Let $A B C D$ and $A E F G$ be unit squares such that the area of their intersection is $\frac{20}{21}$. Given that $\angle B A E<45^{\circ}, \tan \angle B A E$ can be expressed as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$. Compute $100 a+b$.
Solution: Suppose the two squares intersect at a point $X \neq A$. If $\mathcal{S}$ is the region formed by the intersection of the squares, note that line $A X$ splits $\mathcal{S}$ into two congruent pieces of area $\frac{10}{21}$. Each of these pieces is a right triangle with one leg of length 1 , so the other leg must have length $\frac{20}{21}$. Thus, if the two squares are displaced by an angle of $\theta$, then $90-\theta=2 \arctan \frac{20}{21}$. Though there is some ambiguity in how the points are labeled, the fact that $\angle B A F<45^{\circ}$ tells us that $\angle B A F=\theta$. Therefore \[ \tan \angle B A F=\frac{1}{\tan \left(2 \arctan \frac{20}{21}\right)}=\frac{1-\frac{20^{2}}{21^{2}}}{2 \cdot \frac{20}{21}}=\frac{41}{840} \] $\fbox{4940}$.