Competition · AMC preparation · step 4 of 4

AMC 8 · 2018 · #15

Grade 7 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Nested Frame / Concentric Difference
area-circlesratio-proportionformula-substitution area-differenceidentify-subproblems ↑ Prerequisites: area-circlesfraction-arithmetic
📏 Short solution 💡 3 insights 📊 Diagram
Problem
A large circle contains two smaller circles inside it. Each small circle's diameter equals the large circle's radius (so the two small circles fit snugly along a diameter of the large one). The two small circles together have area 1 square unit. Find the area of the shaded part (the large circle minus the two small ones).

Pick an answer.

(A)
$\frac{1}{4}$
(B)
$\frac{1}{3}$
(C)
$\frac{1}{2}$
(D)
1
(E)
$\frac{\pi}{2}$

AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

The shaded region is a compound shape, so Tool #7 (Subproblems) is the natural lead: shaded = (large circle area) - (two small circles' area). To make the radius relationship visible, Tool #1 (Diagram) — label the small radius r and notice the large radius is R = 2r. Tool #9 (Easier Related Problem) is the safety net: instead of carrying π symbolically, observe that the small circles' total area is given as 1, so we only need to compare the large area to that given number — which turns the problem into the simple question "how many times bigger is the large circle than the two small ones combined?"

1STEP 1

Draw and label the circles

Label the small radius r. A small diameter is 2r and that equals the large radius, so R = 2r.

R = 2r
2STEP 2

Write each area

By A = π r², the two small circles total 2π r², and the large circle is π(2r)² = 4π r².

A_small,total = 2π r², A_large = π(2r)² = 4π r²
3STEP 3

Subtract the small circles

Split it: shaded = large - two small = 4π r² - 2π r² = 2π r², the very same expression as the two small circles' combined area.

A_shaded = 4π r² - 2π r² = 2π r²
4STEP 4

Match to the given area

The two small circles give 2π r² = 1, and the shaded area is also 2π r², so the shaded area equals 1.

A_shaded = 2π r² = 1 → (D) 1
Answer
1
Sanity check with pictures: the large circle has radius 2r, so its area is 4 times the area of one small circle (radius doubles → area quadruples). One small circle has area 12\frac{1}{2} (since two of them total 1), so the large circle has area 4 × 12\frac{1}{2} = 2. Shaded = 2 - 1 = 1. ✓ Choice (D). Choice (E) π2\frac{π}{2} is the classic trap for forgetting that the given 1 already absorbed π into a number.
💡Key takeaway

This AMC 8 problem only needs Grade 7 circle-area formula A = π r² — and the cool trick that the shaded ring and the two small circles end up with the exact same expression, so the answer is just the 1 you were already given!

  • Draw and label the circles
  • Write each area
  • Subtract the small circles
  • Match to the given area

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