32. [17] Suppose point $P$ is inside triangle $A B C$. Let $A P, B P$, and $C P$ intersect sides $B C, C A$, and $A B$ at points $D, E$, and $F$, respectively. Suppose $\angle A P B=\angle B P C=\angle C P A, P D=\frac{1}{4}, P E=\frac{1}{5}$, and $P F=\frac{1}{7}$. Compute $A P+B P+C P$.
Answer: $\frac{19}{12}$
Solution:

The key is the following lemma:
Lemma: If $\angle X=120^{\circ}$ in $\triangle X Y Z$, and the bisector of $X$ intersects $Y Z$ at $T$, then
$$
\frac{1}{X Y}+\frac{1}{X Z}=\frac{1}{X T}
$$

Proof of the Lemma. Construct point $W$ on $X Y$ such that $\triangle X W T$ is equilateral. We also have $T W \|$ $X Z$. Thus, by similar triangles,
$$
\frac{X T}{X Z}=\frac{Y T}{Y X}=1-\frac{X T}{X Y}
$$
implying the conclusion.
Now we can write
$$
\begin{array}{l}
\frac{1}{P B}+\frac{1}{P C}=4 \\
\frac{1}{P C}+\frac{1}{P A}=5, \text { and } \\
\frac{1}{P A}+\frac{1}{P B}=7 .
\end{array}
$$

From here we can solve to obtain $\frac{1}{P A}=4, \frac{1}{P B}=3, \frac{1}{P C}=1$, making the answer $\frac{19}{12}$.