# 2018 AMC 12B Problems/Problem 25

## Problem

Circles $\omega_1$, $\omega_2$, and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$, $P_2$, and $P_3$ lie on $\omega_1$, $\omega_2$, and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i=1,2,3$, where $P_4 = P_1$. See the figure below. The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a}+\sqrt{b}$ for positive integers $a$ and $b$. What is $a+b$?

$[asy] unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label("P_1", P1, E*1.5); label("P_2", P2, SW*1.5); label("P_3", P3, N); label("\omega_1", A, W*17); label("\omega_2", B, E*17); label("\omega_3", C, W*17); [/asy]$

$\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554$

## Solution 1

Let $O_i$ be the center of circle $\omega_i$ for $i=1,2,3$, and let $K$ be the intersection of lines $O_1P_1$ and $O_2P_2$. Because $\angle P_1P_2P_3 = 60^\circ$, it follows that $\triangle P_2KP_1$ is a $30-60-90^\circ$ triangle. Let $d=P_1K$; then $P_2K = 2d$ and $P_1P_2 = \sqrt 3d$. The Law of Cosines in $\triangle O_1KO_2$ gives $$8^2 = (d+4)^2 + (2d-4)^2 - 2(d+4)(2d-4)\cos 60^\circ,$$which simplifies to $3d^2 - 12d - 16 = 0$. The positive solution is $d = 2 + \tfrac23\sqrt{21}$. Then $P_1P_2 = \sqrt 3 d = 2\sqrt 3 + 2\sqrt 7$, and the required area is $$\frac{\sqrt 3}4\cdot\left(2\sqrt 3 + 2\sqrt 7\right)^2 = 10\sqrt 3 + 6\sqrt 7 = \sqrt{300} + \sqrt{252}.$$The requested sum is $300 + 252 = \boxed{552}$.

## Solution 2

Let $O_1$ and $O_2$ be the centers of $\omega_1$ and $\omega_2$ respectively and draw $O_1O_2$, $O_1P_1$, and $O_2P_2$. Note than $\angle{OP_1P_2}$ and $\angle{O_2P_2P_3}$ are both right. Furthermore, since $\triangle{P_1P_2P_3}$ is equilateral, $m\angle{P_1P_2P_3} = 60^\circ$ and $m\angle{O_2P_2P_1} = 30^\circ$. Mark $M$ as the base of the altitude from $O_2$ to $P_1P_2$. By special right triangles, $O_2M = 2$ and $P_2M = 2\sqrt{3}$. since $O_1O_2 = 8$ and $O_1P_1 = 4$, we can find find $P_1M = \sqrt{8^2 - (4 + 2)^2} = 2\sqrt{7}$. Thus, $P_1P_2 = P_1M + P_2M = 2\sqrt{3} + 2\sqrt{7}$. This makes $\left[P_1P_2P_3\right] = \frac{\left(2\sqrt{3} + 2\sqrt{7}\right)^2\sqrt{3}}{4} = 10\sqrt{3} + 6\sqrt{7} = \sqrt{252} + \sqrt{300}$. This makes the answer $252 + 300 = 552$. $\boxed{\textbf{D}.}$

## Solution 3

Let $O_i$ be the center of circle $\omega_i$ for $i=1,2,3$. Let $X$ be the centroid of $\triangle{O_1O_2O_3}$, which also happens to be the centroid of $\triangle{P_1P_2P_3}$. Because $m\angle{O_1P_1P_2} = 90^\circ$ and $m\angle{O_1P_1X} = 30^\circ$, $m\angle{O_1P_1X} = 60^\circ$. $O_2M$ is $2/3$ the height of $\triangle{P_1P_2P_3}$, thus $O_2M$ is $8*\sqrt{3}/3$.

Applying cosine law on $\triangle{O_1P_1X}$, one finds that $P_1X = 2 + 2*\sqrt{21}/3$. Multiplying by $3/2$ to solve for the height of $\triangle{P_1P_2P_3}$, one gets $3 + \sqrt{21}$. Simply multiplying by $2/\sqrt{3}$ and then calculating the equilateral triangle's area, one would get the final result of $\sqrt{300} + \sqrt{252}$.

This makes the answer $252 + 300 = 552$. $\boxed{\textbf{D}.}$

~AlbeePach~

## See Also

 2018 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 24 Followed byLast Problem 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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