2018 AMC 10B Problem 7

No lengths are given, so measure everything in "small semicircles."

1. The big radius. The $N$ small diameters lie end to end and exactly cover the big diameter. So the big diameter is $N$ times a small diameter, and the big radius is $N$ times a small radius.
2. The big area. Area grows with the square of the radius. Making the radius $N$ times as big makes the area $N \times N = N^2$ times as big. So the big semicircle has the area of $N^2$ small semicircles.
3. Region $A$. The small semicircles together make $N$ small areas.
4. Region $B$. What is left inside the big semicircle is $N^2 - N$ small areas.
5. Use the ratio. $A : B = 1 : 18$ means $B$ is $18$ times $A$:
$$N^2 - N = 18N.$$
Divide both sides by $N$: $N - 1 = 18$, so $N = 19$.

Check with $N = 19$: $A$ is $19$ small areas and $B$ is $361 - 19 = 342$ small areas, and $342 = 18 \times 19$.

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

When one shape is a scaled copy of another, lengths scale by the scale factor but areas scale by its square. That is why $N$ small semicircles fill only $\frac{1}{N}$ of the big one, and why the answer never needs an actual length.

Test the choices with small radius $1$, so a small semicircle has area $\frac{\pi}{2}$. For $N = 19$ the big radius is $19$, so the big area is $361 \cdot \frac{\pi}{2}$. Then $A = 19 \cdot \frac{\pi}{2}$ and $B = 342 \cdot \frac{\pi}{2}$, and $342 \div 19 = 18$. It works.

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 $\boxed{\textbf{(D)}\ 19}$.