2018 AMC 10B Problem 7

P1-10~2 minPrint

Circles Ratios, Percents & Averages Set up the equation/formula and compute; no special trick needed

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 $N$ ?

Show solution without answering

Key insight

If the big radius is N times the small one, the big semicircle holds N squared small areas, so the leftover is N squared minus N.

Solution

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

Why this works

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.

Alternative approach

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Geometry, Chapter 11: Circles
Introduction to Algebra, Chapter 6: Ratios and Percents
Introduction to Geometry, Chapter 12: Circles and Angles
Introduction to Algebra, Chapter 7: Proportion (§7.1, §7.2, §7.3)