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

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A circle of radius <imath>r</imath> 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 <imath>r</imath>?
<imath>\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}</imath>
==Solution 1==
==Solution 1==
Descartes' circle theorem (curvatures k_i = 1/r_i)
Descartes' circle theorem (curvatures k_i = 1/r_i)

Revision as of 14:49, 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 $r$?

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


Solution 1

Descartes' circle theorem (curvatures k_i = 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 and another person


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’ Theorem 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

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