Art of Problem Solving

2025 AMC 10A Problems/Problem 22

Revision as of 14:32, 6 November 2025 by Jonathanmo (talk | contribs)

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