[Geometry]
Problem: Square $K E N T$ has side length 20 . Point $M$ lies in the interior of $K E N T$ such that $\triangle M E N$ is equilateral. Given that $K M^{2}=a-b \sqrt{3}$, where $a$ and $b$ are integers, compute $b$.

Solution: Let $s$ be the side length of square $K E N T$; then $M E=s$. Let $J$ be the foot of the altitude from $M$ to $\overline{K E}$. Then $\mathrm{m} \angle J E M=30^{\circ}$ and $\mathrm{m} \angle E M J=60^{\circ}$. Hence $M J=\frac{s}{2}, J E=\frac{s \sqrt{3}}{2}$, and $K J=K E-J E=s-\frac{s \sqrt{3}}{2}$. Applying the Pythagorean Theorem to $\triangle K J M$ implies that $K M^{2}=\left(s-\frac{s \sqrt{3}}{2}\right)^{2}+\left(\frac{s}{2}\right)^{2}=2 s^{2}-s^{2} \sqrt{3}$. With $s=20$, the value of $b$ is therefore $s^{2}=\mathbf{4 0 0}$.

Final answer: 400