJUGATE OF X(38) (0.5943, 0.2703) X_184 = INVERSE OF X(125) IN THE BROCARD CIRCLE (0.5788, 0.1496) X_263 = ISOGONAL CONJUGATE OF X(183) (0.6298, 0.3525) X_251 = ISOGONAL CONJUGATE OF X(141) (0.6177, 0.3042) A B C Ma Mb Mc X2 X1369 X6 X22 X32 X37 X82 X184 X263 X251 (a, b, c) = (5, 6, 7) | AB = 1

X251 (ISOGONAL CONJUGATE OF X141) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

Trilinears: \(a^{2}\csc(A + \omega ) : b^{2}\csc(B + \omega ) : c^{2}\csc(C + \omega )\)
Barycentrics: \(a^{3}\csc(A + \omega ) : b^{3}\csc(B + \omega ) : c^{3}\csc(C + \omega )\)
Search Value (6, 9, 13): \(0.572649002745\)
▸ Alternate Coordinates1 form
  • Trilinears: \(a/(b^{2} + c^{2}) : b/(c^{2} + a^{2}) : c/(a^{2} + b^{2})\)

▸ Incident Central Lines9 lines

{2, 32}, {6, 22}, {37, 82}, {110, 694}, {112, 427}, {184, 263}, {308, 385}, {609, 614}, {689, 699}

▸ Functional Relations13 relations

Notes

Let K be the symmedian point of ABC and let A' be the symmedian point of the triangle BCK; define B' and C' cyclically. The lines AA', BB', CC' concur in X251. (Randy Hutson, 9/23/2011)

Let A5'B5'C5' be the 5th anti-Brocard triangle. The radical center of the circumcircles of BCA5', CAB5', ABC5' is X251. (Randy Hutson, July 20, 2016)