X1148

YIU-HATZIPOLAKIS POINT

Suppose LA, LB, LC are lines through a point P, respectively perpendicular to sidelines BC, CA, AB. Let AB be the point where LA meets AB, and let AC be the point where LA meets AC. Define BC, BA, CA, CB cyclically. Then X1148 is the point P, which satisfies
< 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) { 4, 65 }: Central Line L(4, 65) { 46, 243 }: Central Line L(46, 243) { 3, 653 }: Central Line L(3, 653) { 92, 942 }: Central Line L(92, 942) { 1, 1075 }: Central Line L(1, 1075) 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_4 = ORTHOCENTER (0.6122, 0.3958) X_65 = ORTHOCENTER OF THE INTOUCH TRIANGLE (0.6041, 0.2933) X_46 = -CEVA CONJUGATE OF X(1) (0.6917, 0.4543) X_243 = -LINE CONJUGATE OF X(73) (0.7516, 0.9505) X_3 = CIRCUMCENTER (0.5000, 0.1021) X_653 = TRILINEAR POLE OF LINE X(1)X(4) (0.5097, 0.6030) X_92 = CEVAPOINT OF INCENTER AND CLAWSON POINT (0.5949, 0.3639) X_942 = INCIRCLE-INVERSE OF X(36) (0.5878, 0.2633) X_1075 = -CEVA CONJUGATE OF X(4) (0.6630, 1.7036) X_1148 = YIU-HATZIPOLAKIS POINT (0.4716, -1.3697) A B C X1 X4 X65 X46 X243 X3 X653 X92 X942 X1075 X1148 (a, b, c) = (5, 6, 7) | AB = 1

X1148 (YIU-HATZIPOLAKIS 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 \(f(A,B,C) = (\sec B + \sec C - \sec A) \sec A\)
Barycentrics: \((\sin A)f(A,B,C) : (\sin B)f(B,C,A) : (\sin C)f(C,A,B)\)
Search Value (6, 9, 13): \(0.628378654208\)
▸ Incident Central Lines5 lines

{1, 1075}, {3, 653}, {4, 65}, {46, 243}, {92, 942}

▸ Functional Relations3 relations

Notes

|PAB| + |PAC| = |PBC| + |PBA| = |PCA| + |PCB|.

See Hyacinthos messages #4204-4206, 10/01.