Zaslavsky A.A.

Let $O, I$ be the centers of the circumcircle and incircle of a right triangle; $R, r$ be the radii of these circles; $J$ be the point symmetric to the right-angle vertex with respect to $I$. Find $O J$.
Let $ABC$ be a given right triangle, $\angle C=90^{\circ}$. It is obvious that the circle with center $J$ and radius $2r$ touches $AC$ and $BC$. We need to prove that it also touches the circumcircle $\Omega$ of triangle $ABC$; from this it will follow that $OJ=R-2r$.

Consider the circle $\omega$ that touches $AC$, $BC$ at points $P$, $Q$ respectively, and touches $\Omega$ internally at point $T$; we need to prove that $J$ is the center of $\omega$. Since $T$ is the center of homothety of $\omega$ and $\Omega$, the lines $TP$, $TQ$ intersect the circumcircle again at points where the tangents are parallel to $AC$ and $BC$, that is, at the midpoints $B'$, $A'$ of the arcs $AC$ and $C$. Therefore, the lines $AA'$ and $BB'$ intersect at point $I$. By Pascal's theorem (see problem 57105), applied to the broken line $C A A' T B' B$, the points $P$, $I$, $Q$ lie on the same line. Since the line $PQ$ is perpendicular to the angle bisector of $\angle C$, $P$, $Q$ are the projections of $J$ onto $AC$ and $BC$, which means that $J$ is the center of $\omega$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6409e68af5a1ba5e989dg-35.jpg?height=1418&width=1532&top_left_y=800&top_left_x=252)

Answer

$R-2r$.