28. [15] Let $A B C$ be a triangle with $A B=13, B C=14$, and $C A=15$. Pick points $Q$ and $R$ on $A C$ and $A B$ such that $\angle C B Q=\angle B C R=90^{\circ}$. There exist two points $P_{1} \neq P_{2}$ in the plane of $A B C$ such that $\triangle P_{1} Q R, \triangle P_{2} Q R$, and $\triangle A B C$ are similar (with vertices in order). Compute the sum of the distances from $P_{1}$ to $B C$ and $P_{2}$ to $B C$.
Answer:
48

Solution 1: Let $T$ be the foot of the $A$-altitude of $A B C$. Recall that $B T=5$ and $C T=9$.
Let $T^{\prime}$ be the foot of the $P$-altitude of $P Q R$. Since $T^{\prime}$ is the midpoint of the possibilities for $P$, the answer is
$$
\sum_{P} d(P, B C)=2 d\left(T^{\prime}, B C\right)
$$

Since $T^{\prime}$ splits $Q R$ in a $5: 9$ ratio, we have
$$
d\left(T^{\prime}, B C\right)=\frac{9 d(Q, B C)+5 d(R, B C)}{14} .
$$

By similar triangles, $d(Q, B C)=Q B=12 \cdot \frac{14}{9}$, and similar for $d(R, B C)$, giving $d\left(T^{\prime}, B C\right)=24$, and an answer of 48 .

Solution 2: As in the previous solution, let $T$ be the foot from $A$ to $B C$, let $T^{\prime}$ be the foot from $P$ to $Q R$. and recall that
$$
d\left(P_{1}, B C\right)+d\left(P_{2}, B C\right)=2 d\left(T^{\prime}, B C\right) .
$$

Now, notice that since $\triangle P Q R \sim \triangle A B C$, we have $Q T^{\prime}: T^{\prime} R=B T: T C$, so $T T^{\prime}\|B Q\| C R$, implying that $A \in T T^{\prime}$.

However, we recall a well-known fact that $A$ is the midpoint of $T T^{\prime}$ (can be proven by simple similar triangles). Thus, $d\left(T^{\prime}, B C\right)$ is equal to two times the altitude from $A$ to $B C$. Hence, the answer is four times the altitude from $A$ to $B C$, which is 48 .