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 f(A,B,C) : f(B,C,A) : f(C,A,B),
where f(A,B,C) = (sec B + sec C - sec A) sec A
Barycentrics (sin A)f(A,B,C) : (sin B)f(B,C,A) : (sin C)f(C,A,B)Suppose La, Lb, Lc are lines through a point P, respectively perpendicular to sidelines BC, CA, AB. Let Ab be the point where La meets AB, and let Ac be the point where La meets AC. Define Bc, Ba, Ca, Cb cyclically. Then X(1148) is the point P, which satisfies
|PAb| + |PAc| = |PBc| + |PBa| = |PCa| + |PCb|.
See Hyacinthos messages #4204-4206, 10/01.
X(1148) lies on these lines:
1,1075 3,653 4,65 46,243 92,942X(1148) = X(1)-Ceva conjugate of X(4)
X(1148) = X(46)-Hirst inverse of X(243)