Kimberling’s point X(23)

Far-out point X(23)

Let $\triangle A'B'C'$ be the tangential triangle of $\triangle ABC.$

Let $G, \Omega, O, R,$ and $H$ be the centroid, circumcircle, circumcenter, circumradius and orthocenter of $\triangle ABC.$

Prove that the second crosspoint of circumcircles of $\triangle AA'O, \triangle BB'O,$ and $\triangle CC'O$ is point $X_{23}.$ Point $X_{23}$ lies on Euler line of $\triangle ABC, X_{23} = O + \frac {3}{J^2} (H – O), J = \frac {OH}{R}.$

Proof

Denote $I_{\Omega}$ the inversion with respect $\Omega, A_1, B_1, C_1$ midpoints of $BC, AC, AB.$

It is evident that $I_{\Omega}(A') = A_1, I_{\Omega}(B') = B_1, I_{\Omega}(C') = C_1.$

The inversion of circles $AA'O, BB'O, CC'O$ are lines $AA_1, BB_1,CC_1$ which crosses at point $G \implies X_{23} = I_{\Omega}(G).$

Therefore point $X_{23}$ lies on Euler line $OG$ of $\triangle ABC, OG \cdot OX_{23} = R^2 \implies \frac {OX_{23}} {OH} = \frac {R^2}{OG \cdot OH} = \frac {3}{J^2},$ as desired.

vladimir.shelomovskii@gmail.com, vvsss

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Kimberling’s point X(23)

Far-out point X(23)