X1131

ARCTAN(2) KIEPERT POINT

To construct the Vecten point, X485, squares are erected outward on the sides of ABC. If A', B', C' are the centers of these squares, then triangle A'B'C' is perspective to ABC with perspector X485. Now let A″ be the midpoint of the side of the A-square that does not touch line BC, and define B″ and C″ cyclically. Then triangle A″B″C″ is perspective to ABC with perspector X1131. Angle(A″BC) = angle(B″CA) = angle(C″AB), so that X1131 lies on the Kiepert hyperbola. Here, the common angle is arctan(2). (Darij Grinberg 9/22/02)
< 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) { 175, 226 }: Central Line L(175, 226) { 2, 490 }: Central Line L(2, 490) { 20, 485 }: Central Line L(20, 485) { 6, 1132 }: Central Line L(6, 1132) A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6122, 0.5999) X_1151 = ISOGONAL CONJUGATE OF X(1131) (0.5483, 0.1819) X_175 = ISOPERIMETRIC POINT (0.4271, -0.1006) X_226 = -CEVA CONJUGATE OF X(65) (0.5778, 0.2588) X_2 = CENTROID (0.5374, 0.2000) X_490 = CEVAPOINT OF X(20) AND X(488) (0.4037, -0.0241) X_20 = DE LONGCHAMPS POINT (0.3878, -0.1916) X_485 = VECTEN POINT (0.5773, 0.2578) X_6 = SYMMEDIAN POINT (LEMOINE POINT, GREBE POINT) (0.6000, 0.2672) X_1132 = ARCTAN(-2) KIEPERT POINT (0.8864, -0.3774) X_1131 = ARCTAN(2) KIEPERT POINT (0.5904, 0.2888) A B C X1151 X175 X226 X2 X490 X20 X485 X6 X1132 X1131 (a, b, c) = (5, 6, 7) | AB = 1

X1131 (ARCTAN(2) KIEPERT POINT) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

Trilinears: \(1/(\sin A + 2 \cos A) : :\)
Barycentrics: \(1/(2 S_{A} + S) : :\)
Search Value (6, 9, 13): \(-0.147669218497\)
▸ Alternate Coordinates1 form

  • Trilinears: \(\csc(A + t) : :\), where t = arctan(2)

▸ Incident Central Lines5 lines

{2, 490}, {6, 1132}, {4, 3311}, {20, 485}, {175, 226}

▸ Functional Relations5 relations