AMC 12 Mock Exam 5 Problem 20

Place $B=(0,0)$ and $C=(8,0)$. From $AB=5$ and $AC=7$: $x^2+y^2=25$ and $(x-8)^2+y^2=49$, so $x=\frac52$ and $y=\frac{5\sqrt3}{2}$; thus $\angle B=60^\circ$.

Circumcenter $O=(4,y_0)$, with $OB=OA$: $16+y_0^2=\frac94+\left(y_0-\frac{5\sqrt3}{2}\right)^2$, which gives $5\sqrt3\,y_0=5$, so $O=\left(4,\frac{\sqrt3}{3}\right)$ and $R=\sqrt{16+\frac13}=\frac{7\sqrt3}{3}$.

Incenter $I=\frac{a A+bB+cC}{a+b+c}$ with $a=BC=8$, $b=CA=7$, $c=AB=5$: $I=\frac{8\left(\frac52,\frac{5\sqrt3}{2}\right)+5(8,0)}{20}=(3,\sqrt3)$, and indeed $r=\sqrt3$ (the area $10\sqrt3$ divided by $s=10$).

Then $OI^2=(4-3)^2+\left(\frac{\sqrt3}{3}-\sqrt3\right)^2=1+\frac43=\frac73$, so $OI=\boxed{\tfrac{\sqrt{21}}{3}}$. (The coordinate computation above is self-contained. Students who know Euler's formula $OI^2=R^2-2Rr$ can use it as a shortcut: $\frac{49}{3}-14=\frac73$.)