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

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Using Law of Cosines, <imath>2^2=AO^2+BO^2-2AOBO\cos30</imath>. Plugging in the radius yields <imath>2^2=(r+1)^2+(r+1)^2-2(r+1)^2*1/2</imath>. After solving for <imath>r</imath> and simplifying, we get <imath>r=\sqrt6+\sqrt2-1</imath>. Therefore our answer is <imath>7</imath>.
Using Law of Cosines, <imath>2^2=AO^2+BO^2-2AOBO\cos30</imath>. Plugging in the radius yields <imath>2^2=(r+1)^2+(r+1)^2-2(r+1)^2*1/2</imath>. After solving for <imath>r</imath> and simplifying, we get <imath>r=\sqrt6+\sqrt2-1</imath>. Therefore our answer is <imath>7</imath>.


~KMS888
~Kevin Wang


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

Revision as of 16:39, 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(No sin15)

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, $2^2=AO^2+BO^2-2AOBO\cos30$. Plugging in the radius yields $2^2=(r+1)^2+(r+1)^2-2(r+1)^2*1/2$. After solving for $r$ and simplifying, we get $r=\sqrt6+\sqrt2-1$. Therefore our answer is $7$.

~Kevin Wang

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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