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 f(A,B,C) : f(B,C,A) : f(C,A,B), where f(A,B,C) = sqrt(3/4) + sin(A + π/3) - sin(B + π/3) - sin(C + π/3)
= sqrt(3/4) + cos(A - π/6) - cos(B - π/6) - cos(C - π/6)
Barycentrics (sin A)f(A,B,C) : (sin B)f(B,C,A) : (sin C)f(C,A,B)
Let T be the excentral triangle, whose vertices are the A-, B-, C- excenters of triangle ABC. Let U be the equilateral triangle having segment BC as base with vertex A' on the side of BC that does not contain vertex A. Define B' and C' cyclically, and let T' be the triangle A'B'C'. Let V be the equilateral triangle having BC as base with vertex A" on the side of BC that contains A. Define B" and C" cyclically, and let T" = A"B"C". Then T and T' are perspective, and X(1276) is their perspector. (Lawrence Evans, 2/4/2003)
Evans conjectured that X(1), X(484), X(1276), X(1277) are concyclic, and reported that Paul Yiu confirmed this conjecture and noted that the center of this circle is X(1019). (Lawrence Evans, 2/24/2003)
X(1276) lies on these lines: 1,15 4,9 14,484 63,616
X(1276) = inverse-in-Bevan-circle of X(1277) (noted by Peter J. C. Moses, Sept. 8, 2004)