Problem:
Circles Ο1β,Ο2β, and Ο3β each have radius 4 and are placed in the plane so that each circle is externally tangent to the other two. Points P1β, P2β, and P3β lie on Ο1β,Ο2β, and Ο3β, respectively, so that P1βP2β=P2βP3β=P3βP1β and line PiβPi+1β is tangent to Οiβ for each i=1,2,3, where P4β=P1β. See the figure below. The area of β³P1βP2βP3β can be written in the form aβ+bβ, 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 Oiβ be the center of circle Οiβ for i=1,2,3, and let K be the intersection of lines O1βP1β and O2βP2β. Because β P1βP2βP3β=60β, it follows that β³P2βKP1β is a 30β60β90β triangle. Let d=P1βK; then P2βK=2d and P1βP2β=3βd. The Law of Cosines in β³O1βKO2β gives
82=(d+4)2+(2dβ4)2β2(d+4)(2dβ4)cos60β
which simplifies to 3d2β12dβ16=0. The positive solution is d=2+32β21β. Then P1βP2β=3βd=23β+27β, and the required area is
