X3588

CENTER OF THE MYAKISHEV CONIC

X3588 = X8X573∩X(2269,3057), where
< 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) { 42, 181 }: Central Line L(42, 181) { 8, 573 }: Central Line L(8, 573) { 71, 594 }: Central Line L(71, 594) { 37, 1953 }: Central Line L(37, 1953) { 213, 2347 }: Central Line L(213, 2347) { 1824, 2354 }: Central Line L(1824, 2354) A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6122, 0.5999) X_42 = CROSSPOINT OF INCENTER AND SYMMEDIAN POINT (0.5880, 0.2495) X_181 = APOLLONIUS POINT (0.6163, 0.3093) X_8 = NAGEL POINT (0.4694, 0.1333) X_573 = ORTHOHARMONIC OF X(58) (0.4486, 0.0172) X_71 = ISOGONAL CONJUGATE OF X(27) (0.4807, 0.0915) X_594 = ISOGONAL CONJUGATE OF X(593) (0.5025, 0.1672) X_37 = CROSSPOINT OF INCENTER AND CENTROID (0.5567, 0.2158) X_1953 = BICENTRIC SUM OF PU(21) (0.5856, 0.2841) X_213 = -CEVA CONJUGATE OF X(42) (0.6135, 0.2833) X_2347 = -ISOCONJUGATE OF X(1476) (0.5740, 0.2042) X_1824 = ZOSMA TRANSFORM OF X(2) (0.6198, 0.4097) X_2354 = -ISOCONJUGATE OF X(1791) (0.6324, 0.4423) X_3588 = CENTER OF THE MYAKISHEV CONIC (0.4308, -0.0825) A B C X42 X181 X8 X573 X71 X594 X37 X1953 X213 X2347 X1824 X2354 X3588 (a, b, c) = (5, 6, 7) | AB = 1

X3588 (CENTER OF THE MYAKISHEV CONIC) 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 \(f(a,b,c) = a(b + c)[a^{4}(b + c) - b^{4}(c + a) - c^{4}(a + b) + b^{2}c^{2}(b + c) + c^{2}a^{2}(a - c) + a^{2}b^{2}(a - b) - 2ab^{2}c^{2}]\)
Barycentrics: \(af(a,b,c) : bf(b,c,a) : cf(c,a,b)\)
Search Value (6, 9, 13): \(47.864825553817\)
▸ Incident Central Lines8 lines

{8, 573}, {37, 1953}, {42, 181}, {71, 594}, {213, 2347}, {1824, 2354}, {2225, 2264}, {3059, 3198}

▸ Functional Relations4 relations

Notes

In 2008, Alexei Myakishev gave the following construction of a conic. Let CA and CB be points on the line AB satisfying |BCA| = |CB| and |ACB| = |CA| and arranged in this order: CA, B, A, and CB. Define points AB, CB, BC, AC cyclically. The six points CA, BA, AB, CB, BC, AC lie on a conic. Myakishev’s proof is by Carnot’s theorem, since

[c/(c+a)][(a+b)/b][a/(a+b)][(b+c)/c][b/(b+c)][(c+a)/a] = 1.

For details, in Russian, visit Item 4 at Geometry.ru,

Peter Moses found that X3588 is the point of concurrence of three lines constructed from points numbered 1,3,4,6,8,9,10,11, as follows:

The six points Ca, Ba, Ab, Cb, Bc, Ac can be constructed from the extangents triangle as follows. Let A'B'C' be the extangents triangle. Then Ab = BC∩C'A', Ac = BC∩A'B', and Bc, Ba, Ca, Cb are defined cyclically. (Randy Hutson, January 29, 2018)