Kimberling’s point X(22)
Exeter point X(22)
Exeter point is the perspector of the circummedial triangle 
and the tangential triangle 
By another words, let 
be the reference triangle (other than a right triangle). Let the medians through the vertices 
meet the circumcircle 
of triangle 
at 
and 
respectively. Let 
be the triangle formed by the tangents at 
and 
to 
(Let 
be the vertex opposite to the side formed by the tangent at the vertex A). Prove that the lines through 
and 
are concurrent, the point of concurrence lies on Euler line of triangle 
Proof
At first we prove that lines and are concurrent. This follows from the fact that lines 
and 
are concurrent at point 
and Mapping theorem.
Mapping theorem
Let triangle and incircle 
be given.
![\[D = BC \cap \omega, E = AC \cap \omega, F = AB \cap \omega.\]](http://latex.artofproblemsolving.com/e/2/e/e2e94693a56b5f5ff57ba598d083f5fe92f59ded.png)
Let 
be the point in the plane
Let lines 
and 
crossing second time at points 
and 
respectively.
Prove that lines 
and 
are concurrent.
Proof
![\[k_A = \frac {\sin {D_0AE'}}{\sin {D_0AF'}} = \frac {D_0E'}{D_0A} \cdot \frac {D_0A}{D_0F'} = \frac {D_0E'}{D_0F'}.\]](http://latex.artofproblemsolving.com/1/4/d/14dbd6ca0c930f623b2904c5e1011da5d7ec411c.png)
We use Claim and get: 
![\[k_D = \frac {\sin {D_0DE}}{\sin {D_0DF}} = \frac {D_0E}{2R} \cdot \frac {2R}{D_0F} = \frac {D_0E}{D_0F} \implies k_A = k_D^2.\]](http://latex.artofproblemsolving.com/b/a/6/ba66d5bb48cfaceca1e60aa83991c9621a6795d7.png)
Similarly, 
We use the trigonometric form of Ceva's Theorem for point and triangle 
and get
![\[k_D \cdot k_E \cdot k_F = 1 \implies k_A \cdot k_B \cdot k_C = 1^2 = 1.\]](http://latex.artofproblemsolving.com/9/1/2/912314ceabb1baeeeeb9df01d098800ab939f1f2.png)
We use the trigonometric form of Ceva's Theorem for triangle and finish proof that lines and are concurrent.
Claim (Point on incircle)
Let triangle and incircle be given.
![\[D = BC \cap \omega, E = AC \cap \omega, F = AB \cap \omega, P \in \omega, F' \in AB,\]](http://latex.artofproblemsolving.com/1/0/f/10fabc0f6adf1653aaeaa2e90b70bda09f6f7f30.png)
![\[PF' \perp AB, E' \in AC, PE' \perp AC, A' \in EF, PA' \perp EF.\]](http://latex.artofproblemsolving.com/4/7/d/47d871d13e2d442e7ceae01ed919495392cf3221.png)
Prove that 
Proof
![\[AF = AE \implies \angle AFE = \angle AEF = \angle A'EE'.\]](http://latex.artofproblemsolving.com/7/9/2/7921c86e05c6682033f66ba4754b29073544917a.png)
![\[\angle EFP = \angle PEE' \implies \angle PFF' = \angle PEE' \implies\]](http://latex.artofproblemsolving.com/f/6/c/f6ca16247c0782c83ecffbb3b6724365faf850fa.png)
![\[\triangle PFF' \sim \triangle PEA' \implies \frac {PF}{PF'} = \frac {PE}{PA'}.\]](http://latex.artofproblemsolving.com/8/2/b/82b7af65e564dd36998f842e05b581d318ec3ab6.png)
Similarly 
We multiply and divide these equations and get:
![\[PA'^2 = PF' \cdot PE', \frac {PF'}{PE'} = \frac {PF^2}{PE^2}.\]](http://latex.artofproblemsolving.com/1/3/1/131061f99d4a05f88a138dda53cf5674f9be22b6.png)
vladimir.shelomovskii@gmail.com, vvsss
