A geometry problem by Kunal Maan

By homothety, we deduce that $AE = 4 AD$. (The proof can also be executed by similar triangles formed from dropping perpendiculars from the centers of $P$ and $Q$ to $l$.) Therefore, our equality of area condition, or the equality of base times height condition, reduces to the fact that the distance from $B$ to $l$ is four times that from $C$ to $l$. Let the distance from $C$ be $x$ and the distance from $B$ be $4x$. Let $P$ and $Q$ be the centers of their respective circles. Then dropping a perpendicular from $P$ to $Q$ creates a $3-4-5$ right triangle, from which $BC = 4$ and, if $\alpha = \angle{AQC}$, that $\cos \alpha = \dfrac{3}{5}$. Then $\angle{BPA} = 180^\circ - \alpha$, and the Law of Cosines on triangles $APB$ and $AQC$ gives $AB = \dfrac{4}{\sqrt{5}}$ and $AC = \dfrac{8}{\sqrt{5}}.$ Now, using the Pythagorean Theorem to express the length of the projection of $BC$ onto line $l$ gives $\sqrt{\frac{16}{5} - 16x^2} + \sqrt{\frac{64}{5} - x^2} = \sqrt{16 - 9x^2}.$ Squaring and simplifying gives $\sqrt{\left(\frac{1}{5} - x^2\right)\left(\frac{64}{5} - x^2\right)} = x^2,$ and squaring and solving gives $x = \dfrac{8}{5\sqrt{13}}.$ By the Law of Sines on triangle $ABD$, we have $\frac{BD}{\sin A} = 2.$ But we know $\sin A = \dfrac{4x}{AB}$, and so a small computation gives $BD = \dfrac{16}{\sqrt{65}}.$ The Pythagorean Theorem now gives $AD = \sqrt{BD^2 - (4x)^2} + \sqrt{AB^2 - (4x)^2} = \frac{4}{\sqrt{13}},$ and so the common area is $\dfrac{1}{2} \cdot \frac{4}{\sqrt{13}} \cdot \frac{32}{5\sqrt{13}} = \frac{64}{65}.$ The answer is $\boxed{129}.$