Problem 12 · 2023 AMC 8
Hard
Geometry & Measurement
areaarea-decomposition
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Answer: B — 11/36.
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Hint 1 of 2
First use the faint grid to read off every radius (3, 2, 1, and 12). Then notice the big shaded disk has two white circles biting into it — so its shaded area is disk minus those bites.
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Hint 2 of 2
Because every area carries a factor of π, you can drop the π and just compare radius-squared values: outer = 9, big disk = 4, each inner white = 1, each tiny shaded = 14. The π cancels in the final fraction.
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Approach: sum shaded, subtract carved-out whites
- Read the radii off the grid: outer circle 3, big shaded disk 2, the two white circles inside it 1 each, the three tiny shaded circles 12 each. Since we want a fraction, the π will cancel — so really just work with radius-squared.
- Big shaded disk has area 4π, but two white circles (1π each) eat into it: net shaded there = 4π − 2π = 2π.
- Three tiny shaded circles add 3 × π4 = 3π4.
- Total shaded = 2π + 3π4 = 11π4, over the outer 9π: 11π/49π = 1136. Worth keeping: in any ‘what fraction is shaded’ circle problem the π always cancels — compare radius² values and skip π entirely.
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