9. Let $A B C D$ be a trapezoid with $A B \| C D$ and $A D=B D$. Let $M$ be the midpoint of $A B$, and let $P \neq C$ be the second intersection of the circumcircle of $\triangle B C D$ and the diagonal $A C$. Suppose that $B C=27, C D=25$, and $A P=10$. If $M P=\frac{a}{b}$ for relatively prime positive integers $a$ and $b$, compute $100 a+b$.
Answer: 2705
Solution 1: As $\angle P B D=\angle P C D=\angle P A B, D B$ is tangent to $(A B P)$. As $D A=D B, D A$ is also tangent to $(A B P)$. Let $C B$ intersect $(A B P)$ again at $X \neq B$; it follows that $X D$ is the $X$-symmedian of $\triangle A X B$. As $\angle A X C=\angle D A B=\angle A D C, X$ also lies on $(A C D)$. Therefore $\angle P X B=\angle P A B=$ $\angle P C D=\angle A X D$, so $X P$ is the median of $\triangle A X B$, i.e. $X P$ passes through $M$.
Now we have $\triangle M P A \sim \triangle M B X$ and $\triangle A B X \sim \triangle B C D$. Therefore
$$
\frac{M P}{A P}=\frac{M B}{X B}=\frac{A B}{2 X B}=\frac{B C}{2 D C} \Longrightarrow M P=\frac{B C \cdot A P}{2 D C}=\frac{27 \cdot 10}{2 \cdot 25}=\frac{27}{5} .
$$

Solution 2: Let $E$ be the point such that $\square B D C E$ is a parallelogram. Then $\square A D C E$ is an isosceles trapezoid, therefore $\angle P A B=\angle C A B=\angle B E D=\angle C D E$; this angle is also equal to $\angle P B D=\angle P C D$. Now, $\angle A B P=\angle A B D-\angle P B D=\angle B E C-\angle B E D=\angle D E C$, therefore $\triangle P A B \sim \triangle C D E$. Let $F$ be the midpoint of $D E$, which lies on $B C$ because $\square B D C E$ is a parallelogram. It follows that $(\triangle P A B, M) \sim(\triangle C D E, F)$, therefore
$$
\frac{M P}{A P}=\frac{F C}{D C}=\frac{B C}{2 D C}
$$
and again this gives $M P=\frac{27}{5}$.