Take a look at the third problem here:
Let $ABCD$ and $AEFG$ be unit squares such that the area of their intersection is $\frac{20}{21}$. Given that $\angle BAE<45^\circ$, $\tan\angle BAE=\frac ab$ for relatively prime positive integers $a$ and $b$. Compute $100a+b$
Is my solution correct? It is a bit lengthy and I am wondering if I messed up somewhere (I did get the right answer but just to be sure).
Let $\angle BAE=\theta$ for clarity. Let the second point of intersection of the two squares be $A'$. We know that $AA'E\cong ABA'$ by the HL theorem. Since the areas of these triangles must be $\frac{10}{21}$, $A'B$ and $A'E$ equal $\frac{20}{21}$. So we have that $$\tan\left(\frac{\theta}2\right)=\frac{20}{21}$$Since $\tan(\frac x2)=\sqrt{\frac{1-\cos x}{1+\cos x}}$, we have $$\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}=\frac{20}{21}$$Let $\cos\theta=a$. Then $$\frac{1-a}{1+a}=\frac{400}{441}\implies1-a=\frac{400}{441}+\frac{400}{441}a\implies\frac{41}{441}=\frac{841}{441}a\implies a=\frac{41}{841}$$By the Pythagorean identity, $b=\sin\theta=\frac{840}{841}$. So: $$\tan\theta=\frac{41}{840}$$The rest is trivial. Multiply the numerator by $100$ and add that to the denominator to get $4940$