Let $ABCD$ and $AEFG$ be unit squares such that the area of their intersection is $\frac{20}{21}$. Given that $\angle BAE < 45^\circ$, $\tan \angle BAE$ can be expressed as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$. Compute $100a + b$.

Let's think step by step and output the final answer within \boxed{}.
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