Problem:
Circles ω1,ω2, and ω3 each have radius 4 and are placed in the plane so that each circle is externally tangent to the other two. Points P1, P2, and P3 lie on ω1,ω2, and ω3, respectively, so that P1P2=P2P3=P3P1 and line PiPi+1 is tangent to ωi for each i=1,2,3, where P4=P1. See the figure below. The area of △P1P2P3 can be written in the form a+b, where a and b are positive integers. What is a+b ?
Answer Choices:
A. 546
B. 548
C. 550
D. 552
E. 554 Solution:
Let Oi be the center of circle ωi for i=1,2,3, and let K be the intersection of lines O1P1 and O2P2. Because ∠P1P2P3=60∘, it follows that △P2KP1 is a 30−60−90∘ triangle. Let d=P1K; then P2K=2d and P1P2=3d. The Law of Cosines in △O1KO2 gives
82=(d+4)2+(2d−4)2−2(d+4)(2d−4)cos60∘
which simplifies to 3d2−12d−16=0. The positive solution is d=2+3221. Then P1P2=3d=23+27, and the required area is
