Trilinears sec(A - π/4) : sec(B - π/4) : sec(C - π/4)
=1/(sin A + cos A) : 1/(sin B + cos B) : 1/(sin C + cos C)
Trilinears sin A + cos(B - C) : sin B + cos(C - A) : sin C + cos(A - B) (Peter J. C. Moses, 8/22/03)
Barycentrics sin A sec(A - π/4) : sin B sec(B - π/4) : sin C sec(C - π/4)
Erect a square outwardly from each side of triangle ABC. Let A'B'C' be the triangle formed by the respective centers of the squares. The lines AA', BB', CC' concur in X(485). For details, visit Floor van Lamoen's site, Vierkanten in een driehoek: 1. Omgeschreven vierkanten (van Lamoen, 4/26/98) and his article "Friendship Among Triangle Centers," Forum Geometricorum, 1 (2001) 1-6. See also Paul Yiu's papers
"Squares Erected on the Sides of a Triangle", and
"On the Squares Erected Externally on the Sides of a Triangle".
X(485) lies on these lines: 2,372 3,590 4,371 5,6 69,639 76,491 226,481 489,671
X(485) = reflection of X(488) in X(641)
X(485) = isogonal conjugate of X(371)
X(485) = isotomic conjugate of X(492)
X(485) = complement of X(488)
X(485) = anticomplement of X(641)
X(485) = X(3)-cross conjugate of X(486)
X(485) = internal center of similitude of nine-point circle and 2nd Lemoine circle
, select the center name from the list and, then, click on the vertices A, B and C successively.