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2025 AMC 10B Problem 20

Problem 20 of 25HarderAlgebraGeometry

Four congruent semicircles are inscribed in a square of side length 11 so that their diameters are on the sides of the square, one endpoint of each diameter is at a vertex of the square, and adjacent semicircles are tangent to each other. A small circle centered at the center of the square is tangent to each of the four semicircles, as shown below. The diameter of the small circle can be written as (a+b)(c+d),(\sqrt{a} + b)(\sqrt{c} + d), where a,a, b,b, c,c, and dd are integers. What is a+b+c+d?a + b + c + d?

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Solution

Let each semicircle have radius ρ,\rho, with centers like (ρ,0)(\rho, 0) and (1,ρ).(1, \rho). Adjacent semicircles are tangent, so these centers are 2ρ2\rho apart: (1−ρ)2+ρ2=4ρ2.(1 - \rho)^2 + \rho^2 = 4\rho^2. This gives 2ρ2+2ρ−1=0,2\rho^2 + 2\rho - 1 = 0, so ρ=3−12.\rho = \tfrac{\sqrt3 - 1}{2}. The small circle of radius tt sits at (12,12),\left(\tfrac12, \tfrac12\right), and it’s tangent to a semicircle when its distance to that center equals ρ+t.\rho + t. That distance is 2−3,\sqrt{2 - \sqrt3}, so t=2−3−ρ,t = \sqrt{2 - \sqrt3} - \rho, and the diameter is 2t=6−22t = \sqrt6 - \sqrt2 −3+1=(3−1)(2−1).- \sqrt3 + 1 = (\sqrt3 - 1)(\sqrt2 - 1). So a+b+c+d=3a + b + c + d = 3 +(−1)+2+(−1)=3.+ (-1) + 2 + (-1) = 3. Therefore, the answer is A.

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Concepts: tangent circles · coordinate geometry · radical

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.