X199

X10-CEVA CONJUGATE OF X6

X199 = X10-Ceva conjugate of X6
< 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) { 2, 3 }: Euler Line { 42, 172 }: Central Line L(42, 172) { 51, 572 }: Central Line L(51, 572) { 184, 573 }: Central Line L(184, 573) { 55, 1030 }: Central Line L(55, 1030) A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6122, 0.5999) X_10 = SPIEKER CENTER (0.5204, 0.1833) X_6 = SYMMEDIAN POINT (LEMOINE POINT, GREBE POINT) (0.6000, 0.2672) X_2 = CENTROID (0.5374, 0.2000) X_3 = CIRCUMCENTER (0.5000, 0.1021) X_42 = CROSSPOINT OF INCENTER AND SYMMEDIAN POINT (0.5880, 0.2495) X_172 = TRILINEAR PRODUCT X(6)*X(171) (0.6080, 0.2866) X_51 = CENTROID OF ORTHIC TRIANGLE (0.6093, 0.3186) X_572 = ORTHOGONAL OF X(58) (7.2953, -0.6907) X_184 = INVERSE OF X(125) IN THE BROCARD CIRCLE (0.5788, 0.1496) X_573 = ORTHOHARMONIC OF X(58) (0.4583, 0.2009) X_55 = INSIMILICENTER(CIRCUMCIRCLE, INCIRCLE) (0.5490, 0.1921) X_1030 = ISOGONAL CONJUGATE OF X(1029) (0.5271, 0.1468) X_199 = -CEVA CONJUGATE OF X(6) (0.4796, 0.0487) A B C X10 X6 X2 X3 X42 X172 X51 X572 X184 X573 X55 X1030 X199 (a, b, c) = (5, 6, 7) | AB = 1

X199 (X10-CEVA CONJUGATE OF X6) 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 \(f(a,b,c) = a[b^{4} + c^{4} - a^{4} + (b^{2} + c^{2} - a^{2})(bc + ca + ab)]\)
Barycentrics: \(af(A,B,C) : bf(B,C,A) : cf(C,A,B)\)
Search Value (6, 9, 13): \(13.467812089845\)
▸ Incident Central Lines5 lines

{2, 3}, {42, 172}, {51, 572}, {55, 1030}, {184, 573}

▸ Functional Relations7 relations

Notes

As a point on the Euler line, X199 has Shinagawa coefficients \((E + 2F + \$bc\$, -2E - 2F + \$bc\$)\).

Let IA, IB, IC be the excenters. Let (OA) be the circle tangent to the circumcircle at A and passing through IA. Let A' be the antipode of A in (OA). Let LA be the tangent to (OA) at A'. Define LB and LC cyclically. Let TA = LB∩LC, and define TB and TC cyclically. Triangle TATBTC is homothetic to the tangential triangle at X199. (Randy Hutson, June 7, 2019)