🔥 New! 2025 AMC 10 B

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Problem 20 Easy

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?

  • A.

    3

  • B.

    5

  • C.

    8

  • D.

    9

  • E.

    11

Answer:A
Solution

Let r be the radius of the small circle and R be the larger radius.

Let O, E, F be the centers, then OE = OF = r + R and EF = 2R.

Since CF = R and \angle ECF = 90^\circ, we have \angle CFE = 60^\circ and CE = \sqrt{3}R.

Then:

R + \sqrt{3}R = 1 \implies R = \frac{1}{\sqrt{3}+1}

Since \angle FOE = 90^\circ by symmetry, EF = \sqrt{2} OE, i.e., 2R = \sqrt{2}(R+r).

r = (\sqrt{2}-1)R = \frac{\sqrt{2}-1}{\sqrt{3}+1} = \frac{1}{2}(\sqrt{2}-1)(\sqrt{3}-1)

\implies 2r = (\sqrt{2}-1)(\sqrt{3}-1)

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