X1285¶
LEMOINE HOMOTHETIC CENTER
X1285 (LEMOINE HOMOTHETIC CENTER) constructed on acute reference triangle \((a, b, c) = (5, 6, 7)\) with unit baseline AB = \(1\).¶
Constructed Elements
Central Lines: {4, 32}, {6, 376}
Related Centers: X4 (ORTHOCENTER), X32 (3rd POWER POINT), X6 (SYMMEDIAN POINT), X376 (CENTROID OF THE ANTIPEDAL TRIANGLE OF X)
Incident Central Lines9 lines
Notes¶
Let T denote the antipedal triangle of X2, and let T(m) denote X(m)-of-T. Then T(m) is a triangle center of the reference triangle ABC. The appearance of (m,n) in the following list means that T(m) = Xn. Note the special case m = 1285.
(2,376), (6,2), (115,99), (125,1296), (597,3), (599,20), (1285,1285), (1992,4), (2030,187), (2549,69)
Let U denote the pedal triangle of X6, and let U(m) denote X(m)-of-U. Then U(m) is a triangle center of the reference triangle ABC. The appearance of (m,n) in the following list means that U(m) = Xn. (This list and the one just above were contributed by Peter Moses, 11/15/2007.)
(2,6), (3,597), (4,1992), (20,599), (69,2549), (99,115), (187,2030), (376,2), (1285, 1285), (1296, 125)
Let A'B'C' be the antipedal triangle of X2. The centroid of A'B'C' is X376. Let A″B″C″ be the antipedal triangle, wrt A'B'C', of X376. A″B″C″ is homothetic to ABC, and the center of homothety is X1285. (Randy Hutson, March 21, 2019)
Let A'B'C' be the pedal triangle of X6. Let K' be X6 of A'B'C'. Let A″B″C″ be the pedal triangle, wrt A'B'C', of K'. A″B″C″ is homothetic to ABC, and the center of homothety is X1285. (Randy Hutson, March 21, 2019)