Problem (Geometry):
Two square tiles of area 9 are placed with one directly on top of the other. The top tile is then rotated about its center by an acute angle $\theta$. If the area of the overlapping region is 8 , compute $\sin \theta+\cos \theta$.

Solution:
In the diagram below, $O$ is the center of both squares $A_{1} A_{2} A_{3} A_{4}$ and $B_{1} B_{2} B_{3} B_{4}$. Let $P_{1}, P_{2}, P_{3}, P_{4}$ and $Q_{1}, Q_{2}, Q_{3}, Q_{4}$ be the intersections of the sides of the squares as shown. Let $H_{A}$ be on $\overline{A_{3} A_{4}}$ so that $\angle A_{3} H_{A} O$ is right. Similarly, let $H_{B}$ be on $\overline{B_{3} B_{4}}$ such that $\angle B_{3} H_{B} O$ is right. Then the angle by which $B_{1} B_{2} B_{3} B_{4}$ was rotated is $\angle H_{A} O H_{B}$. Extend $\overline{O H_{B}}$ to meet $\overline{A_{3} A_{4}}$ at $M$.



Both $\triangle H_{A} O M$ and $\triangle H_{B} P_{3} M$ are right triangles sharing acute $\angle M$, so $\triangle H_{A} O M \sim \triangle H_{B} P_{3} M$. By an analogous argument, both triangles are similar to $\triangle B_{3} P_{3} Q_{3}$. Thus $\mathrm{m} \angle Q_{3} P_{3} B_{3}=\theta$. Now let $B_{3} P_{3}=x, B_{3} Q_{3}=y$, and $P_{3} Q_{3}=z$. By symmetry, notice that $B_{3} P_{3}=B_{2} P_{2}$ and that $P_{3} Q_{3}=P_{2} Q_{3}$. Thus

$$
x+y+z=B_{3} Q_{3}+Q_{3} P_{2}+P_{2} B_{2}=B_{2} B_{3}=3 .
$$

By the Pythagorean Theorem, $x^{2}+y^{2}=z^{2}$. Therefore

$$
\begin{aligned}
x+y & =3-z \\
x^{2}+y^{2}+2 x y & =9-6 z+z^{2} \\
2 x y & =9-6 z .
\end{aligned}
$$

The value of $x y$ can be determined from the areas of the four triangles $\triangle B_{i} P_{i} Q_{i}$. By symmetry, these four triangles are congruent to each other. Their total area is the area not in both squares, i.e., $9-8=1$. Thus $\frac{x y}{2}=\frac{1}{4}$, so $2 x y=1$. Applying this result to the above equation,

$$
\begin{aligned}
1 & =9-6 z \\
z & =\frac{4}{3}
\end{aligned}
$$



The desired quantity is $\sin \theta+\cos \theta=\frac{x}{z}+\frac{y}{z}$, and

$$
\begin{aligned}
\frac{x}{z}+\frac{y}{z} & =\frac{x+y+z}{z}-\frac{z}{z} \\
& =\frac{3}{z}-1 \\
& =\frac{\mathbf{5}}{\mathbf{4}}
\end{aligned}
$$

Final answer: $\frac{5}{4}$