8. Let $A B C$ be an acute triangle with circumcircle $\Gamma$. Let the internal angle bisector of $\angle B A C$ intersect $B C$ and $\Gamma$ at $E$ and $N$, respectively. Let $A^{\prime}$ be the antipode of $A$ on $\Gamma$ and let $V$ be the point where $A A^{\prime}$ intersects $B C$. Given that $E V=6, V A^{\prime}=7$, and $A^{\prime} N=9$, compute the radius of $\Gamma$.
Answer: $\square$
Solution 1: Let $H_{a}$ be the foot of the altitude from $A$ to $B C$. Since $A E$ bisects $\angle H_{a} A V$, by the angle bisector theorem $\frac{A H_{a}}{H_{a} E}=\frac{A V}{V E}$. Note that $\triangle A H_{a} E \sim \triangle A N A^{\prime}$ are similar right triangles, so $\frac{A N}{N A^{\prime}}=\frac{A H_{a}}{H_{a} E}$.

Let $R$ be the radius of $\Gamma$. We know that $A A^{\prime}=2 R$, so $A N=\sqrt{A A^{\prime 2}-N A^{\prime 2}}=\sqrt{4 R^{2}-81}$ and $A V=A A^{\prime}-V A^{\prime}=2 R-7$. Therefore
$$
\frac{\sqrt{4 R^{2}-81}}{9}=\frac{A N}{N A^{\prime}}=\frac{A H_{a}}{H_{a} E}=\frac{A V}{V E}=\frac{2 R-7}{6} .
$$

The resulting quadratic equation is
$$
0=9(2 R-7)^{2}-4\left(4 R^{2}-81\right)=20 R^{2}-252 R+765=(2 R-15)(10 R-51) .
$$

We are given that $A B C$ is acute so $V A^{\prime}<R$. Therefore $R=\frac{15}{2}$.
Solution 2: Let $\Psi$ denote inversion about $A$ with radius $\sqrt{A B \cdot A C}$ composed with reflection about $A E$. Note that $\Psi$ swaps the pairs $\{B, C\},\{E, N\}$, and $\left\{H_{a}, A^{\prime}\right\}$. Let $K=\Psi(V)$, which is also the second intersection of $A H_{a}$ with $\Gamma$. Since $A E$ bisects $\angle K A A^{\prime}$, we have $N K=N A^{\prime}=9$. By the inversion distance formula,
$$
N K=\frac{A B \cdot A C \cdot V E}{A E \cdot A V}=\frac{A E \cdot A N \cdot V E}{A E \cdot A V}=\frac{A N \cdot V E}{A V} .
$$

This leads to the same equation as the previous solution.