2025 AMC 10A Problems/Problem 22: Difference between revisions

Revision as of 20:38, 6 November 2025

A circle of radius $r$ is surrounded by three circles, whose radii are 1, 2, and 3, all externally tangent to the inner circle and externally tangent to each other, as shown in the diagram below.

What is $r$ ?

$\textbf{(A) }\frac{1}{4}\qquad\textbf{(B) }\frac{6}{23}\qquad\textbf{(C) }\frac{3}{11}\qquad\textbf{(D) }\frac{5}{17}\qquad\textbf{(E) }\frac{3}{10}$

Diagram

$[asy] import olympiad; size(260); // radii (outer ones specified: 1,2,3; inner r0 = 6/23) real r0 = 6/23.0; real r1 = 1.0; real r2 = 2.0; real r3 = 3.0; // distances from origin real d1 = r0 + r1; real d2 = r0 + r2; real d3 = r0 + r3; // angles found numerically (radians) real t1 = 0; real t2 = 1.9857887796653; real t3 = -2.0480149718113; // centers pair O = (0,0); pair C1 = d1*dir(degrees(t1)); pair C2 = d2*dir(degrees(t2)); pair C3 = d3*dir(degrees(t3)); // draw the circles draw(circle(O, r0), black+0.9); draw(circle(C1, r1), black+0.9); draw(circle(C2, r2), black+0.9); draw(circle(C3, r3), black+0.9); // tight crop, no box currentpicture.fit(); [/asy]$ ~Avs2010

Solution 1

Descartes' Circle Formula (curvatures $k_i = \frac{1}{r_i}$ ) $k_4 = k_1 + k_2 + k_3 \pm 2\sqrt{k_1k_2 + k_2k_3 + k_3k_1}.$

For radii 1, 2, 3 we have $k_1 = 1,\quad k_2 = \frac{1}{2},\quad k_3 = \frac{1}{3}.$

Compute the sum and the square-root term $k_1+k_2+k_3 = \frac{11}{6},\qquad k_1k_2+k_2k_3+k_3k_1 = 1.$

Therefore $k_4 = \frac{11}{6} \pm 2.$

Choose the plus sign for the small circle tangent externally to the three given circles $k_4 = \frac{11}{6} + 2 = \frac{23}{6}, \qquad r_4 = \frac{1}{k_4} = \frac{6}{23}.$ ~Jonathanmo

Solution 2 (Descartes’ Theorem, Detailed Derivation)

We are given the radii of three circles and are asked to find an external tangent to the radius of the circle.

We can use Descartes’ Theorem to find it using curvatures (reciprocals of the radii). The curvatures are $k_1 = 1, \quad k_2 = \tfrac{1}{2}, \quad k_3 = \tfrac{1}{3}.$

Descartes' Circle Formula states $(k_1 + k_2 + k_3 + k_4)^2 = 2(k_1^2 + k_2^2 + k_3^2 + k_4^2).$

Plugging in, we get $\left(1 + \tfrac{1}{2} + \tfrac{1}{3} + k_4\right)^2 = 2\left(1 + \tfrac{1}{4} + \tfrac{1}{9} + k_4^2\right).$

Simplifying, $\left(\tfrac{11}{6} + k_4\right)^2 = 2\left(\tfrac{49}{36} + k_4^2\right).$ $\tfrac{121}{36} + \tfrac{11}{3}k_4 + k_4^2 = \tfrac{49}{18} + 2k_4^2.$ $k_4^2 - \tfrac{11}{3}k_4 - \tfrac{23}{36} = 0.$

Using the quadratic formula, $k_4 = \frac{\tfrac{11}{3} \pm \sqrt{\tfrac{121}{9} - (-\tfrac{23}{9})}}{2} = \frac{\tfrac{11}{3} \pm \sqrt{\tfrac{144}{9}}}{2} = \frac{\tfrac{11}{3} \pm \sqrt{\tfrac{12}{3}}}{2} = \tfrac{23}{6}, -\tfrac{1}{6}.$

Since the curvature is the reciprocal of the radius, the two tangent circles possible are $r = \frac{1}{|k_4|} = \frac{6}{23} \text{ and } 6.$

The circle with radius 6 is the one that contains all of the three tangent circles inside it, so the answer is $\boxed{(B)\ \tfrac{6}{23}}.$ ~AlgeBruh16

Solution 3(Heron's)

Let the center of the circle radius 2 be $A$ , radius 1 be $B$ , radius 3 be $C$ . Let the unknown circle's center be $P$ .

We know $\overline{AB}=3+1=4, \overline{AC}=3+2=5, \overline{BC}=2+1=3.$

Thus, $\triangle ABC$ is a 3-4-5 triangle, where $\angle B=90^\circ$ .

Now, the area of $\triangle BPC=\sqrt{s(s-\overline{BP})(s-\overline{CP})(s-\overline{BC})}=\sqrt{(4+r)\cdot 1\cdot 3\cdot r}=\sqrt{3r^2+12r}.$

Thus, the height from $P$ to $\overline{BC}$ is $\frac{2[ABC]}{\overline{BC}}=\frac{\sqrt{3r^2+12r}}{2}.$ Now, since $\angle B=90^\circ$ , we know the height from $P$ to $\overline{AB}$ squared is $\overline{BP}^2-($ height from $P$ to $\overline{BC})^2=(1+r)^2-\frac{3r^2+12r}{4}=\frac{r^2-4r+4}{4}.$

$\Rightarrow$ height from $P$ to $\overline{AB}$ is $\frac{r-2}{2}$ . It's also

$\frac{2[ABP]}{\overline{AB}}=\frac{2\sqrt{2r^2+6r}}{3}$

Now, solve the equation:

\begin{align*} \frac{r-2}{2}&=\frac{2\sqrt{2r^2+6r}}{3} \\ \Rightarrow 3(r-2)&=4\sqrt{2r^2+6r} \\ \Rightarrow 9r^2-36r+36&=32r^2+96r \\ \Rightarrow 23r^2+132r-36&=0 \end{align*}

Solving, you get the only positive root is $r=\frac{6}{23}=\boxed{B}.$

Note: To solve the final quadratic, alternatively, you can see that $\frac{6}{23}$ is the only answer choice with denominator 23, and if you test it, it works.

~bluedolphin36,eggon Edited by GarudS

Solution 4 (Variation of Solution 3)

Follow the same starting steps as solution 3 (drawing right triangle ABC connecting the centers of the three larger circles and connecting points A, B, and C to the center of the smaller circle). Using Heron's formula, we can easily find the areas of the three smaller triangles that make up triangle ABC in terms of $r$ . Since the sum of these terms is an integer, or 6, we can theorize that the square roots have to simplify out into rational numbers. Out of all the answer choices, $\boxed{(B)}$ , or $\dfrac{6}{23}$ , is the only one that works. (will add more later)

~stjwyl

Video Solution (In 3 Mins)

https://youtu.be/BD-AUw_m65U?si=f8deq2OpR5LdOpr9 ~ Pi Academy

Video Solution

https://www.youtube.com/watch?v=OGr0NVDt9lI ~ ABIRGH

- This video was posted 2 years ago as an explanation of Descartes' theorem, and it was coincidentally used on the test. There were no known leaks.

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

@@ Line 164: / Line 164: @@
 == Video Solution (In 3 Mins) ==
 https://youtu.be/BD-AUw_m65U?si=f8deq2OpR5LdOpr9 ~ Pi Academy
+== Video Solution ==
+https://www.youtube.com/watch?v=OGr0NVDt9lI ~ ABIRGH
+- This video was posted 2 years ago as an explanation of Descartes' theorem, and it was coincidentally used on the test. There were no known leaks.
 ==Video Solution==

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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2025 AMC 10A Problems/Problem 22: Difference between revisions

Revision as of 20:38, 6 November 2025

Contents

Diagram

Solution 1

Solution 2 (Descartes’ Theorem, Detailed Derivation)

Solution 3(Heron's)

Solution 4 (Variation of Solution 3)

Video Solution (In 3 Mins)

Video Solution

Video Solution

See Also