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 $252 + 300 = 552$ . $\boxed{\textbf{D}.}$

2018 AMC 12B (Problems • Answer Key • Resources)
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2018 AMC 12B Problems/Problem 25

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also