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) = cos B + cos C - 2 cos A;
= g(a,b,c) : g(b,c,a) : g(c,a,b), where g(a,b,c) = (b - c)2 + a(b + c - 2a)
Barycentrics (sin A)f(A,B,C) : (sin B)f(B,C,A) : (sin C)f(C,A,B)
= ag(a,b,c) : bg(b,c,a) : cg(c,a,b)
Let XYZ be the intouch triangle of ABC; i.e., the pedal triangle of the incenter, I. The circles AIX, BIY, CIZ concur in two points. One of them is I; the other is X(1155). This result is obtain by inversion in
Heinz Schröder, "Die Inversion und ihre Anwendung im Unterricht der Oberstufe," Der Mathematikunterricht 1 (1957) 59-80.
Each vertex of the tangential triangle of any triangle T is the inverse-in-the-circumcircle-of-T of the midpoints of the sides of T. Applying this to triangle XYZ shows that X(1155) is the inverse-in-the-incircle of the centroid of XYZ; i.e., X(1155) is X(23)-of-the-intouch-triangle. (Darij Grinberg, #6319, 1/11/03; coordinates by Jean-Pierre Ehrmann, #6320, 1/11/03)
X(1155) lies on these lines:
1,3 10,535 11,516 37,750 44,513 47,582 63,210 88,105 89,1002 100,518 227,603 238,1054 243,653 244,902 404,960
X(1155) = midpoint of X(36) and X(484)
X(1155) = reflection of X(1319) in X(36)
X(1155) = isogonal conjugate of X(1156)
X(1155) = inverse-in-circumcircle of X(55)
X(1155) = inverse-in-incircle of X(354)
X(1155) = inverse-in-Bevan-circle of X(57)
X(1155) = crosspoint of X(I) and X(J) for these (I,J): (1,1156), (527,1323)
X(1155) = crosssum of X(1) and X(1155)
X(1155) = crossdifference of any two points on line X(1)X(650)