2018 AMC 10B Problem 7

Question

In the figure below,

N

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

A

be the combined area of the small semicircles and

B

be the area of the region inside the large semicircle but outside the small semicircles. The ratio

A : B

is

1 : 18.

What is

N?

16

17

18

19

36

Answer: D

Concepts:circle area ratio and proportion

Difficulty rating: 1310

Small hint:

If each small semicircle has radius

r,

the large semicircle has radius

Nr

Big hint:

Show

A : B = 1 : (N-1),

then set this equal to

1 : 18

Solution:

Let each small semicircle have radius

r.

The

N

diameters cover the big diameter, so the large radius is

Nr.

Then

A = N \cdot \tfrac12 \pi r^2,

and the large semicircle has area

\tfrac12 \pi (Nr)^2,

so the leftover region is

B = \tfrac12 \pi r^2(N^2 - N).

This gives

A : B = N : N(N-1)

= 1 : (N-1).

Set

N - 1 = 18,

and

N = 19.

Thus, D is the correct answer.