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

Revision as of 14:26, 24 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 $P_1P_2.$ Since $\triangle P_2O_2M$ is a 30-60-90 triangle, $O_2M = 2$ and $P_2M = 2\sqrt{3}$ . Also, 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 $\left[P_1P_2P_3\right] = \frac{\sqrt 3}4\cdot\left(2\sqrt 3 + 2\sqrt 7\right)^2 = 10\sqrt 3 + 6\sqrt 7 = \sqrt{300} + \sqrt{252}.$ So, our answer is $252 + 300 = \boxed{\textbf{D) }552}$ .

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

Solution 5 (Pure Coordinate Bash)

Let's forget the lengths of the radius for a moment and focus instead on the ratio of the circles' radius to $\triangle P_1P_2P_3$ 's side length.

WLOG let $P_1P_2 = 2$ . Place triangle $P_1P_2P_3$ on the coordinate plane with $P_1(0,\sqrt3)$ , $P_2(1,0)$ , and $P_3(-1,0)$ . Let $r$ be the radius of the circles.

Now, we find the coordinates of the centers of circles $\omega_1$ and $\omega_2$ . Since $\omega_2$ is tangent to the x-axis at $(1,0)$ , the center of $\omega_2$ is $(1,r)$ . Draw a right triangle with legs parallel to the x and y axes, and with hypotenuse as the segment from the center of $\omega_1$ to $P_1$ . Since the slope of $\overline{P_1P_2}$ is $-\sqrt3$ , the slope of the hypotenuse is $\frac{1}{\sqrt{3}}$ , so the right triangle is $30-60-90$ . It's easy to see that the center of $\omega_1$ is $\left(-\frac{r\sqrt{3}}{2}, \sqrt{3} - \frac{r}{2}\right)$ .

Since $\omega_1$ and $\omega_2$ are tangent, the distance between the centers is $2r$ , so we have $\sqrt{\left(1 + \frac{r\sqrt{3}}{2}\right)^2 + \left(\frac{3r}{2} - \sqrt{3}\right)^2} = 2r$ $\left(1 + \frac{r\sqrt{3}}{2}\right)^2 + \left(\frac{3r}{2} - \sqrt{3}\right)^2 = 4r^2$ $1 + r\sqrt{3} + \frac{3r^2}{4} + \frac{9r^2}{4} - 3r\sqrt{3} + 3 = 4r^2$ $4 - 2r\sqrt{3} + 3r^2 = 4r^2$ $r^2 + 2r\sqrt{3} - 4 = 0$ By the Quadratic Formula, $r = \frac{-2\sqrt{3} \pm \sqrt{12 + 16}}{2} = -\sqrt{3}\pm\sqrt{7}$ . We take the positive value to get the radius of the circle is $\sqrt{7} - \sqrt{3}$ .

Therefore, the ratio of the radius to the side length of the equilateral triangle is $\sqrt{7} - \sqrt{3} : 2 = 4 : \frac{8}{\sqrt{7} - \sqrt{3}} = 4 : 2(\sqrt{7} + \sqrt{3})$ .

The side length of the equilateral triangle is $2(\sqrt{7} + \sqrt{3})$ , so its area is $(\sqrt{7} + \sqrt{3})^2\sqrt{3} = 10\sqrt{3} + 6\sqrt{7} = \sqrt{300} + \sqrt{252} \implies 300 + 252 = \boxed{\textbf{(D) }552}$ .

@@ Line 88: / Line 88: @@
 ==Solution 5 (Pure Coordinate Bash)==
-Let's forget the lengths of the radius for a moment and focus instead on the ratio of the circle's radius to <math>\triangle P_1P_2P_3</math>'s side length.
+Let's forget the lengths of the radius for a moment and focus instead on the ratio of the circles' radius to <math>\triangle P_1P_2P_3</math>'s side length.
 WLOG let <math>P_1P_2 = 2</math>. Place triangle <math>P_1P_2P_3</math> on the coordinate plane with <math>P_1(0,\sqrt3)</math>, <math>P_2(1,0)</math>, and <math>P_3(-1,0)</math>. Let <math>r</math> be the radius of the circles.

2018 AMC 12B (Problems • Answer Key • Resources)
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