jmc

Let $A$, $B$, and $C$ be the vertices of the triangle so that $AB=5$, $AC=12$, and $BC=13$. Let $I$ and $O$ be the incenter and circumcenter of triangle $ABC$, respectively. Let the incircle of triangle $ABC$ be tangent to sides $BC$, $AC$, and $AB$ at $D$, $E$, and $F$, respectively.



Since $\angle BAC=90^{\circ }$, the circumcenter $O$ of triangle $ABC$ is the midpoint of hypotenuse $BC$.

Since $AE$ and $AF$ are tangents from $A$ to the same circle, $AE=AF$. Let $x=AE=AF$. Similarly, let $y=BD=BF$ and $z=CD=CE$. Then $x+y=AF+BF=AB=5$, $x+z=AE+CE=AC=12$, $y+z=BD+CD=BC=13$. Solving this system of equations, we find $x=2$, $y=3$, and $z=10$. Then $DO=BO-BD=BC/2-y=13/2-3=7/2$.

The inradius $r$ of triangle $ABC$ is given by $r=K/s$, where $K$ is the area of triangle $ABC$, and $s$ is the semi-perimeter. We see that $K=[ABC]=1/2\cdot AB\cdot AC=1/2\cdot 5\cdot 12=30$, and $s=(AB+AC+BC)/2=(5+12+13)/2=15$, so $r=30/15=2$.

Hence, by the Pythagorean theorem on right triangle $IDO$, $$IO=\sqrt{I{D}^2+D{O}^2}=\sqrt{2^2+{\left(\frac{7}{2}\right)}^2}=\sqrt{\frac{65}{4}}=\boxed{\frac{\sqrt{65}}{2}}.$$