Problem:
Circles ω and γ, both centered at O, have radii 20 and 17 , respectively. Equilateral triangle ABC, whose interior lies in the interior of ω but in the exterior of γ, has vertex A on ω, and the line containing side BC is tangent to γ. Segments AO and BC intersect at P, and CPBP=3. Then AB can be written in the form nm−qp for positive integers m,n,p,q with gcd(m,n)=gcd(p,q)=1. What is m+n+p+q?
Answer Choices:
A. 42
B. 86
C. 92
D. 114
E. 130 Solution:
Let S be the point of tangency between BC and γ, and M be the midpoint of BC. Note that AM⊥BS and OS⊥BS. This implies that ∠OAM≅∠AOS, and ∠AMP≅∠OSP. Thus, △PMA∼△PSO.
If we let s be the side length of △ABC, then it follows that AM=23s and PM=4s. This implies that AP=413s, so APAM=1323. Furthermore, AOAM+SO=APAM (because △PMA∼△PSO ) so this gives us the equation
