Interactive Applet |
You can move the points A, B and C (click on the point and drag it).
Press the keys “+” and “−” to zoom in or zoom out the visualization window and use the arrow keys to translate it.
You can also construct all centers related with this one (as described in ETC) using the “Run Macro Tool”. To do this, click on the icon , select the center name from the list and, then, click on the vertices A, B and C successively.
Information from Kimberling's Encyclopedia of Triangle Centers |
Trilinears (cos B/2 + cos C/2) sec A/2 : (cos C/2 + cos A/2) sec B/2 : (cos A/2 + cos B/2) sec C/2
Barycentrics g(A,B,C) : g(B,C,A) : g(C,A,B), where g(A,B,C) = (sin A)(cos B/2 + cos C/2) sec A/2Let A',B',C' be the first points of intersection of the angle bisectors of ABC with its incircle. The tangents at A', B', C' form a triangle A"B"C", and the lines AA",BB",CC" concur in X(177). Also, X(177) = X(1) of the intouch triangle.
Clark Kimberling and G. R. Veldkamp, Problem 1160 and Solution, Crux Mathematicorum 13 (1987) 298-299 [proposed 1986].
X(177) is the perspector of ABC and the Yff central triangle, and X(177) is X(65)-of-the-Yff-central-triangle . (Darij Grinberg, Hyacinthos #7689, 8/25/2003)
X(177) lies on these lines: 1,167 7,555 8,556 9,173 57,164
X(177) = isogonal conjugate of X(260)
X(177) = X(7)-Ceva conjugate of X(234)
X(177) = crosspoint of X(7) and X(174)
X(177) = crosssum of X(55) and X(259)