View source for Kimberling’s point X(26)

==CIRCUMCENTER OF THE TANGENTIAL TRIANGLE X(26)==
[[File:X26.png|500px|right]]
The circumcenter of the tangential triangle <math>\triangle A'B'C'</math> of <math>\triangle ABC</math> (Kimberling’s point <math>X(26))</math> lies on the Euler line of <math>\triangle ABC</math> (The tangential triangle of a reference triangle (other than a right triangle) is the triangle whose sides are on the tangent lines to the reference triangle's circumcircle at the reference triangle's vertices).

<i><b>Proof</b></i>

Let <math>A_0, B_0,</math> and <math>C_0</math> be midpoints of <math>BC, AC,</math> and <math>AB,</math> respectively.

Let <math>\omega</math> be circumcircle of <math>\triangle A_0B_0C_0.</math> It is nine-points circle of the <math>\triangle ABC.</math>

Let <math>\Omega</math> be circumcircle of <math>\triangle ABC.</math> Let <math>\Omega'</math> be circumcircle of <math>\triangle A'B'C'.</math>

<math>A'B</math> and <math>A'C</math> are tangents to <math>\Omega \implies</math> inversion with respect <math>\Omega</math> swap <math>B'</math> and <math>B_0.</math> Similarly, this inversion swap  <math>A'</math> and <math>A_0,  C'</math> and <math>C_0.</math>

Therefore this inversion swap <math>\omega</math> and <math>\Omega'.</math> 
The center <math>N</math> of  <math>\omega</math> and the center <math>O</math> of <math>\Omega</math> lies on Euler line, so the center  <math>O'</math> of <math>\Omega'</math> lies on this line, as desired.

After some calculations in the case, shown on diagram, we can get 

<math>OO'= \frac {2R}{t^{–1}–t}, </math> where <math>t = \sqrt {9 – \frac{a^2 +b^2 + c^2}{R^2}}, R = OA, a = BC, b = AC, c = AB.</math>

Similarly, one can find position of point <math>X(26)</math> on Euler line in another cases.

'''vladimir.shelomovskii@gmail.com, vvsss'''