Answer: 192
Solution: The point $P$ has many well-known properties, including the property that $\angle B A P=\angle A C P$ and $\angle C A P=\angle B A P$. We prove this for completeness.
Invert at $A$ with radius $\sqrt{A B \cdot A C}$ and reflect about the $A$-angle bisector. Let $P^{\prime}$ be the image of $P$. The angle conditions translate to
- $P^{\prime}$ lies on line $A M$
- $P^{\prime}$ lies on the line parallel to $B C$ that passes through the reflection of $A$ about $B C$ (since $P$ lies on the circle with diameter $\overline{A O}$ )

In other words, $P^{\prime}$ is the reflection of $A$ about $M$. Then $B P^{\prime} \| A C$ and $C P^{\prime} \| A B$, so the circumcircles of $\triangle A B P$ and $\triangle A C P$ are tangent to $A C$ and $A B$, respectively. This gives the desired result.
Extend $B P$ and $C P$ to meet the circumcircle of $\triangle A B C$ again at $B^{\prime}$ and $C^{\prime}$, respectively. Then $\angle C^{\prime} B A=\angle A C P=\angle B A P$, so $B C^{\prime} \| A P$. Similarly, $C B^{\prime} \| A P$, so $B C B^{\prime} C^{\prime}$ is an isosceles trapezoid. In particular, this means $B^{\prime} P=C P$, so $B P+P C=B B^{\prime}$. Now observe that $\angle A B P=\angle C A P=$ $\angle B A M$, so if $A M$ meets the circumcircle of $\triangle A B C$ again at $A^{\prime}$, then $A A^{\prime}=B B^{\prime}$. Thus the perimeter of $\triangle B P C$ is $B P+P C+B C=B B^{\prime}+B C=A A^{\prime}+B C$.
Now we compute. We have
$$
B C=2 \sqrt{A O^{2}-O M^{2}}=2 \sqrt{81 \cdot 25}=90
$$
and Power of a Point gives
$$
M A^{\prime}=\frac{B M^{2}}{A M}=\frac{45^{2}}{75}=27 .
$$

Thus $A A^{\prime}+B C=75+27+90=192$.
32. [18] Let acute triangle $A B C$ have circumcenter $O$, and let $M$ be the midpoint of $B C$. Let $P$ be the unique point such that $\angle B A P=\angle C A M, \angle C A P=\angle B A M$, and $\angle A P O=90^{\circ}$. If $A O=53, O M=28$, and $A M=75$, compute the perimeter of $\triangle B P C$.