Problem:
Circles Ο and Ξ³, both centered at O, have radii 20 and 17 , respectively. Equilateral triangle ABC, whose interior lies in the interior of Ο but in the exterior of Ξ³, has vertex A on Ο, and the line containing side BC is tangent to Ξ³. Segments AO and BC intersect at P, and CPBPβ=3. Then AB can be written in the form nβmββqβpβ 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 Ξ³, and M be the midpoint of BC. Note that AMβ₯BS and OSβ₯BS. This implies that β OAMβ β AOS, and β AMPβ β OSP. Thus, β³PMAβΌβ³PSO.
If we let s be the side length of β³ABC, then it follows that AM=23ββs and PM=4sβ. This implies that AP=413ββs, so APAMβ=13β23ββ. Furthermore, AOAM+SOβ=APAMβ (because β³PMAβΌβ³PSO ) so this gives us the equation
