X3397

SEC(A + 2π/5) POINT

X3397 and related points occupy Bernard Gibert’s Table 38. See the note at X3364.
< 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) { 2, 1140 }: Central Line L(2, 1140) A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6122, 0.5999) X_3394 = SIN(A + π/5) POINT (0.5720, 0.2210) X_1139 = OUTER PENTAGON POINT (0.5970, 0.3096) X_2 = CENTROID (0.5374, 0.2000) X_1140 = INNER PENTAGON POINT (0.5248, 1.3312) X_3397 = SEC(A + 2π/5) POINT (0.5013, 0.1676) A B C X3394 X1139 X2 X1140 X3397 (a, b, c) = (5, 6, 7) | AB = 1

X3397 (SEC(A + 2π/5) POINT) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

Trilinears: \(f(A,B,C) : f(B,C,A) : f(C,A,B)\), where \(f(A,B,C) = \sec(A + 2\pi /5)\)
Barycentrics: \((\sin A)f(A,B,C) : (\sin B)f(B,C,A) : (\sin C)f(C,A,B)\)
Search Value (6, 9, 13): \(4.674850901461\)

X3397 lies on the Kiepert hyperbola.

▸ Incident Central Lines1 line

{2, 1140}

▸ Functional Relations2 relations

Notes

Let A' be the center of the regular pentagon \(AA_{2}\)A3A4A5, with the same orientation than ABC and such that B ∈ A2A3 and C ∈ A4A5. Build B' and C' cyclically. Then the lines AA', BB', CC' concur at X3397. (César E. Lozada, May 14, 2019, Hyacinthos #29013).