8. Let $A B C$ be a triangle with circumradius $R=17$ and inradius $r=7$. Find the maximum possible value of $\sin \frac{A}{2}$.
Answer:
$$
\frac{17+\sqrt{51}}{34}
$$

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 $O$.