AMC 8 · 2023 · #12

Grade 7 geometry-2d

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

area-circlesratio-proportionfraction-arithmetic area-differenceidentify-subproblems ↑ Prerequisites: area-circlesfraction-arithmetic

📏 Long solution 💡 3 insights 📊 Diagram

Problem

A large white circle is sitting on a unit grid. Inside it there is one big shaded circle, two white circles cut out of the shaded one, and three small shaded circles floating elsewhere. Using the grid to read off each radius, what fraction of the area of the large white circle is actually shaded?

Pick an answer.

(A)

$\frac{1}{4}$

(B)

$\frac{11}{36}$

(C)

$\frac{1}{3}$

(D)

$\frac{19}{36}$

(E)

$\frac{5}{9}$

AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The figure is the problem. Tool #1 (Draw a Diagram) tells us to read radii straight off the grid — every circle's diameter is a whole or half number of grid squares, so no algebra is needed. Tool #7 (Identify Subproblems) splits the shaded region into three easy pieces: (a) the big shaded disk, (b) MINUS the two inner white disks that punch holes in it, (c) PLUS the three little shaded disks elsewhere. Compute the three pieces, combine, then divide by the area of the large white circle.

1STEP 1

Use the grid as a ruler: the big white circle spans 6 units so r=3, the shaded disk r=2, inner whites r=1, tiny shaded r= $\frac{1}{2}$ .

r_big=3, r_shaded=2, r_white=1, r_small=

\frac{1}{2}

2STEP 2

Big white circle is the denominator: A=πr² with r=3 gives 9π, the whole we compare against.

A_big white=π(3)²=9π

3STEP 3

Big shaded disk: same formula with r=2 gives π·2²=4π — the shading before the holes are cut.

A_shaded disk=π(2)²=4π

4STEP 4

Two inner white circles: each r=1 gives π, so 2π gets subtracted from the shaded disk.

2×π(1)²=2π

5STEP 5

Three tiny shaded circles: each r= $\frac{1}{2}$ gives $\frac{\pi}{4}$ , so $\frac{3\pi}{4}$ is added — they overlap nothing.

3×π(

\frac{1}{2}

)²=3×

\frac{\pi}{4}

\frac{3\pi}{4}

6STEP 6

Combine: 4π − 2π + $\frac{3\pi}{4}$ = 2π + $\frac{3\pi}{4}$ = $\frac{11\pi}{4}$ of shaded area.

A_shaded=4π-2π+

\frac{3\pi}{4}

\frac{8\pi+3\pi}{4}

\frac{11\pi}{4}

7STEP 7

Divide shaded by whole: (11π/4)/(9π); π cancels to give $\frac{11}{36}$ , choice (B).

A_shaded/(A_big white)=11π/4/9π=

\frac{11}{4· 9}

\frac{11}{36}

→ (B)

Answer

\frac{11}{36}

Does

\frac{11}{36}

make sense? The big shaded disk alone covers

\frac{4\pi}{9\pi}

\frac{4}{9}

≈ 0.44 of the big white circle, but the two inner white holes punch out half of that, dropping us close to

\frac{2}{9}

≈ 0.22. Adding three small shaded circles (

\frac{3\pi}{4}

, a smallish amount) nudges us back up to around 0.30 — and

\frac{11}{36}

≈ 0.306 lands exactly there. So (B) is the only choice between

\frac{1}{4}

and

\frac{1}{3}

that fits, which matches our visual estimate.

💡Key takeaway

This AMC 8 problem only needs the Grade 7 circle-area formula π r² you already know — once you have it, the rest is just adding and subtracting circle areas like puzzle pieces!

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