X65¶
ORTHOCENTER OF THE INTOUCH TRIANGLE
X65 (ORTHOCENTER OF THE INTOUCH TRIANGLE) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).¶
Constructed Elements
Derived Triangle: Intouch (Contact) Triangle △TaTbTc — vertices Ta, Tb, Tc (Contact triangle of Incircle touchpoints on BC, CA, AB)
Circle: Incircle (incircle) — Inscribed circle tangent to sidelines at T_a, T_b, T_c; centered at X1 INCENTER
Concurrence Lines: Ta–X65, Tb–X65, Tc–X65 concurring at X65
Related Centers: X1 (INCENTER)
Alternate Coordinates3 forms
Trilinears: \((b + c)/(b + c - a) : (c + a)/(c + a - b) : (a + b)/(a + b - c)\)
Trilinears: \(\sin(A/2) \cos(B/2 - C/2) : \sin(B/2) \cos(C/2 - A/2) : \sin(C/2) \cos(A/2 - B/2)\)
Trilinears: \(R_{a} + r : R_{b} + r : R_{c} + r\), where \(R_{a}\), \(R_{b}\), \(R_{c}\) are the exradii
Incident Central Lines40 lines
{1, 3}, {2, 959}, {4, 158}, {6, 19}, {7, 8}, {10, 12}, {11, 117}, {29, 296}, {31, 1104}, {33, 64}, {37, 71}, {41, 910}, {42, 73}, {44, 374}, {58, 109}, {63, 958}, {68, 91}, {74, 108}, {77, 969}, {79, 80}, {81, 961}, {110, 229}, {169, 218}, {172, 248}, {224, 1004}, {225, 407}, {243, 412}, {257, 894}, {278, 387}, {279, 1002}, {386, 994}, {409, 1098}, {474, 997}, {497, 938}, {516, 950}, {519, 553}, {604, 1100}, {651, 895}, {1039, 1041}, {1061, 1063}
Functional Relations43 relations
X65 = reflection of Xi in Xj for these (i,j): (1, 942), (72, 10)
X65 = isogonal conjugate of X21
X65 = isotomic conjugate of X314
X65 = complement of X3869
X65 = anticomplement of X960
X65 = circumcircle-inverse of X5172
X65 = Xi-Ceva conjugate of Xj for these (i,j): (1, 73), (4, 225), (7, 226), (10, 227), (109, 513), (226, 37)
X65 = X42-cross conjugate of X37
X65 = crosspoint of Xi and Xj for these (i,j): (1, 4), (7, 57)
X65 = crosssum of Xi and Xj for these (i,j): (1, 3), (9, 55), (56, 1394), (1805, 1806)
X65 = crossdifference of every pair of points on line X521X650
X65 = X1284-Hirst inverse of X1400
X65 = Xi-beth conjugate of Xj for these (i,j): (1, 65), (8, 72), (10, 10), (65, 1042), (80, 65), (100, 65), (101, 213), (291, 65), (668, 65), (1018, 65)
X65 = bicentric sum of PU(15)
X65 = trilinear product of PU(81)
X65 = trilinear pole of line X647X661
X65 = perspector of intouch triangle and inverse(n(hexyl triangle))
X65 = orthologic center of inverse(n(hexyl triangle)) to hexyl triangle; the reciprocal orthologic center is X84
X65 = perspector of ABC and the extangents triangle
X65 = homothetic center of intangents triangle and reflection of extangents triangle in X40
X65 = homothetic center of extangents triangle and reflection of intangents triangle in X1
X65 = {P,Q}-harmonic conjugate of X1463, where P and Q are the intersections of the incircle and line X7X8
X65 = pairwise perspector of: intouch triangle, 4th extouch triangle, 5th extouch triangle
X65 = perspector of [reflection of incentral triangle in X1] and tangential triangle, wrt incentral triangle, of circumconic of incentral triangle centered at X1 (bicevian conic of X1 and X57)
X65 = X5 of reflection triangle of X1
X65 = radical trace of circumcircle and circumcircle of reflection triangle of X1
X65 = perspector of ABC and cross-triangle of ABC and 4th extouch triangle
X65 = perspector of ABC and cross-triangle of ABC and 5th extouch triangle
X65 = polar conjugate of X31623
X65 = pole wrt polar circle of trilinear polar of X31623 (line X521X1948)
X65 = perspector of ABC and anti-tangential midarc triangle
X65 = homothetic center of extangents triangle and anti-tangential midarc triangle
X65 = excentral-to-intouch similarity image of X1
Notes¶
Let A' be the intersections of the tangents to the Yiu conic at the points where they meet the A-excircle. Define B' and C' similarly. The lines AA', BB', CC' concur in X65. (Randy Hutson, July 20, 2016)
Let Ab, Ac be the points where the A-excircle touches lines CA and AB resp., and define Bc, Ba, Ca, Cb cyclically. Let Ta be the intersection of the tangents to the Yiu conic (defined at X478) at Bc and Ca, and define Tb, Tc cyclically. Let Ta' be the intersection of the tangents to the Yiu conic at Ba and Cb, and define Tb', Tc' cyclically. Let Sa = TbTc∩Tb'Tc', Sb = TcTa∩Tc'Ta', Sc = TaTb∩Ta'Tb'. The lines ASa, BSb, CSc concur in X65. (See also X1903.) (Randy Hutson, July 20, 2016)
Let A' be the homothetic center of the orthic triangles of the intouch and A-extouch triangles, and define B' and C' cyclically. The triangle A'B'C' is perspective to the extouch triangle at X65. (Randy Hutson, July 20, 2016)
Let HaHbHc be the orthic triangle. Let HbHcA″ be the triangle similar to ABC such that segment HaA″ crosses the line HbHc. Define B″ and C″ cyclically. Equivalently, A″ is the reflection of A in HbHc, and cyclically for B″ and C″. Let Ia be the incenter of HbHcA″, and define Ib and Ic cyclically. The circumcenter of triangle IaIbIc is X65. Let A* be the intersection of lines A″Ia and HbHc, and define B* and C* cyclically. The lines HaA*, HbB*, HcC* concur in X65. (Randy Hutson, July 20, 2016)
Let Ab, Ac be the points where the A-excircle touches lines CA and AB resp., and define Bc, Ba, Ca, Cb cyclically. Let IaIbIc be the intouch triangle. Let Oa be the circle through Ab, Ac, Ib, Ic, and define Ob, Oc cyclically. X65 is the radical center of Oa, Ob, Oc. (Randy Hutson, July 20, 2016)
Let A'B'C' be the intouch triangle of the extangents triangle, if ABC is acute. Then A'B'C' is perspective to the intouch triangle and 4th and 5th extouch triangles at X65. (Randy Hutson, December 2 2017)
Let OA be the circle centered at the A-vertex of the Wasat triangle and passing through A; define OB and OC cyclically. X65 is the radical center of OA, OB, OC. (Randy Hutson, August 30, 2020)
Let A' be the isogonal conjugate of A wrt the A-extouch triangle. Define B' and C' cyclically. The lines AA', BB', CC' concur in X65. (Randy Hutson, August 30, 2020)
In the plane of a triangle ABC, let
Ba = reflection of A in the external angular bisector of angle B, and define Cb and Ac cyclically;
Ca = reflection of A in the external angular bisector of angle C, and define Ab and Bc cyclically;
Pa = AcBc∩AbCb, and define Pb and Pc cyclically;
Ka = AbBa∩AcCa, and define Kb and Kc cyclically.
Then PaPbPc and ABC, and also KaKbKc and ABC, are perspective, and the perspector is X65. (Dasari Naga Vijay Krishna, June 19, 2021)