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2018 AMC 12B Problem 25

Problem 25 of 25HarderGeometry

Circles ω1,\omega_1, ω2,\omega_2, and ω3\omega_3 each have radius 44 and are placed in the plane so that each circle is externally tangent to the other two. Points P1,P_1, P2,P_2, and P3P_3 lie on ω1,\omega_1, ω2,\omega_2, and ω3,\omega_3, respectively, so that P1P2=P2P3=P3P1P_1P_2=P_2P_3=P_3P_1 and line PiPi+1P_iP_{i+1} is tangent to ωi\omega_i for each i=1,i=1, 2,2, 3,3, where P4=P1.P_4=P_1. See the figure below. The area of P1P2P3\triangle P_1P_2P_3 can be written in the form a+b,\sqrt{a}+\sqrt{b}, where aa and bb are positive integers. What is a+b?a+b?

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Solution

Let OiO_i be the center of ωi,\omega_i, and let KK be the intersection of lines O1P1O_1P_1 and O2P2.O_2P_2. Because P1P2P3=60,\angle P_1P_2P_3=60^\circ, triangle P2KP1P_2KP_1 is a 3030-6060-9090^\circ triangle. With d=P1K,d=P_1K, we get P2K=2dP_2K=2d and P1P2=3d.P_1P_2=\sqrt3\,d. The Law of Cosines in O1KO2\triangle O_1KO_2 (with O1O2=8O_1O_2=8) gives 82=(d+4)2+(2d4)22(d+4)(2d4)cos60, \begin{gathered} 8^2=(d+4)^2+(2d-4)^2 \\ {}-2(d+4)(2d-4)\cos60^\circ, \end{gathered} which simplifies to 3d212d16=0,3d^2-12d-16=0, so d=2+2321.d=2+\tfrac23\sqrt{21}. Then P1P2=3d=23+27,P_1P_2=\sqrt3\,d=2\sqrt3+2\sqrt7, and the area is 34(23+27)2=103+67=300+252. \begin{gathered} \dfrac{\sqrt3}{4}\left(2\sqrt3+2\sqrt7\right)^2 \\ =10\sqrt3+6\sqrt7 \\ =\sqrt{300}+\sqrt{252}. \end{gathered} So a+b=300+252=552.a+b=300+252=552. Thus, the correct answer is D.

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Concepts: tangent circles · special right triangle · law of cosines · equilateral triangle

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.