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 $\frac{1}{2}$ π 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 $\frac{1}{2}$ π r², and with N of them the combined small area is A = N· $\frac{1}{2}$ π r².

A = N·

\frac{1}{2}

π r²

3STEP 3

Area of the big semicircle

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

A + B =

\frac{1}{2}

π (Nr)² =

\frac{1}{2}

π N² r²

4STEP 4

Find the leftover region B

Subtract the small area from the big one and factor out $\frac{1}{2}$ π r²: B = $\frac{1}{2}$ π r² (N² - N).

B =

\frac{1}{2}

π r² (N² - N)

5STEP 5

Form the ratio and cancel

Dividing A by B, the $\frac{1}{2}$ π r² cancels and N(N-1) reduces, leaving $\frac{A}{B}$ = $\frac{1}{N-1}$ .

\frac{A}{B}

\frac{N· \frac{1}{2}π r²}{\frac{1}{2}π r² (N²-N)}

\frac{N}{N(N-1)}

\frac{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

\frac{1}{2}

π r² and the big one is

\frac{1}{2}

π N² r², so one small piece is

\frac{1}{N²}

of the big region. With N small pieces, A is

\frac{N}{N²}

\frac{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 $\frac{1}{N}$ 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