X3615

2nd HUTSON-FEUERBACH POINT

X3615 = cevapoint of X1 and X5
< 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) { 5, 49 }: Central Line L(5, 49) { 21, 36 }: Central Line L(21, 36) { 11, 60 }: Central Line L(11, 60) { 12, 59 }: Central Line L(12, 59) { 1, 564 }: Central Line L(1, 564) { 2, 582 }: Central Line L(2, 582) 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_5 = NINE-POINT CENTER (0.5561, 0.2489) X_49 = CENTER OF SINE-TRIPLE-ANGLE CIRCLE (0.6275, 0.1789) X_21 = SCHIFFLER POINT (0.5287, 0.1771) X_36 = INVERSE-IN-CIRCUMCIRCLE OF INCENTER (1.3333, 1.6330) X_11 = FEUERBACH POINT (0.7347, 0.0667) X_60 = ISOGONAL CONJUGATE OF X(12) (0.5790, 0.2258) X_12 = {X(1),X(5)}-HARMONIC CONJUGATE OF X(11) (0.5641, 0.2407) X_59 = ISOGONAL CONJUGATE OF X(11) (0.5333, 0.4355) X_564 = INTERSECTION OF LINES X(1)X(1048) AND X(47,91) (0.4993, 0.3100) X_2 = CENTROID (0.5374, 0.2000) X_582 = HARMONIC OF X(500) (0.8750, 0.7214) X_3615 = 2nd HUTSON-FEUERBACH POINT (0.5641, 0.2411) A B C X1 X5 X49 X21 X36 X11 X60 X12 X59 X564 X2 X582 X3615 (a, b, c) = (5, 6, 7) | AB = 1

X3615 (2nd HUTSON-FEUERBACH POINT) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

Trilinears: \(1/(\cos(B - A) + \cos(C - A)) : 1/((\cos(C - B) + \cos(A - B)) : 1/(\cos(A - C) + \cos(B - C))\)
Barycentrics: \((\sin(A))/(\cos(B - A) + \cos(C - A)) : (\sin(B))/((\cos(C - B) + \cos(A - B)) : (\sin(C))/(\cos(A - C) + \cos(B - C))\)
Search Value (6, 9, 13): \(0.242505726184\)
▸ Alternate Coordinates1 form
  • Trilinears: \(f(a,b,c) : f(b,c,a) : f(c,a,b)\), where \(f(a,b,c) = bc(b + c - a)/[(b + c)(b^{2} + c^{2} - a^{2} + bc)]\) (Peter Moses, September 5, 2011)

▸ Incident Central Lines14 lines

{1, 564}, {2, 582}, {5, 49}, {11, 60}, {12, 59}, {21, 36}, {29, 1870}, {58, 3582}, {86, 1443}, {476, 953}, {501, 3583}, {655, 2595}, {946, 1325}, {2287, 2323}

▸ Functional Relations5 relations

Notes

Let EuaEubEuc be the Feuerbach triangle, L the line through Eua and X5, and A″ = L∩BC; define B″ and C″ cyclically. Then the lines AA″, BB″, CC″ concur in X3615. (Randy Hutson, August 30, 2011.)

Let EuaEubEuc be the Feuerbach triangle. Let A″ be the trilinear pole of line EubEuc, and define B″ and C″ cyclically. (The lines AA″, BB″, CC″ concur in X1.) Let A* be the trilinear pole of line B″C″, and define B*and C* cyclically. The lines AA*, BB*, CC* concur in X3615. (Randy Hutson, July 11, 2019)