4. Let $A B C D$ be a trapezoid with $A B \| C D, A B=5, B C=9, C D=10$, and $D A=7$. Lines $B C$ and $D A$ intersect at point $E$. Let $M$ be the midpoint of $C D$, and let $N$ be the intersection of the circumcircles of $\triangle B M C$ and $\triangle D M A$ (other than $M$ ). If $E N^{2}=\frac{a}{b}$ for relatively prime positive integers $a$ and $b$, compute $100 a+b$.
Answer: 90011
Solution: From $\triangle E A B \sim \triangle E D C$ with length ratio $1: 2$, we have $E A=7$ and $E B=9$. This means that $A, B, M$ are the midpoints of the sides of $\triangle E C D$. Let $N^{\prime}$ be the circumcenter of $\triangle E C D$. Since $N^{\prime}$ is on the perpendicular bisectors of $E C$ and $C D$, we have $\angle N^{\prime} M C=\angle N^{\prime} B C=90^{\circ}$. Thus $N^{\prime}$ is on the circumcircle of $\triangle B M C$. Similarly, $N^{\prime}$ is on the circumcircle of $\triangle D M A$. As $N^{\prime} \neq M$, we must have $N^{\prime}=N$. So it suffices to compute $R^{2}$, where $R$ is the circumradius of $\triangle E C D$.
We can compute $K=[\triangle E C D]$ to be $21 \sqrt{11}$ from Heron's formula, giving
$$
R=\frac{10 \cdot 14 \cdot 18}{4 K}=\frac{30}{\sqrt{11}} .
$$

So $R^{2}=\frac{900}{11}$, and the final answer is 90011 .