2018 AMC 10B Problem 7

Let each small semicircle have radius $r$. The $N$ small diameters, each $2r$, tile the large diameter, so the large radius is $Nr$.

Combined small area: $A = N \cdot \frac{\pi r^2}{2}$.

Large semicircle: $\frac{\pi (Nr)^2}{2} = N^2 \cdot \frac{\pi r^2}{2}$, so the leftover region is
$$B = (N^2 - N)\frac{\pi r^2}{2}.$$

Therefore
$$\frac{A}{B} = \frac{N}{N^2 - N} = \frac{1}{N - 1}.$$
Setting $N - 1 = 18$ gives $N = 19$.

The answer is $\boxed{\textbf{(D)}\ 19}$.

Area scales with the square of the radius, so $N$ semicircles of radius $r$ cover only $\frac{1}{N}$ of a semicircle of radius $Nr$. Everything cancels except $N$, which is why the problem never needs an actual length. Whenever similar figures are nested, work with the ratio of a linear dimension and square it.

Take $r = 1$ and test the choices: for $N = 19$, $A = 19 \cdot \frac{\pi}{2}$ and $B = (361 - 19)\frac{\pi}{2} = 342 \cdot \frac{\pi}{2}$, and $342 / 19 = 18$.

• Setting A:(A+B) or A:B equal to 1:N and answering 18, or comparing radii instead of areas.
• Using the small diameter as the radius (radius $2r$ each, large radius $2Nr$), which still cancels correctly but invites arithmetic errors.