Problem (Combinatorics):
Suppose that, for some angles $x$ and $y$,

$$
\begin{aligned}
& \sin ^{2} x+\cos ^{2} y=\frac{3}{2} a \\
& \cos ^{2} x+\sin ^{2} y=\frac{1}{2} a^{2}
\end{aligned}
$$

Determine the possible value(s) of $a$.

Solution:
Adding the two equations, we obtain

$$
\begin{aligned}
\sin ^{2} x+\cos ^{2} x+\sin ^{2} y+\cos ^{2} y & =\frac{3}{2} a+\frac{1}{2} a^{2} \\
2 & =\frac{3}{2} a+\frac{1}{2} a^{2} \\
4 & =3 a+a^{2} \\
0 & =a^{2}+3 a-4 \\
0 & =(a+4)(a-1)
\end{aligned}
$$

and so $a=-4$ or $a=1$.

However, $a=-4$ is impossible, since this would give $\sin ^{2} x+\cos ^{2} y=-6$, whose left side is non-negative and whose right side is negative.

Therefore, the only possible value for $a$ is $a=1$.

(We can check that angles $x=90^{\circ}$ and $y=45^{\circ}$ give $\sin ^{2} x+\cos ^{2} y=\frac{3}{2}$ and $\cos ^{2} x+\sin ^{2} y=$ $\frac{1}{2}$, so $a=1$ is indeed possible.)

Final answer: 1