Kimberling’s point X(22)

Exeter point X(22)

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

Proof

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

Let $A_1, B_1,$ and $C_1$ be the midpoints of $BC, AC,$ and $AB,$ respectively. The points $A, G, A_1,$ and $A_0$ are collinear. Similarly the points $B, G, B_1,$ and $B_0$ are collinear.

Denote $I_{\Omega}$ the inversion with respect $\Omega.$ It is evident that $I_{\Omega}(A_0) = A_0, I_{\Omega}(A') = A_1, I_{\Omega}(B_0) = B_0, I_{\Omega}(B') = B_1.$

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

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

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

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

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

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

Mapping theorem

Let triangle $ABC$ and incircle $\omega$ be given. $D = BC \cap \omega, E = AC \cap \omega, F = AB \cap \omega.$ Let $P$ be the point in the plane $ABC.$ Let lines $DP, EP,$ and $FP$ crossing $\omega$ second time at points $D_0, E_0,$ and $F_0,$ respectively.

Prove that lines $AD_0, BE_0,$ and $CF_0$ are concurrent.

Proof

$k_A = \frac {\sin {D_0AE'}}{\sin {D_0AF'}} = \frac {D_0E'}{D_0A} \cdot \frac {D_0A}{D_0F'} = \frac {D_0E'}{D_0F'}.$ We use Claim and get: $k_A = \frac {D_0E^2}{D_0F^2}.$ $k_D = \frac {\sin {D_0DE}}{\sin {D_0DF}} = \frac {D_0E}{2R} \cdot \frac {2R}{D_0F} = \frac {D_0E}{D_0F} \implies k_A = k_D^2.$ Similarly, $k_B = k_E^2, k_C = k_F^2.$

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

Claim (Point on incircle)

Let triangle $ABC$ and incircle $\omega$ be given. $D = BC \cap \omega, E = AC \cap \omega, F = AB \cap \omega, P \in \omega, F' \in AB,$ $PF' \perp AB, E' \in AC, PE' \perp AC, A' \in EF, PA' \perp EF.$ Prove that $\frac {PF'}{PE'} = \frac {PF^2}{PE^2}, PA'^2 = PF' \cdot PE'.$

Proof

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

Similarly $\triangle PEE' \sim \triangle PFA' \implies \frac {PE}{PE'} = \frac {PF}{PA'}.$

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

vladimir.shelomovskii@gmail.com, vvsss

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Kimberling’s point X(22)

Exeter point X(22)