Competition · AMC preparation · step 4 of 4

AMC 8 · 2005 · #25

Grade 8 geometry-2d
Archetype Overlap / Complement Counting └ Sub-archetype Geometric Area via Inclusion-Exclusion or Area Balance
area-circlesarea-rectanglesarea-differenceset-partition identify-subproblemsarea-difference ↑ Prerequisites: area-circlesarea-rectangles
📏 Short solution 💡 3 insights 📊 Diagram
Problem
A square with side 2 and a circle share the same center. The area of the part inside the circle but outside the square equals the area of the part inside the square but outside the circle. Find the radius r of the circle.

Pick an answer.

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

AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Find an Invariant

The overlap region between the square and the circle has a messy shape, and the problem never asks for its area. That is the hint to use Tool #11 (Find an Invariant): call the overlap area I, then notice it appears on both sides of the equation and cancels. Tool #1 (Draw a Diagram) sets up the picture — circle and square sharing a center create three pieces (circle-only, square-only, overlap) — so that subtracting the overlap from the whole circle and from the whole square is obvious. After cancellation, circle area equals square area, and one area formula gives r.

1STEP 1

Label the three regions

Draw both shapes sharing a center and call the overlap I; the circle-only region is π r² - I and the square-only region is 4 - I.

circle-only = π r² - I, square-only = 4 - I
2STEP 2

Set the two areas equal

Set the two crescents equal; the unknown I cancels from both sides, leaving π r² = 4 — the circle and square must have equal area.

π r² - I = 4 - I → π r² = 4
3STEP 3

Solve for the radius

Solve π r² = 4 for r: divide by π to get r² = 4π\frac{4}{π}, then take the positive square root, matching choice (A).

r² = 4/π → r = √(4/π) = 2/√(π) → (A)
Answer
2/√(π)
Check the size. π ≈ 3.14, so r = 23.14\frac{2}{\sqrt{3.14}} ≈ 21.77\frac{2}{1.77} ≈ 1.13. The square's inscribed circle has radius 1 and its circumscribed circle has radius √(2) ≈ 1.41, so the answer 1.13 sits between them — exactly what equal crescent areas should give. Choice (C) 32\frac{3}{2} = 1.5 is too big (circle would swallow the square), and (B) 1+22\frac{1+\sqrt{2}}{2} ≈ 1.21 is close but does not give equal areas: π(1.21)² ≈ 4.59 ≠ 4.
💡Key takeaway

The messy overlap area never has to be computed — it appears on both sides and cancels. Once it does, circle area equals square area, and one square root gives r = 2π\frac{2}{\sqrt{π}}.

  • Label the three regions
  • Set the two areas equal
  • Solve for the radius

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