29. [15] In acute triangle $A B C$, let $H$ be the orthocenter and $D$ the foot of the altitude from $A$. The circumcircle of triangle $B H C$ intersects $A C$ at $E \neq C$, and $A B$ at $F \neq B$. If $B D=3, C D=7$, and $\frac{A H}{H D}=\frac{5}{7}$, the area of triangle $A E F$ can be expressed as $\frac{a}{b}$, where $a, b$ are relatively prime positive integers. Compute $100 a+b$.
Answer:
12017

Solution:
Let $A H$ intersect the circumcircle of $\triangle A B C$ again at $P$, and the circumcircle of $\triangle B H C$ again at $Q$. Because $\angle B H C=180-\angle A=\angle B P C, P$ is the reflection of $H$ over $D$. Thus, we know that $P D=H D$. From power of a point and $A D=\frac{12 H D}{7}$,
$$
B D \cdot C D=A D \cdot P D=\frac{12 H D^{2}}{7} .
$$

From this, $H D=\frac{7}{2}$ and $A H=\frac{5}{2}$. Furthermore, because $\triangle B H C$ is the reflection of $\triangle B P C$ over $B C$, the circumcircle of $\triangle B H C$ is the reflection of the circumcircle of $\triangle A B C$ over $B C$. Then, $A Q=$ $2 A D=12$. Applying Power of a Point,
$$
A C \cdot A E=A B \cdot A F=A H \cdot A Q=30 .
$$

We can compute $A C=\sqrt{85}$ and $A B=3 \sqrt{5}$, which means that $A E=\frac{6 \sqrt{85}}{17}$ and $A F=2 \sqrt{5}$. Also, $[A B C]=\frac{B C \cdot A D}{2}=30$. Therefore,
$$
[A E F]=\frac{A E \cdot A F}{A C \cdot A B} \cdot[A B C]=\frac{4}{17} \cdot 30=\frac{120}{17} .
$$