Convex pentagon ABCDE is inscribed in circle $γ$. Suppose that $AB = 14$, $BE = 10$, $BC = CD = DE$, and $[ABCDE] = 3[ACD]$. Then there are two possible values for the radius of $γ$. The sum of these two values is $\sqrt n$ for some positive integer $n$. Compute $n$.
I don't know how to solve such questions. Please tell how to do such problems.
For any polygon $P_1P_2\ldots P_n$, we will use $[P_1P_2\ldots P_n]$ to denote its area.
Let $\ell$ be a line through $A$ parallel to $BE$. let $A'$ be the other intersection of $\gamma$ and $\ell$.
Let $M$ be the midpoint of $AA'$. When $[ABCDE] = 3[ACD]$, we have
$$\begin{align}[MBCDE] &= [MBE] + [BCDE] = [ABE] + [BCDE]\\ &= [ABCDE] = 3[ACD] = 3[MCD]\end{align}$$
Since $[MBCDE] = [MBC] + [MCD] + [MDE]$ and by symmetry $[MBC] = [MDE]$, we obtain $$[MBC] = [MCD] = [MDE]$$ Since $BC = CD = DE$, the distance of $M$ to lines $BC$, $CD$ and $DE$ is the same. So $M$ lies on the angular bisectors for $\angle BCD$ and $\angle CDE$.
Since $B, C, D, E$ lies on $\gamma$ and $BC = CD = DE$, these two angular bisectors intersect at $O$, the center of $\gamma$. As a result, $M = O$ and $AA'$ is a diameter of $\gamma$.
Let $N$ be the midpoint of $BE$ and $R$ be the radius of $\gamma$.
When $AB < AE$, as illustrated in the diagram above, there are several parallel and perpendicular lines that allow us to apply Pythagorean theorem. Choose a coordinate system so that $O = M$ is the origin and $\ell$ is the $y$-axes. It is not hard to see
$$\begin{align}14^2 = AB^2 &= (AM - BN)^2 + MN^2 = (AM - BN)^2 + BM^2 - BN^2\\ &= (R-5)^2 + R^2 - 5^2\end{align} $$ This leads to $$R^2 - 5R - 98 = 0\quad \implies \quad R = \frac{\sqrt{417}+5}{2}$$ (the other root of the quadratic equation is negative).
Similarly, when $AB > AE$, we have:
$$R^2 + 5R - 98 = 0 \quad\implies\quad R = \frac{\sqrt{417}-5}{2}$$
Combine these, we find the sum of the two solutions of $R$ is $\sqrt{417}$ and hence $n = 417$.