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Difference between revisions of "2024 AMC 12A Problems/Problem 22"

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#redirect[[2024 AMC 10A Problems/Problem 25]]
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==Problem==
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Three circles of radius <math>6</math> are mutually externally tangent, as shown below. For each pair of circles, construct the lines through their point of tangency that are tangent to the third circle. In total, this creates six new tangency points. If the area of the convex hexagon formed by these six points can be expressed as <math>a\sqrt{2} + b\sqrt{3}</math> for integers <math>a</math> and <math>b</math>, what is <math>a + b</math>?
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<asy>
 +
import olympiad;
 +
size(150);
 +
defaultpen(linewidth(0.6) + fontsize(10));
 +
pen dotstyle = black;
 +
real xmin = -8.903758953234952, xmax = 14.962621077942625, ymin = -5.903115312371685, ymax = 10.567409309904837; /* image dimensions */
 +
/* draw figures */
 +
draw(circle((0,0), 3), linewidth(1));
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draw(circle((6,0), 3), linewidth(1));
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draw(circle((3,5.196152422706632), 3), linewidth(1));
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draw((0.5505102572168217,3.4641016151377544)--(3,0), linewidth(1));
 +
draw((3,0)--(5.449489742783179,3.4641016151377544), linewidth(1));
 +
/* dots */
 +
dot((0.5505102572168217,3.4641016151377544),linewidth(4pt) + dotstyle);
 +
dot((5.449489742783179,3.4641016151377544),linewidth(4pt) + dotstyle);
 +
dot((5.724744871391589,2.987345747344081),linewidth(4pt) + dotstyle);
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dot((3.275255128608411,-1.2552949397752038),linewidth(4pt) + dotstyle);
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dot((0.27525512860841106,2.987345747344081),linewidth(4pt) + dotstyle);
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dot((2.724744871391589,-1.2552949397752038),linewidth(4pt) + dotstyle);
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
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</asy>
 +
 
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<math>\textbf{(A)}~18\qquad\textbf{(B)}~24\qquad\textbf{(C)}~36\qquad\textbf{(D)}~42\qquad\textbf{(E)}~45</math>
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==Solution==
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Let <math>A</math> and <math>B</math> be the tangency points on the top circle and <math>C</math> be the center of the top circle. Further, let <math>D</math> be the tangency point between the bottom two circles and <math>O</math> be the center of the hexagon. Additionally, extend the chords formed by the tangency points to create a larger equilateral triangle. Note that this triangle also has center <math>O</math>.
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Triangle <math>ACD</math> is right with <math>AC = 6</math>. We can find <math>CD</math> using the fact that it is the height of the triangle formed by the centers of the three circles, which means <math>CD = 6\sqrt{3}</math>. This will give us <math>AD = \sqrt{\left(6\sqrt{3}\right)^{2} - 6^{2}} = 6\sqrt{2}</math>. Let <math>M</math> be the midpoint of <math>\overline{AB}</math>. Note that <math>\triangle ACD \sim \triangle MCA</math>, and <math>\tfrac{AC}{CD} = \tfrac{\sqrt{3}}{3}</math>. Therefore, <math>AM = \tfrac{\sqrt{3}}{3} \cdot 6\sqrt{2} = 2\sqrt{6}</math>, and <math>AB = 4\sqrt{6}</math>.
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Now, we want to find the side length of the larger equilateral triangle formed by the chord extensions. We can do this
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by finding <math>OM</math>, which is the distance from the center of the triangle to a side, so the side length in question is <math>2\sqrt{3} \cdot OM</math>. Note that <math>OM = CD - CM - OD</math>.
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From the similarity described earlier, <math>CM = \tfrac{\sqrt{3}}{3} \cdot AC = \tfrac{\sqrt{3}}{3} \cdot 6 = 2\sqrt{3}</math>, and <math>OD = 2\sqrt{3}</math> because it is the distance from <math>O</math> to a side of the equilateral triangle of side length <math>12</math> formed by the centers of the three circles. Therefore, <math>OM = 6\sqrt{3} - 2\sqrt{3} - 2\sqrt{3} = 2\sqrt{3}</math>, and the side length of the large equilateral triangle is <math>2\sqrt{3} \cdot 2\sqrt{3} = 12</math>.
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The area of the hexagon is equal to the area of this triangle minus the area of the three smaller equilateral triangles, each of which have side length <math>\tfrac{1}{2}\left(12 - 4\sqrt{6}\right) = 6 - 2\sqrt{6}</math>. This is equal to <cmath>\tfrac{\sqrt{3}}{4} \cdot 12^{2} - 3 \cdot\tfrac{\sqrt{3}}{4} \cdot \left(6 - 2\sqrt{6}\right)^{2} = 36\sqrt{3} - \tfrac{3\sqrt{3}}{4}\left(60 - 24\sqrt{6}\right) = 36\sqrt{3} - 45\sqrt{3} + 54\sqrt{2} = 54\sqrt{2} - 9\sqrt{3},</cmath> so <math>a = 54</math>, <math>b = -9</math> and the answer is <math>\boxed{\textbf{(E)}~45}</math>.
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==See also==
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{{AMC12 box|year=2024|ab=A|num-b=21|num-a=23}}
 +
{{MAA Notice}}

Revision as of 21:35, 20 March 2025

Problem

Three circles of radius $6$ are mutually externally tangent, as shown below. For each pair of circles, construct the lines through their point of tangency that are tangent to the third circle. In total, this creates six new tangency points. If the area of the convex hexagon formed by these six points can be expressed as $a\sqrt{2} + b\sqrt{3}$ for integers $a$ and $b$, what is $a + b$?

[asy] import olympiad; size(150); defaultpen(linewidth(0.6) + fontsize(10)); pen dotstyle = black; real xmin = -8.903758953234952, xmax = 14.962621077942625, ymin = -5.903115312371685, ymax = 10.567409309904837; /* image dimensions */ /* draw figures */ draw(circle((0,0), 3), linewidth(1)); draw(circle((6,0), 3), linewidth(1)); draw(circle((3,5.196152422706632), 3), linewidth(1)); draw((0.5505102572168217,3.4641016151377544)--(3,0), linewidth(1)); draw((3,0)--(5.449489742783179,3.4641016151377544), linewidth(1)); /* dots */ dot((0.5505102572168217,3.4641016151377544),linewidth(4pt) + dotstyle); dot((5.449489742783179,3.4641016151377544),linewidth(4pt) + dotstyle); dot((5.724744871391589,2.987345747344081),linewidth(4pt) + dotstyle); dot((3.275255128608411,-1.2552949397752038),linewidth(4pt) + dotstyle); dot((0.27525512860841106,2.987345747344081),linewidth(4pt) + dotstyle); dot((2.724744871391589,-1.2552949397752038),linewidth(4pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]

$\textbf{(A)}~18\qquad\textbf{(B)}~24\qquad\textbf{(C)}~36\qquad\textbf{(D)}~42\qquad\textbf{(E)}~45$

Solution

Let $A$ and $B$ be the tangency points on the top circle and $C$ be the center of the top circle. Further, let $D$ be the tangency point between the bottom two circles and $O$ be the center of the hexagon. Additionally, extend the chords formed by the tangency points to create a larger equilateral triangle. Note that this triangle also has center $O$.

Triangle $ACD$ is right with $AC = 6$. We can find $CD$ using the fact that it is the height of the triangle formed by the centers of the three circles, which means $CD = 6\sqrt{3}$. This will give us $AD = \sqrt{\left(6\sqrt{3}\right)^{2} - 6^{2}} = 6\sqrt{2}$. Let $M$ be the midpoint of $\overline{AB}$. Note that $\triangle ACD \sim \triangle MCA$, and $\tfrac{AC}{CD} = \tfrac{\sqrt{3}}{3}$. Therefore, $AM = \tfrac{\sqrt{3}}{3} \cdot 6\sqrt{2} = 2\sqrt{6}$, and $AB = 4\sqrt{6}$.

Now, we want to find the side length of the larger equilateral triangle formed by the chord extensions. We can do this by finding $OM$, which is the distance from the center of the triangle to a side, so the side length in question is $2\sqrt{3} \cdot OM$. Note that $OM = CD - CM - OD$.

From the similarity described earlier, $CM = \tfrac{\sqrt{3}}{3} \cdot AC = \tfrac{\sqrt{3}}{3} \cdot 6 = 2\sqrt{3}$, and $OD = 2\sqrt{3}$ because it is the distance from $O$ to a side of the equilateral triangle of side length $12$ formed by the centers of the three circles. Therefore, $OM = 6\sqrt{3} - 2\sqrt{3} - 2\sqrt{3} = 2\sqrt{3}$, and the side length of the large equilateral triangle is $2\sqrt{3} \cdot 2\sqrt{3} = 12$.

The area of the hexagon is equal to the area of this triangle minus the area of the three smaller equilateral triangles, each of which have side length $\tfrac{1}{2}\left(12 - 4\sqrt{6}\right) = 6 - 2\sqrt{6}$. This is equal to \[\tfrac{\sqrt{3}}{4} \cdot 12^{2} - 3 \cdot\tfrac{\sqrt{3}}{4} \cdot \left(6 - 2\sqrt{6}\right)^{2} = 36\sqrt{3} - \tfrac{3\sqrt{3}}{4}\left(60 - 24\sqrt{6}\right) = 36\sqrt{3} - 45\sqrt{3} + 54\sqrt{2} = 54\sqrt{2} - 9\sqrt{3},\] so $a = 54$, $b = -9$ and the answer is $\boxed{\textbf{(E)}~45}$.

See also

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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

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