X3548

EULER LINE INTERCEPT OF LINE X68X_125

< A B C >: Reference Triangle △ABC < M_a M_b M_c >: Medial Triangle △M_a M_b M_c [ A B ]: Sideline AB (Base c = 1) [ B C ]: Sideline BC (Side a) [ C A ]: Sideline CA (Side b) { 2, 3542 }: Euler Line { 2, 3 }: Euler Line { 68, 125 }: Central Line L(68, 125) { 498, 1060 }: Central Line L(498, 1060) { 499, 1062 }: Central Line L(499, 1062) { 1204, 1568 }: Central Line L(1204, 1568) A = Vertex A (0.0000, 0.0000) B = Vertex B (1.0000, 0.0000) C = Vertex C (0.6122, 0.5999) M_a = Midpoint M_a of BC (0.8061, 0.2999) M_b = Midpoint M_b of CA (0.3061, 0.2999) M_c = Midpoint M_c of AB (0.5000, 0.0000) X_2 = CENTROID (0.5374, 0.2000) X_3542 = EULER LINE INTERCEPT OF LINE X(33)X(498) (0.6774, 0.5661) X_3 = CIRCUMCENTER (0.5000, 0.1021) X_68 = PRASOLOV POINT (-0.0535, -0.0055) X_125 = CENTER OF JERABEK HYPERBOLA (0.6803, 0.0260) X_498 = YFF CONCURRENT CONGRUENT CIRCLES POINT (0.5518, 0.2140) X_1060 = INTERSECTION OF LINES X(1)X(3) AND X(5)X(34) (0.5181, 0.1353) X_499 = {X(1),X(2)}-HARMONIC CONJUGATE OF X(498) (0.4458, 0.1102) X_1062 = INTERSECTION OF LINES X(1)X(3) AND X(5)X(33) (0.4633, 0.0347) X_1204 = 3rd SARAGOSSA POINT OF X(64) (0.6497, 0.1923) X_1568 = ORTHOJOIN OF X(800) (0.1362, -0.3019) X_3548 = EULER LINE INTERCEPT OF LINE X(68)X(125) (0.4674, 0.0169) A B C Ma Mb Mc X2 X3542 X3 X68 X125 X498 X1060 X499 X1062 X1204 X1568 X3548 (a, b, c) = (5, 6, 7) | AB = 1

X3548 (EULER LINE INTERCEPT OF LINE X68X_125) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).

Constructed Elements

Trilinears: \(\sec B \sec C - 4 \sin B \sin C : :\)
Search Value (6, 9, 13): \(3.969030629239\)
▸ Incident Central Lines7 lines

{2, 3}, {68, 125}, {254, 2970}, {498, 1060}, {499, 1062}, {1147, 1899}, {1204, 1568}

▸ Functional Relations1 relation

Notes

As a point on the Euler line, X3548 has Shinagawa coefficients \((E - 4F, -E)\).

Let La be the polar of X4 wrt the circle centered at A and passing through X3, and define Lb and Lc cyclically. (Note that X4 is the perspector of any circle centered at a vertex of ABC.) Let A″ = Lb∩Lc, and define B″ and C″ cyclically. Triangle A″B″C″ is homothetic to the medial triangle, and the center of homothety is X3548. (Randy Hutson, December 11, 2015)