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

Latest revision as of 20:49, 11 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 $a+b+c?$

$[asy] defaultpen(fontsize(12)+linewidth(1)); size(200); real r=2.925, x=360/12; pair O=origin; draw(CR(O,r),black+1.5); for (int i = 0; i<12; ++i) { draw(CR((r+1)*dir(i*x),1)); } dot(O); dot((r+1)*right); draw(O--(r,0)^^(r+1,0)--(r+2,0), linewidth(0.5)); label("$r$",(r/2,0),up); label("$1$",(r+3/2,0),up); [/asy]$

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

Solution 1 (Sin 15)

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 $\sin{(A-B)} = \sin{A} \cos{B} - \sin{B} \cos{A}$ , 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

Clarification Request: How would triangle $OA_1A_2$ have a side length of $2$ ? The line connecting $A_1$ and $A_2$ doesn't go through the point of tangency of the respective circles.

Answer to Clarification: The segment $A_1A_2$ does go through the point of tangency of the two circles, by definition.

Solution 2 (Law of Cosines Bash)

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 $2^2=(r+1)^2+(r+1)^2-2(r+1)^2\cdot\frac{\sqrt{3}}{2}$ . After solving for $r$ and simplifying, we get $r=\sqrt6+\sqrt2-1$ . Therefore our answer is $7$ .

~Kevin Wang

Worked Out Solution

$4=(2-\sqrt{3})(r+1)^2$ . Divide both sides to get $\frac{4}{2-\sqrt{3}}=(r+1)^2$ . Rationalize and expand to get $8+4\sqrt{3}=r^2+2r+1$ . Then $r^2+2r+(-7-4\sqrt{3})=0$ . Use the quadratic formula to get $r=\frac{-2+\sqrt{4+28+16\sqrt{3}}}{2}$ , or $r=-1+\sqrt{8+4\sqrt{3}}$ (Note: We only take the plus instead of plus-minus because r must be positive). Clearly $c=-1$ , and then we need to find some a and b such that $\sqrt{a} + \sqrt{b} = \sqrt{8+4\sqrt{3}}$ . If we square both sides, $a+b+2\sqrt{ab} = 8+4\sqrt{3}$ , so $a+b=8$ . Plug in our previously found values to find $a+b+c=8+(-1)$ . Therefore our answer is $7$ .

~Samuel and Jungbin

$\textbf{Remark}$

We could've also noticed that $\sqrt{2-\sqrt{3}} = \sqrt{\frac{1}{2}\cdot(4-2\sqrt{3})} = \frac{1}{\sqrt{2}}*\sqrt{3-2\sqrt{3}+1} = \frac{1}{\sqrt{2}}*(\sqrt{3}-1)$ , and therefore directly gotten $r+1 = \sqrt{\frac{4}{2-\sqrt{3}}} = \frac{2}{\frac{1}{\sqrt{2}}*(\sqrt{3}-1)} = \frac{2\sqrt{2}}{\sqrt{3}-1}$ and solved from there.

Solution 3 (Easy Geo Method)

We connect the centers of the smaller circles at $0^\circ$ and $90^\circ$ as follows: $[asy] import graph; size(250); real R = -1 + sqrt(2) + sqrt(6); real r = 1; pair O = (0,0); label("$O$", 0, W); pair C0 = (R + r) * dir(0); label("$A$", C0, E); pair C15 = (R + r) * dir(15); label("2", C15, NE); pair C30 = (R + r) * dir(30); label("$B$", C30, E); pair C45 = (R + r) * dir(45); label("2", C45, NE); pair C60 = (R + r) * dir(60); label("$C$", C60, NE); pair C75 = (R + r) * dir(75); label("2", C75, NE); pair C90 = (R + r) * dir(90); label("$D$", C90, N); pair project(pair P, pair A, pair B) { pair AB = B - A; real t = dot(P - A, AB) / dot(AB, AB); return A + t * AB; } pair foot30 = project(C30, C0, C90); pair foot60 = project(C60, C0, C90); draw(Circle(O, R), black + linewidth(1bp)); for(int i = 0; i < 12; ++i) { real theta = i * 360 / 12; pair center = (R + r) * dir(theta); draw(Circle(center, r), blue + linewidth(1bp)); } draw(O -- C0, black + linewidth(1bp)); draw(O -- C90, black + linewidth(1bp)); draw(C0 -- C30, red + linewidth(1bp)); draw(C30 -- C60, red + linewidth(1bp)); draw(C60 -- C90, red + linewidth(1bp)); draw(C0 -- C90, red + linewidth(1bp)); draw(C30 -- foot30, blue + linewidth(1bp)); draw(C60 -- foot60, blue + linewidth(1bp)); [/asy]$

Note that connecting the centers of the 12 smaller circles forms a regular 12-gon. A regular 12-gon has angles

$\frac{10\cdot 180}{12} = 150^\circ,$

so $\angle OAB = \angle ODC = 75^\circ.$ Furthermore, since $OA=OD$ and $\angle AOD = 90^\circ$ , we have $\angle OAD = 45^\circ$ , so $\angle DAB = 30^\circ$ . Next, we drop the perpendiculars from $B$ and $C$ to $AD$ . This creates a $1 \times 2$ rectangle and two $30-60-90$ triangles. Thus,

$AD = (r+1)\sqrt{2} = 2+2\sqrt{3},$

and manipulating yields

$r = -1 + \sqrt{2}+\sqrt{6},$

so the desired answer is $-1+2+6 = \boxed{\text{(C) } 7.}$

P.S. I posted the asymptote code here. Anyone that wishes to use it can steal~ :)

~ABC09090927

Video Solution 1 by OmegaLearn

https://youtu.be/nR67IIw_kgg

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

2025 AMC 12A (Problems • Answer Key • Resources)
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