X847

X5-CROSS CONJUGATE OF X4

X847 = barycentric product X91*X92
< 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, 52 }: Central Line L(4, 52) { 24, 96 }: Central Line L(24, 96) { 2, 254 }: Central Line L(2, 254) { 91, 225 }: Central Line L(91, 225) { 3, 925 }: Central Line L(3, 925) { 378, 1105 }: Central Line L(378, 1105) A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6122, 0.5999) X_5 = NINE-POINT CENTER (0.5561, 0.2489) X_4 = ORTHOCENTER (0.6122, 0.3958) X_52 = ORTHOCENTER OF ORTHIC TRIANGLE (0.7155, 0.4580) X_24 = PERSPECTOR OF ABC AND ORTHIC-OF-ORTHIC TRIANGLE (0.6627, 0.5278) X_96 = ISOGONAL CONJUGATE OF X(52) (-0.0943, -0.0197) X_2 = CENTROID (0.5374, 0.2000) X_254 = -CROSS CONJUGATE OF X(4) (1.4506, 0.5587) X_91 = ISOGONAL CONJUGATE OF X(47) (-0.0865, -0.0205) X_225 = -CEVA CONJUGATE OF X(65) (0.6246, 0.4451) X_3 = CIRCUMCENTER (0.5000, 0.1021) X_925 = ANTICOMPLEMENT OF X(136) (0.0074, -0.0312) X_378 = REFLECTION OF X(22) IN X(3) (0.5857, 0.3262) X_1105 = CEVAPOINT OF X(3) AND X(4) (0.5529, 0.3087) X_847 = -CROSS CONJUGATE OF X(4) (-0.1806, -0.0821) A B C X5 X4 X52 X24 X96 X2 X254 X91 X225 X3 X925 X378 X1105 X847 (a, b, c) = (5, 6, 7) | AB = 1

X847 (X5-CROSS CONJUGATE OF X4) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

Trilinears: \(\sec A \sec 2A : \sec B \sec 2B : \sec C \sec 2C\)
Barycentrics: \(\tan A \sec 2A : \tan B \sec 2B : \tan C \sec 2C\)
Search Value (6, 9, 13): \(0.746590301756\)
▸ Alternate Coordinates1 form

  • Barycentrics: \(1/(a^{2} (a^{2} - b^{2} - c^{2}) (a^{4} + b^{4} + c^{4} - 2 a^{2} b^{2} - 2 a^{2} c^{2})) : :\)

X847 lies on the McCay orthic cubic; see McCay orthic cubic.

▸ Incident Central Lines7 lines

{2, 254}, {3, 925}, {4, 52}, {24, 96}, {91, 225}, {378, 1105}, {403, 1093}

▸ Functional Relations6 relations

Notes

This point solves a problem posed by Antreas Hatzipolakis in Hyacinthos #3130, June 25, 2001; see also Jean-Pierre Ehrmann, #3135, June 26, 2001. The problem and solution may be stated as follows. Let ABC be a triangle, LA, LB, LC the perpendicular bisectors of sides BC, CA, AB, and AA', BB', CC' the altitudes of ABC, respectively. Let AB be the point of intersection of AA' and LB, and let AC be the point of intersection of AA' and LC. Let A″ be the point of intersection of BAB and CAC. Define B″ and C″ cyclically. Then triangle A″B″C″ is perspective to triangle ABC, with perspector X847.

Let A'B'C' be the X3-cevian triangle of the orthic triangle of triangle ABC. The lines AA', BB', CC' concur in X847. (Randy Hutson, 9/23/2011)