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Difference between revisions of "2018 AMC 12B Problems/Problem 25"

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

Revision as of 18:01, 14 February 2021

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 $30-60-90$ triangle. Let $x=P_1K$; then $P_2K = 2x$ and $P_1P_2 = x \sqrt 3$. The Law of Cosines in $\triangle O_1KO_2$ gives \[8^2 = (x+4)^2 + (2x-4)^2 - 2(x+4)(2x-4)\cos 60^\circ,\]which simplifies to $3x^2 - 12x - 16 = 0$. The positive solution is $x = 2 + \tfrac23\sqrt{21}$. Then $P_1P_2 = x\sqrt 3 = 2\sqrt 3 + 2\sqrt 7$, and so the area of $\triangle P_1P_2P_3$ 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~

Solution 4

[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*30); label("$\omega_2$", B, E*17); label("$\omega_3$", C, W*17); label("$\Gamma$", A, W*15); draw(Circle(A,2),red); pair X=foot(A,P1,P3); dot(X,blue); draw(A--X,blue); label("$2\sqrt{7}$", X--P3); label("$2\sqrt{3}$",X--P1); label("$X$",X,dir(-80),blue); [/asy]

First, note that because the $\angle P_1=\angle P_2=\angle P_3=\pi/3$, the arcs inside the shaded equilateral triangle are each $2\pi/3$. Also, the distances between the centers of any two of the $3$ given circles are each $8$. Draw the circle $\Gamma$ concentric with $\omega_1$ with radius $2$. Because the arc of $\omega_1$ inside said triangle is $2\pi/3$, $\Gamma$ touches $P_1P_3$, say at a point $X$. Thus, $P_1P_3$ is a common tangent of $\omega_3$ and $\Gamma$, and it can be seen from inspection of the given diagram that the line is an common internal tangent. The length of the common internal tangent segment $XP_3$ of $\Gamma$ and $\omega_3$ is then $\sqrt{8^2-(2+4)^2}=2\sqrt{7}$, and it is easily seen that $XP_1=4\sin \pi/3=2\sqrt{3}$. Because $P_1P_3=2(\sqrt{3}+\sqrt{7})$, the area of the shaded equilateral triangle is $\sqrt{3}(\sqrt{3}+\sqrt{7})^2=10\sqrt{3}+6\sqrt{7}$. We get $\sqrt{300}+\sqrt{252}\Rightarrow\boxed{\textbf{(D) }552}.$


~crazyeyemoody907

See Also

2018 AMC 12B (Problems • Answer Key • Resources)
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Problem 24 		Followed by
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All AMC 12 Problems and Solutions

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