View source for Kimberling’s point X(22)

==Exeter point X(22)==
[[File:Exeter X22.png|500px|right]]
Exeter point is the perspector of the circummedial triangle <math>A_0B_0C_0</math> and the tangential triangle <math>A'B'C'.</math> By another words, let <math>\triangle ABC</math> be the reference triangle (other than a right triangle). Let the medians through the vertices <math>A, B, C</math> meet the circumcircle <math>\Omega</math> of triangle <math>ABC</math> at <math>A_0, B_0,</math> and <math>C_0</math> respectively. Let <math>A'B'C'</math> be the triangle formed by the tangents at <math>A, B,</math> and <math>C</math> to <math>\Omega.</math> (Let <math>A'</math> be the vertex opposite to the side formed by the tangent at the vertex A). Prove that the lines through <math>A_0A', B_0B',</math> and <math>C_0C'</math> are concurrent, the point of concurrence lies on Euler line of triangle <math>ABC,</math> the point of concurrence <math>X_{22}</math> lies on Euler line of triangle <math>ABC, \vec {X_{22}} = \vec O + \frac {2}{J^2–3} (\vec H – \vec O), J = \frac {|OH|}{R},</math> where <math>O</math> - circumcenter, <math>H</math> - orthocenter, <math>R</math> - circumradius.

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

At first we prove that lines <math>A_0A', B_0B',</math> and <math>C_0C'</math> are concurrent. This follows from the fact that lines <math>AA_0, BB_0,</math> and <math>CC_0</math> are concurrent at point <math>G</math> and <i><b>Mapping theorem</b></i>.

Let <math>A_1, B_1,</math> and <math>C_1</math> be the midpoints of <math>BC, AC,</math> and <math>AB,</math> respectively. The points <math>A, G, A_1,</math> and <math>A_0</math> are collinear. Similarly the points <math>B, G, B_1,</math> and <math>B_0</math> are collinear.
 
Denote <math>I_{\Omega}</math> the inversion with respect <math>\Omega.</math>  It is evident that <math>I_{\Omega}(A_0) = A_0, I_{\Omega}(A') = A_1, I_{\Omega}(B_0) = B_0, I_{\Omega}(B') = B_1.</math>

Denote <math>\omega_A = I_{\Omega}(A'A_0), \omega_B =  I_{\Omega}(B'B_0) \implies</math>
<cmath>A_0 \in \omega_A, A_1 \in \omega_A, O \in \omega_A, B_0 \in \omega_B, B_1 \in \omega_B, O \in \omega_B \implies O = \omega_A \cap \omega_B.</cmath>

The power of point <math>G</math> with respect <math>\omega_A</math> is <math>GA_1 \cdot GA_0 = \frac {1}{2} AG \cdot GA_0.</math>

Similarly the power of point <math>G</math> with respect <math>\omega_B</math> is <math>GB_1 \cdot GB_0 = \frac {1}{2} BG \cdot GB_0.</math>

<math>G = BB_0 \cap AA_0 \implies AG \cdot GA_0 = BG \cdot GB_0 \implies G</math> lies on radical axis of <math>\omega_A</math> and <math>\omega_B.</math>

Therefore second crosspoint of  <math>\omega_A</math> and <math>\omega_B</math> point <math>D</math> lies on line <math>OG</math> which is the Euler line of <math>\triangle ABC</math> as desired.

Last we will find the length of <math>OX_{22}.</math>
<cmath>A_1 = BC \cap AA_0 \implies AA_1 \cdot A_1A_0 = BA_1 \cdot CA_1 = \frac {BC^2}{4}.</cmath>
<cmath>GO \cdot GD =GO \cdot (GO+ OD) =  GA_1 \cdot GA_0</cmath>
<cmath>GA_1 \cdot GA_0 = \frac {AA_1}{3} \cdot ( \frac {AA_1}{3} + A_1A_0) =  \frac {AA_1^2}{9} + 
 \frac {BC^2}{3 \cdot 4} = \frac {AB^2 + BC^2 +AC^2}{18}= \frac {R^2 – GO^2} {2}.</cmath>
<cmath>2GO^2 + 2 GO \cdot OD =  R^2 – GO^2 \implies  2 GO \cdot OD =  R^2 – 3GO^2.</cmath>
<cmath> I_{\Omega}(D) = X_{22} \implies  OX_{22} = \frac {R^2} {OD}  
 = \frac {R^2 \cdot 2 GO}{R^2 – 3 GO^2} = \frac {2 HO}{3 – \frac {HO^2}{R^2}} = \frac {2}{3 – J^2} HO</cmath> as desired.

<i><b>Mapping theorem</b></i>
[[File:Transformation.png|500px|right]]
Let triangle <math>ABC</math> and incircle <math>\omega</math> be given.
<cmath>D = BC \cap \omega, E = AC \cap \omega, F  = AB \cap \omega.</cmath>
Let <math>P</math> be the point in the plane <math>ABC.</math>
Let lines <math>DP, EP,</math> and <math>FP</math> crossing <math>\omega</math> second time at points <math>D_0, E_0,</math> and <math>F_0,</math> respectively.

Prove that lines <math>AD_0, BE_0, </math> and <math>CF_0</math> are concurrent.

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

<cmath>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'}.</cmath>
We use Claim and get: <math>k_A =  \frac {D_0E^2}{D_0F^2}.</math>
<cmath>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.</cmath>
Similarly, <math>k_B = k_E^2, k_C = k_F^2.</math>

We use the trigonometric form of Ceva's Theorem for point <math>P</math> and triangle <math>\triangle DEF</math> and get 
<cmath>k_D \cdot k_E \cdot k_F = 1 \implies k_A \cdot k_B \cdot k_C = 1^2 = 1.</cmath>
We use the trigonometric form of Ceva's Theorem for triangle <math>\triangle ABC</math> and finish proof that lines <math>AD_0, BE_0, </math> and <math>CF_0</math> are concurrent.
 

<i><b>Claim (Point on incircle)</b></i>
[[File:Point on incircle.png|450px|right]]
Let triangle <math>ABC</math> and incircle <math>\omega</math> be given.  
<cmath>D = BC \cap \omega, E = AC \cap \omega, F  = AB \cap \omega, P \in \omega, F' \in AB,</cmath>
<cmath>PF' \perp AB, E' \in AC, PE' \perp AC, A' \in EF, PA' \perp EF.</cmath>
Prove that <math>\frac {PF'}{PE'} = \frac {PF^2}{PE^2}, PA'^2 = PF' \cdot PE'.</math>

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

<cmath>AF = AE \implies \angle AFE = \angle AEF = \angle A'EE'.</cmath>
<cmath>\angle EFP = \angle PEE' \implies \angle PFF' = \angle PEE' \implies</cmath>
<cmath>\triangle PFF' \sim \triangle PEA' \implies \frac {PF}{PF'} = \frac {PE}{PA'}.</cmath>

Similarly <math>\triangle PEE' \sim \triangle PFA' \implies \frac {PE}{PE'} = \frac {PF}{PA'}.</math>

We multiply and divide these equations and get:
<cmath>PA'^2 = PF' \cdot PE', \frac {PF'}{PE'} = \frac {PF^2}{PE^2}.</cmath>

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