Problem:
Circles C1,C2, and C3 have their centers at (0,0),(12,0), and (24,0), and have radii 1,2, and 4, respectively. Line t1 is a common internal tangent to C1 and C2 and has a positive slope, and line t2 is a common internal tangent to C2 and C3 and has a negative slope. Given that lines t1 and t2 intersect at (x,y), and that x=p−qr, where p,q, and r are positive integers and r is not divisible by the square of any prime, find p+q+r.
Solution:
Let O1,O2, and O3 be the centers of C1,C2, and C3, respectively, let A and B be the points where t1 is tangent to C1 and C2, respectively, and let D and E be the points where t2 is tangent to C2 and C3, respectively. Radii O1A and O2B are perpendicular to line AB. Let P be the intersection of AB and O1O2. Then △O1AP∼△O2BP with similarity ratio 1:2. Therefore O1P=4 and O2P=8, so PB=82−22=215. The slope of line t1 is equal to tan∠BPO2=1/15, so line t1 has equation y=(1/15)(x−4). Similarly, let Q be the intersection of DE and O2O3, and conclude that O2Q=4 and O3Q=8, and then that DQ=42−22=23. The slope of line t2 is equal to tan∠DQO3=−tan∠DQO2=−1/3, so line t2 has equation y=(−1/3)(x−16). Now (1/15)(x−4)=(−1/3)(x−16) implies x−4=−5(x−16), so x=5+1165+4=19−35, and p+q+r=27.