Question

Four congruent semicircles are inscribed in a square of side length $1$ 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 $(\sqrt{a} + b)(\sqrt{c} + d),$ where $a, b, c,$ and $d$ are integers. What is $a + b + c + d?$

$3$

$5$

$8$

$9$

$11$

Solution:

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

2025 AMC 10B Problem 20

Solution:

Problem 20 in Other Years