Problem (Geometry):
Some portions of the line $y=4 x$ lie below the curve $y=10 \pi \sin ^{2} x$, and other portions lie above the curve. Compute the sum of the lengths of all the segments of the graph of $y=4 x$ that lie in the first quadrant, below the graph of $y=10 \pi \sin ^{2} x$.

Solution:
Notice first that all intersections of the two graphs occur in the interval $0 \leq x \leq \frac{5 \pi}{2}$, because the maximum value of $10 \pi \sin ^{2} x$ is $10 \pi$ (at odd multiples of $\frac{\pi}{2}$ ), and $4 x>10 \pi$ when $x>\frac{5 \pi}{2}$. The graphs are shown below.



Within that interval, both graphs are symmetric about the point $A=\left(\frac{5 \pi}{4}, 5 \pi\right)$. For the case of $y=10 \pi \sin ^{2} x$, this symmetry can be seen by using the power-reducing identity $\sin ^{2} x=$ $\frac{1-\cos 2 x}{2}$. Then the equation becomes $y=5 \pi-5 \pi \cos 2 x$, which has amplitude $5 \pi$ about the line $y=5 \pi$, and which crosses the line $y=5 \pi$ for $x=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \ldots$ Label the points of intersection $A, B, C, D, E, F$, and $O$ as shown. Then $\overline{A B} \cong \overline{A C}, \overline{B D} \cong \overline{C E}$, and $\overline{O D} \cong \overline{E F}$. Thus

$$
\begin{aligned}
B D+A C+E F & =O D+D B+B A \\
& =O A .
\end{aligned}
$$

By the Pythagorean Theorem,

$$
\begin{aligned}
O A & =\sqrt{\left(\frac{5 \pi}{4}\right)^{2}+(5 \pi)^{2}} \\
& =\frac{5 \pi}{4} \sqrt{1^{2}+4^{2}} \\
& =\frac{5 \pi}{\mathbf{4}} \sqrt{\mathbf{1 7}}
\end{aligned}
$$

Final answer: $\frac{5 \pi}{4} \sqrt{17}$