HMMT-Feb 2017 geo, Problem 8

Let $A B C$ be a triangle with circumradius $R=17$ and inradius $r=7$. Find the maximum possible value of $\sin \frac{A}{2}$.

Solution: Letting $I$ and $O$ denote the incenter and circumcenter of triangle $A B C$ we have by the triangle inequality that \[ A O \leq A I+O I \Longrightarrow R \leq \frac{r}{\sin \frac{A}{2}}+\sqrt{R(R-2 r)} \] and by plugging in our values for $r$ and $R$ we get \[ \sin \frac{A}{2} \leq \frac{17+\sqrt{51}}{34} \] as desired. Equality holds when $A B C$ is isosceles and $I$ lies between $A$ and $O$. $\fbox{\frac{17+\sqrt{51}}{34}}$.

Answer: \frac{17+\sqrt{51}}{34}