AMC 8 · 2010 · #23

Grade 8 geometry-2d

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

area-circlespythagorean-theoremcoordinate-geometryratio-proportion identify-subproblemscoordinate-geometry ↑ Prerequisites: area-circlespythagorean-theorem

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A circle O is centered at the origin and passes through the four points P(-1,1), Q(1,1), R(-1,-1), and S(1,-1). Two semicircles are drawn: one on chord PQ as its diameter (passing through center O) and one on chord RS as its diameter (also passing through O). What is the ratio of the combined area of the two semicircles to the area of circle O?

Pick an answer.

(A)

$\frac{\sqrt{2}}{4}$

(B)

$\frac{1}{2}$

(C)

$\frac{2}{\pi}$

(D)

$\frac{2}{3}$

(E)

$\frac{\sqrt{2}}{2}$

AMC 8 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Use Coordinates

The figure is already drawn on the coordinate plane with the four corner points labeled, so Tool #11 (Use Coordinates) turns every length into a distance calculation. Tool #7 (Identify Subproblems) splits the question into two clean pieces — find the big circle's radius from O to P, and find each semicircle's radius from the chord length PQ — so the area ratio at the end is just $\frac{2 · \frac{1}{2}π r\_small²}{π r\_big²}$ .

1STEP 1

O=(0,0) and P(-1,1) lies on the circle, so by the Pythagorean Theorem the big radius is OP = √(2).

r_big = √((-1-0)² + (1-0)²) = √(1 + 1) = √(2)

2STEP 2

Plug r=√(2) into A=π r² to get circle O's area = 2π.

A_big = π (√(2))² = 2π

3STEP 3

PQ is horizontal from (-1,1) to (1,1), so PQ=2; halving that diameter gives each semicircle radius 1.

PQ = 1 - (-1) = 2 → r_small =

\frac{PQ}{2}

= 1

4STEP 4

Each semicircle has area $\frac{1}{2}$ π(1)² = $\frac{π}{2}$ , so the two together make π — one unit circle's worth.

A_semis = 2 ·

\frac{1}{2}

π (1)² = π

5STEP 5

Divide the semicircle area by circle O's area: $\frac{π}{2π}$ = $\frac{1}{2}$ , choice (B).

\frac{A\_semis}{A\_big}

\frac{π}{2π}

\frac{1}{2}

→ (B)

Answer

\frac{1}{2}

The big circle has radius √(2) ≈ 1.41, so its area 2π is exactly twice the area of a unit circle. The two semicircles glue together (in area) into one unit circle of area π. So the ratio must be π : 2π = 1 : 2, matching answer (B). The other choices fail simple sanity checks:

\frac{2}{π}

≈ 0.637 would be a dimensional mismatch (no stray π survives the ratio), and

\frac{√(2)}{4}

\frac{√(2)}{2}

would require some length, not area, in the ratio.

💡Key takeaway

This AMC 8 geometry problem only needs the Grade 8 distance formula and the Grade 7 circle-area formula that you already know!

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