Problem:
Circles P and Q have radii 1 and 4, respectively, and are externally tangent at point A. Point B is on P and point C is on Q so that line BC is a common external tangent of the two circles. A line ℓ through A intersects P again at D and intersects Q again at E. Points B and C lie on the same side of ℓ, and the areas of △DBA and △ACE are equal. This common area is nm​, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let P and Q be the centers of the circles P and Q, respectively. Let F be on CQ​ so that CBPF is a rectangle. Note that in right △PFQ,PQ=1+4=5 and QF=4−1=3, so BC=PF=4.
Let G and H be on ℓ so that BG and CH are altitudes of △ABD and △ACE, respectively, as shown, and let ℓ intersect line BC at I. Because the sectors of the two circles cut off by ℓ are similar with a 1:4 ratio, it follows that AE=4AD. Because △ABD and △ACE have the same areas, it follows that BG=4CH. Because △IGB is similar to △IHC, it follows that 4IC=IB=IC+4 and IC=34​.
Calculate AI by letting J be the projection of A onto line BC. Because PA=51​PQ, it follows that AJ=54​PB+51​QC=58​ and BJ=51​BC=54​. Then
AI=AJ2+IJ2​=(58​)2+(1568​)2​=34​13​
Now calculate the area of △DBA by finding BG and AD. For the former, by similarity △BGI∼△AJI, it follows that BIBG​=AIAJ​, giving BG=6532​13​. For the latter, the Power of a Point Theorem gives IA⋅ID=IB2, so ID=3964​13​ and AD=ID−IA=134​13​. So the area of △DBA is
