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2018 AMC 10B Problem 7

Problem 7 of 25EasierGeometryArithmetic

In the figure below, NN congruent semicircles are drawn along a diameter of a large semicircle, with their diameters covering the diameter of the large semicircle with no overlap. Let AA be the combined area of the small semicircles and BB be the area of the region inside the large semicircle but outside the small semicircles. The ratio A:BA : B is 1:18.1 : 18. What is N?N?

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Solution

Let each small semicircle have radius r.r. The NN diameters cover the big diameter, so the large radius is Nr.Nr. Then A=N⋅12πr2,A = N \cdot \tfrac12 \pi r^2, and the large semicircle has area 12π(Nr)2,\tfrac12 \pi (Nr)^2, so the leftover region is B=12πr2(N2−N).B = \tfrac12 \pi r^2(N^2 - N). This gives A:B=N:N(N−1)A : B = N : N(N-1) =1:(N−1).= 1 : (N-1). Set N−1=18,N - 1 = 18, and N=19.N = 19. Thus, D is the correct answer.
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