AMC 10 · 2018 · #7

Grade 7 geometry-2d

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

area-circlesratio-proportionsimilar-figures convert-to-algebraidentify-subproblems ↑ Prerequisites: area-circles

📏 Long solution 💡 2 insights 📊 Diagram

Problem

A row of N equal small semicircles sits on the flat diameter of one big semicircle, their diameters filling that diameter exactly with no gaps or overlaps. A is the total area of the small semicircles; B is the area inside the big semicircle but outside the small ones. Given A:B = 1:18, find N.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Convert to Algebra

Tool #1 (Draw a Diagram): the figure tells us the key fact — N small diameters laid end to end equal the big diameter, so the big radius is exactly N times the small radius. Tool #13 (Convert to Algebra): name the small radius r and write every area in terms of r and N; the messy 1/2π r² factors cancel when we form the ratio, leaving a clean equation in N. Tool #7 (Identify Subproblems): split the work into three pieces — total small area A, big-semicircle area, and the leftover B — then combine. The ratio condition then becomes a one-line equation to solve.

1STEP 1

Name the radii from the picture

Let each small radius be r; N diameters of 2r fill the big diameter, so the big radius is R = Nr.

R = N· 2r ÷ 2 = Nr

2STEP 2

Area of all the small semicircles

One small semicircle is 1/2π r², and with N of them the combined small area is A = N· 1/2π r².

A = N· 1/2π r²

3STEP 3

Area of the big semicircle

The big radius Nr gives big area 1/2π N² r², and since it holds the small ones plus the rest, A + B = 1/2π N² r².

A + B = 1/2π (Nr)² = 1/2π N² r²

4STEP 4

Find the leftover region B

Subtract the small area from the big one and factor out 1/2π r²: B = 1/2π r² (N² - N).

B = 1/2π r² (N² - N)

5STEP 5

Form the ratio and cancel

Dividing A by B, the 1/2π r² cancels and N(N-1) reduces, leaving A/B = 1/(N-1).

A/B = (N· 1/2π r²)/(1/2π r² (N²-N)) = N/(N(N-1)) = 1/(N-1)

6STEP 6

Solve for N

Set 1:(N-1) equal to the given 1:18, so N - 1 = 18 and N = 19, choice (D).

N - 1 = 18 → N = 19 → (D)

Answer

Check the slick version: each small semicircle is 1/2π r² and the big one is 1/2π N² r², so one small piece is 1/N² of the big region. With N small pieces, A is N/N² = 1/N of the whole. So A:(A+B) = 1:N, meaning A:B = 1:(N-1). Setting N-1=18 gives N=19, agreeing with the algebra. The trap answer (C) 18 comes from reading A:B as 1:N instead of 1:(N-1), and (E) 36 = 2× 18 comes from a doubling slip — only (D) survives.

💡Key takeaway

Since the big radius is N times a small radius, the big semicircle is N² times one small piece, so N small pieces fill 1/N of it — making A:B = 1:(N-1), and 1:18 forces N=19.

Name the radii from the picture
Area of all the small semicircles
Area of the big semicircle
Find the leftover region B
Form the ratio and cancel
Solve for N