AMC 8 · 2025 · #18

Grade 7 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Nested Frame / Concentric Difference

area-circlessimilar-figuresarea-trianglesperfect-squares easier-related-problemidentify-subproblemsarea-difference ↑ Prerequisites: area-circlesarea-triangleslinear-equations-one-var

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Two circles each contain an inscribed square (corners on the circle). The left circle has radius 1, and the whole region inside the circle but outside the square is shaded. The right circle has radius R, and only one quarter of that same in-between region is shaded. The two shaded areas are equal — find R.

Pick an answer.

(A)

$\sqrt{2}$

(B)

(C)

$2\sqrt{2}$

(D)

(E)

$4\sqrt{2}$

AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The shaded "in-between" region is awkward as a single shape, but tool #7 (Subproblems) turns it into a clean difference: (circle area) - (inscribed-square area). Tool #1 (Diagram) — specifically drawing the two diagonals of the inscribed square — splits the square into four right isoceles triangles whose legs are the radius, so we get the square's area as 2r² without needing Pythagoras. Tool #9 (Easier Problem) is the key insight: both circles have the same shape, so both "between" regions follow the same formula (π - 2)r². That reduces the whole problem to solving (π - 2)(1)² = $\frac{1}{4}$ (π - 2)R², which is just R² = 4.

1STEP 1

Draw the square's two diagonals: they split it into 4 right isosceles triangles with legs equal to radius r, so its area is 2r².

square area = 4 ×

\frac{1}{2}

· r · r = 2r²

2STEP 2

The between-region is (circle area) - (square area), which cleans up to (π - 2) r².

between area(r) = π r² - 2r² = (π - 2) r²

3STEP 3

Left shades the whole region, giving π - 2; right shades a quarter, giving $\frac{1}{4}$ (π - 2) R².

left shaded = (π - 2)(1)² = π - 2 right shaded =

\frac{1}{4}

(π - 2) R²

4STEP 4

Set them equal: π - 2 = $\frac{1}{4}$ (π - 2) R². Cancel the positive (π - 2) to get R² = 4, so R = 2.

π - 2 =

\frac{1}{4}

(π - 2) R² → 1 = R²/4 → R² = 4 → R = 2

5STEP 5

R = 2 matches answer choice (B).

R = 2 → (B)

Answer

Intuition check: the right between-region only contributes one quarter of itself to the shaded area, so it has to be 4 times bigger overall than the left between-region. Area scales with the square of the radius, so the radius only needs to be √(4) = 2 times bigger — not 4 times. That rules out (D) 4 and (E) 4√(2) immediately and lands exactly on (B) 2, matching our algebra. A numerical sanity check: left shaded = π - 2 ≈ 1.14; right shaded =

\frac{1}{4}

(π - 2)(2)² = π - 2 ≈ 1.14. They match.

💡Key takeaway

This AMC 8 problem only needs Grade 7 "area of a circle is π r²" plus the idea that when you scale a shape, its area grows by the square of the scale — that you already know!

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