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: We begin by extending $\overline{AB}$ upwards until it intersects Circle $\mathcal{Q}$ . We can call this point of intersection $F$ . Connect $F$ with $E$ $C$ , and $A$ for future use.
Create a trapezoid with points $B$ $C$ , and the origins of Circles $\mathcal{P}$ and $\mathcal{Q}$ . After quick inspection, we can conclude that the distance between the origins is 5 and that $\overline{BC}$ is 4.
(Note: It is important to understand that we could simply use the formula for the distance between the tangent points on a line and two different circles, which is $2\cdot \sqrt{a\cdot b}$ , where $a$ and $b$ are the two respective radii of the circles. In our case, we get $2\cdot \sqrt{4} = 4$
Using similar triangles or homotheties, $AE=4\cdot AD$ and $AF=4\cdot BA$ $BC^2 = BA\cdot BF$ \[16 = BA\cdot (5\cdot BA)\] \[BA = \frac{4}{\sqrt{5}}\] \[AF = \frac{16}{\sqrt{5}}\] \[BF = 4\cdot \sqrt{5}\]
Inspecting $\triangle{BFC}$ , we recognize that it is a right triangle ( $\angle{BCF} = 90$ ) as the final length ( $\overline{FC}$ ) being 8 would allow for an $x-2x-x\sqrt{5}$ triangle. Hence, the diameter of circle $\mathcal{Q}$ $CF$ . This also means that $\angle{CAF} = \angle{CEF} = 90$
From the fact that $\triangle{ABC}$ is a right triangle: \[AB^2 + AC^2 = BC^2\] \[AC = \frac{8}{\sqrt{5}}\] (Note: We could have also  used $\triangle{FAC}$ .)
Our next step is to start angle chasing to find any other similar triangles or shared angles. Label the intersection between $\overline{FC}$ and $\overline{DE}$ as $M$ . Label $\angle{CAE} = \angle{CFE} = \theta$ . Since $\angle{BAC} = 90$ $\angle{EAF} = \angle{DAB} = 90-\theta$ . Now, use the sine formula and the fact that the areas of $\triangle{DBA}$ and $\triangle{ACE}$ are equal to get: \[\frac{1}{2}\cdot AC\cdot AE\cdot \sin{\theta} = \frac{1}{2}\cdot AD\cdot AB\cdot \sin{(90-\theta)}\] Since $\sin{(90-\theta)} = \cos{\theta}$ \[\tan{\theta} = \frac{AD}{2\cdot AE}\]
Using right $\triangle{FCE}$ , since $\angle{ACM} = \angle{CFE}$ $\tan{\theta} = \frac{CE}{FE}$ . Hence, plugging into the previous equation: \[\frac{CE}{FE} = \frac{AD}{2\cdot AE}\] Using the Pythagorean theorem on $\triangle{FCE}$ $FE = \sqrt{(64 - CE^2)}$ . We also know that $AE=4\cdot AD$ Plugging back in: \[\frac{CE}{\sqrt{(64 - CE^2)}} = \frac{AD}{2\cdot (4\cdot AD)}\] \[\frac{CE}{\sqrt{(64 - CE^2)}} = \frac{1}{8}\] .
From here, we can square both sides and bring everything to one side to get: \[EF^2 + 64EF - 64 = 0\] \[EF = \frac{64}{\sqrt{65}}\] \[CE = \frac{8}{\sqrt{65}}\] We should also return to the fact that $\sin{\theta} = \frac{CE}{CF}$ from $\triangle{FCE}$ , so \[\sin{\theta} = \frac{1}{\sqrt{65}}\]
From the fact that $\angle{CAF} = \angle{CEF} = 90$ , we can use Ptolemy's Theorem on quadrilateral $ACEF$ $AC\cdot EF + CE\cdot FA = CF\cdot AE$ . Plugging in and solving, we get that $AE = \frac{16}{\sqrt{13}}$
We now have all of our pieces to use the Sine Formula on $\triangle{ACE}$ \[\frac{1}{2}\cdot AC\cdot AF\cdot \sin{\theta}\] \[\frac{1}{2}\cdot \frac{8}{\sqrt{5}}\cdot \frac{16}{\sqrt{13}}\cdot \frac{1}{\sqrt{65}} = \frac{64}{65} = \boxed{129}\]

Answer:  129