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2025 AMC 12A Problems/Problem 24: Difference between revisions

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Let the center of the large circle be <imath>O</imath> and the centers of any two circles be <imath>A</imath> and <imath>B</imath>. Triangle <imath>OAB</imath> has side lengths <imath>r+1, r+1, 2</imath>, with the angle opposite <imath>2</imath> being <imath>360/12 = 30</imath>.
Let the center of the large circle be <imath>O</imath> and the centers of any two circles be <imath>A</imath> and <imath>B</imath>. Triangle <imath>OAB</imath> has side lengths <imath>r+1, r+1, 2</imath>, with the angle opposite <imath>2</imath> being <imath>360/12 = 30</imath>.


Using Law of Cosines, <imath>2^2=AO^2+BO^2-2AOBO</imath>
Using Law of Cosines, <imath>2^2=AO^2+BO^2-2AOBO\cos30</imath>
 


== Solution 2 ==
== Solution 2 ==

Revision as of 15:17, 6 November 2025

Problem

A circle of radius $r$ is surrounded by $12$ circles of radius $1,$ externally tangent to the central circle and sequentially tangent to each other, as shown. Then $r$ can be written as $\sqrt a + \sqrt b + c,$ where $a, b, c$ are integers. What is $a+ b+c?$

[center][img width=60]https://wiki.randommath.com/amc/2025_11_05_e415ed6e80d65fa550edg-7.jpg[/img][/center]

$\textbf{(A) } 3 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 7 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 11$

Solution 1

Let the center of the large circle be $O$ and the centers of any two circles be $A$ and $B$. Triangle $OAB$ has side lengths $r+1, r+1, 2$, with the angle opposite $2$ being $360/12 = 30$.

Using Law of Cosines, $2^2=AO^2+BO^2-2AOBO\cos30$

Solution 2

Let the center of the large circle be $O$ and the centers of the $12$ circles be $A_1, A_2, A_3, \dots, A_{12}$. Triangle $OA_1A_2$ has side lengths $r+1, r+1, 2$, with the angle opposite $2$ being $360/12 = 30$.

Drawing the angle bisector of the $30$ degree angle, we split $OA_1A_2$ into two congruent right triangles, each with hypotenuse $r+1$ and side opposite the $15$ degree angle $1$.

From here, note that $\sin{15} = \frac{\sqrt{6}-\sqrt{2}}{4}$, which be derived using the trigonometric identity $\sin{(A-B)}$, with $A=45$ and $B=30$.

In our right triangle, $\sin{15} = \frac{1}{r+1} = \frac{\sqrt{6}-\sqrt{2}}{4}$. Let $x=r+1$. Solving for $x$, we get $x = \frac{4}{\sqrt{6}-\sqrt{2}}$. Rationalizing, we get that $x = \sqrt{6}+\sqrt{2}$.

Remember $x = r+1 = \sqrt{6}+\sqrt{2}$, so $r = \sqrt{6}+\sqrt{2} - 1$. Therefore, our answer is $6+2-1 = \boxed{7}.$

~lprado

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