[Geometry]
Problem: Point $P$ is on the hypotenuse $\overline{E N}$ of right triangle $B E N$ such that $\overline{B P}$ bisects $\angle E B N$. Perpendiculars $\overline{P R}$ and $\overline{P S}$ are drawn to sides $\overline{B E}$ and $\overline{B N}$, respectively. If $E N=221$ and $P R=60$, compute $\frac{1}{B E}+\frac{1}{B N}$.

Solution: We observe that $\frac{1}{B E}+\frac{1}{B N}=\frac{B E+B N}{B E \cdot B N}$. The product in the denominator suggests that we compare areas. Let $[B E N]$ denote the area of $\triangle B E N$. Then $[B E N]=\frac{1}{2} B E \cdot B N$, but because $P R=P S=60$, we can also write $[B E N]=[B E P]+[B N P]=\frac{1}{2} \cdot 60 \cdot B E+\frac{1}{2} \cdot 60 \cdot B N$. Therefore $B E \cdot B N=60(B E+B N)$, so $\frac{1}{B E}+\frac{1}{B N}=\frac{B E+B N}{B E \cdot B N}=\frac{1}{\mathbf{6 0}}$. Note that this value does not depend on the length of the hypotenuse $\overline{E N}$; for a given location of point $P, \frac{1}{B E}+\frac{1}{B N}$ is invariant.
Using similar triangles, we have $\frac{E R}{P R}=\frac{P S}{S N}=\frac{B E}{B N}$, so $\frac{B E-60}{60}=$ $\frac{60}{B N-60}=\frac{B E}{B N}$ and $B E^{2}+B N^{2}=221^{2}$. Using algebra, we find that $B E=204, B N=85$, and $\frac{1}{204}+\frac{1}{85}=\frac{1}{60}$.

Final answer: $\frac{1}{60}$