1. Let $\odot O, \odot I$ be the circumcircle and incircle of Rt $\triangle ABC$, $R, r$ be the radii of these circles, and $J$ be the point symmetric to the right-angle vertex $C$ with respect to $I$. Find the length of $OJ$.
1. Obviously, the circle $\odot J$ with radius $2r$ is tangent to $AC$ and $BC$.
Below, we prove that $\odot J$ is tangent to $\odot O$.
As shown in Figure 1, consider the circle $\omega$ that is tangent to $AC$ and $BC$ at points $P$ and $Q$, respectively, and is internally tangent to $\odot O$ at point $T$.
Next, we prove that $J$ is the center of circle $\omega$.
Since $T$ is the center of homothety of circles $\omega$ and $\odot O$, the lines $TP$ and $TQ$ intersect $\odot O$ at points $B'$ and $A'$, which are the midpoints of arcs $\overparen{AC}$ and $\overparen{BC}$, respectively.
Therefore, the lines $AA'$ and $BB'$ intersect at point $I$.
Applying Pascal's theorem to the hexagon $CA'ATB'B$, we know that points $P$, $I$, and $Q$ are collinear:

Since line $PQ$ is perpendicular to the angle bisector of $\angle C$, $P$ and $Q$ are the projections of point $J$ onto $AC$ and $BC$, respectively, meaning that $J$ is the center of circle $\omega$.
Therefore, $OJ = R - 2r$.