AMC 12 Mock Exam 1 Problem 22

Let $R$ and $r$ be the circumradius and inradius of $\triangle ABC$, with angles $A,B,C$ at the vertices. The center $P$ of $\omega$ lies on the bisector of $\angle A$, say $AP=t$, and $\omega$ has radius $\rho=t\sin\frac A2$.

Let $O$ be the circumcenter. The line $AO$ makes an angle $\left|\frac{B-C}{2}\right|$ with the bisector of $\angle A$ (the bisector bisects the angle between $AO$ and the altitude from $A$, and the latter makes the angle $|B-C|$ with $AO$). Internal tangency means $OP=R-\rho$. By the law of cosines in $\triangle AOP$ (with $AO=R$), $$t^2+R^2-2tR\cos\tfrac{B-C}{2}=(R-t\sin\tfrac A2)^2.$$ Expanding, $t^2\left(1-\sin^2\tfrac A2\right)=2tR\left(\cos\tfrac{B-C}2-\sin\tfrac A2\right)$. Because $\sin\frac A2=\cos\frac{B+C}2$, $$\cos\tfrac{B-C}2-\cos\tfrac{B+C}2=2\sin\tfrac B2\sin\tfrac C2,$$ so $t\cos^2\frac A2=4R\sin\frac B2\sin\frac C2$. Multiplying by $\sin\frac A2$ and using $r=4R\sin\frac A2\sin\frac B2\sin\frac C2$, $$\rho=\frac{r}{\cos^2\frac A2}.$$

For the $13$–$14$–$15$ triangle, the area is $84$ and $s=21$, so $r=\frac{84}{21}=4$. With $a=14$, $b=15$, $c=13$: $\cos A=\frac{15^2+13^2-14^2}{2\cdot15\cdot13}=\frac{33}{65}$, so $\cos^2\frac A2=\frac{1+\cos A}{2}=\frac{49}{65}$. Therefore $\rho=4\cdot\frac{65}{49}=\boxed{\tfrac{260}{49}}$.

(Why the other values appear: the analogous circles at $B$ and $C$ have radii $\frac{4}{\cos^2(B/2)}=\frac{52}{9}$ and $\frac{4}{\cos^2(C/2)}=5$.)