X1140

INNER PENTAGON POINT

Let A' be the innermost vertex of the regular pentagon erected inward on side BC of ABC. Define B' and C' cyclically. Then triangle A'B'C' is perspective to ABC, and the perspector is X1140. See references at X1140.
< 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) { 6, 1139 }: Central Line L(6, 1139) { 5, 3370 }: Central Line L(5, 3370) { 2, 3396 }: Central Line L(2, 3396) { 4, 3395 }: Central Line L(4, 3395) { 3, 3397 }: Central Line L(3, 3397) A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6122, 0.5999) X_3369 = SIN(A - 2π/5) POINT (0.3449, -0.1541) X_6 = SYMMEDIAN POINT (LEMOINE POINT, GREBE POINT) (0.6000, 0.2672) X_1139 = OUTER PENTAGON POINT (0.5970, 0.3096) X_5 = NINE-POINT CENTER (0.5561, 0.2489) X_3370 = SEC(A - 2π/5) POINT (0.5569, 0.2237) X_2 = CENTROID (0.5374, 0.2000) X_3396 = SIN(A + 2π/5) POINT (0.5378, 0.1645) X_4 = ORTHOCENTER (0.6122, 0.3958) X_3395 = COS(A + 2π/5) POINT (0.6210, 0.3019) X_3 = CIRCUMCENTER (0.5000, 0.1021) X_3397 = SEC(A + 2π/5) POINT (0.5013, 0.1676) X_1140 = INNER PENTAGON POINT (0.5248, 1.3312) A B C X3369 X6 X1139 X5 X3370 X2 X3396 X4 X3395 X3 X3397 X1140 (a, b, c) = (5, 6, 7) | AB = 1

X1140 (INNER PENTAGON POINT) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

Trilinears: \((\csc A)/(\cot A - \cot 2\pi /5) : (\csc B)/(\cot B - \cot 2\pi /5) : (\csc C)/(\cot C - \cot 2\pi /5)\)
Barycentrics: \(1/(\cot A - \cot 2\pi /5) : 1/(\cot B - \cot 2\pi /5) : 1/(\cot C - \cot 2\pi /5)\)
Search Value (6, 9, 13): \(10.504284984735\)
▸ Alternate Coordinates1 form

  • Trilinears: \(\csc(A + 3\pi /5) : \csc(B + 3\pi /5) : \csc(C + 3\pi /5)\) (Joe Goggins, Oct. 19, 2005)

▸ Incident Central Lines5 lines

{2, 3396}, {3, 3397}, {4, 3395}, {5, 3370}, {6, 1139}

▸ Functional Relations1 relation