Problem: Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] import cse5; pathpen=black; pointpen=black; size(6cm);  pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689);  filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy]

Solution: Let $M$ be the intersection of $\overline{BC}$ and the common internal tangent of $\mathcal P$ and $\mathcal Q.$ We claim that $M$ is the circumcenter of right $\triangle{ABC}.$ Indeed, we have $AM = BM$ and $BM = CM$ by equal tangents to circles, and since $BM = CM, M$ is the midpoint of $\overline{BC},$ implying that $\angle{BAC} = 90.$ Now draw $\overline{PA}, \overline{PB}, \overline{PM},$ where $P$ is the center of circle $\mathcal P.$ Quadrilateral $PAMB$ is cyclic, and by Pythagorean Theorem $PM = \sqrt{5},$ so by Ptolemy on $PAMB$ we have \[AB \sqrt{5} = 2 \cdot 1 + 2 \cdot 1 = 4 \iff AB = \dfrac{4 \sqrt{5}}{5}.\] Do the same thing on cyclic quadrilateral $QAMC$ (where $Q$ is the center of circle $\mathcal Q$ and get $AC = \frac{8 \sqrt{5}}{5}.$
Let $\angle A = \angle{DAB}.$ By Law of Sines, $BD = 2R \sin A = 2 \sin A.$ Note that $\angle{D} = \angle{ABC}$ from inscribed angles, so \begin{align*} [ABD] &= \dfrac{1}{2} BD \cdot AB \cdot \sin{\angle B} \\ &= \dfrac{1}{2} \cdot \dfrac{4 \sqrt{5}}{5} \cdot 2 \sin A \sin{\left(180 - \angle A - \angle D\right)} \\ &= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \sin{\left(\angle A + \angle D\right)} \\ &= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \left(\sin A \cos D + \cos A \sin D\right) \\ &= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \left(\sin A \cos{\angle{ABC}} + \cos A \sin{\angle{ABC}}\right) \\ &= \dfrac{4 \sqrt{5}}{5} \cdot \sin A \cdot \left(\dfrac{\sqrt{5} \sin A}{5} + \dfrac{2 \sqrt{5} \cos A}{5}\right) \\ &= \dfrac{4}{5} \cdot \sin A \left(\sin A + 2 \cos A\right) \end{align*} after angle addition identity.
Similarly, $\angle{EAC} = 90 - \angle A,$ and by Law of Sines $CE = 8 \sin{\angle{EAC}} = 8 \cos A.$ Note that $\angle{E} = \angle{ACB}$ from inscribed angles, so \begin{align*} [ACE] &= \dfrac{1}{2} AC \cdot CE \sin{\angle C} \\ &= \dfrac{1}{2} \cdot \dfrac{8 \sqrt{5}}{5} \cdot 8 \cos A \sin{\left[180 - \left(90 - \angle A\right) - \angle E\right]} \\ &= \dfrac{32 \sqrt{5}}{5} \cdot \cos A \sin{\left[\left(90 - \angle A\right) + \angle{ACB}\right]} \\ &= \dfrac{32 \sqrt{5}}{5} \cdot \cos A \left(\dfrac{2 \sqrt{5} \cos A}{5} + \dfrac{\sqrt{5} \sin A}{5}\right) \\ &= \dfrac{32}{5} \cdot \cos A \left(\sin A + 2 \cos A\right) \end{align*} after angle addition identity.
Setting the two areas equal, we get \[\tan A = \frac{\sin A}{\cos A} = 8 \iff \sin A = \frac{8}{\sqrt{65}}, \cos A = \frac{1}{\sqrt{65}}\] after Pythagorean Identity. Now plug back in and the common area is $\frac{64}{65} \iff \boxed{129}.$

Answer:  129