X1489

3rd STEVANOVIC POINT

< 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, 1143 }: Central Line L(2, 1143) { 1, 188 }: Central Line L(1, 188) { 174, 558 }: Central Line L(174, 558) { 258, 483 }: Central Line L(258, 483) 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_1143 = 2nd MALFATTI-RABINOWITZ POINT (0.5857, 0.2649) X_1 = INCENTER (0.5714, 0.2333) X_188 = 2nd MID-ARC POINT OF ANTICOMPLEMENTARY TRIANGLE (0.5245, 0.1802) X_174 = YFF CENTER OF CONGRUENCE (0.5813, 0.2549) X_558 = (X(1),X(57))-ANSWER TO QUESTION A (0.6164, 0.3210) X_258 = CONGRUENT INCIRCLES ISOSCELIZER POINT (0.5464, 0.1945) X_483 = RADICAL CENTER OF AJIMA-MALFATTI CIRCLES (0.5762, 0.2431) X_1489 = 3rd STEVANOVIC POINT (0.5675, 0.2288) A B C Ma Mb Mc X2 X1143 X1 X188 X174 X558 X258 X483 X1489 (a, b, c) = (5, 6, 7) | AB = 1

X1489 (3rd STEVANOVIC 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 \(f(A,B,C) = (\csc A)[1 - \cos(A/2) - \sin(A/2)]\)
Barycentrics: \(1 - \cos(A/2) - \sin(A/2) : 1 - \cos(B/2) - \sin(B/2) : 1 - \cos(C/2) - \sin(C/2)\)
Search Value (6, 9, 13): \(1.775305425860\)
▸ Incident Central Lines4 lines

{1, 188}, {2, 1143}, {174, 558}, {258, 483}

▸ Functional Relations3 relations

Notes

Let U be the A-excenter of triangle ABC; let A' be the incenter of triangle UBC, and define B', C' cyclically. Let I be the incenter of ABC, let A″ be the incenter of triangle IBC, and define B″, C″ cyclically. Let X = A'A″∩BC, and define Y, Z cyclically. Then AX, BY, CZ concur in X1489. (Milorad R. Stevanovic, Hyacinthos #7185, 5/21/03. See also X1130 and X1489.)