Problem 18 · 2024 AMC 8
Hard
Geometry & Measurement
areaarea-fraction
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Answer: A — 108°.
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Hint 1 of 2
"Shaded = unshaded" is the gift: each must be exactly HALF the whole disk. So you don't compare two messy regions — you just set shaded = half the total.
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Hint 2 of 2
Break the disk into three rings (inner disk, ring 1–2, ring 2–3) and find each area from πr2. The only part the angle controls is the θ-sector of the outer ring.
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Approach: shaded equals half the disk; only the outer sector depends on the angle
- The phrase "shaded = unshaded" means each is exactly half the disk — so skip comparing regions and just set shaded = half the total. Total disk area = π·32 = 9π, so shaded must be 4.5π.
- Now tally the shaded parts in rings. The fully-shaded inner annulus (radii 1 to 2) has area π(22 − 12) = 3π, no angle involved. The outer annulus (radii 2 to 3) has area π(32 − 22) = 5π, but only the θ-slice of it is shaded: θ360(5π).
- Set shaded = half: 3π + θ360(5π) = 4.5π. The 3π already covers most of the half, leaving the sector to supply just 1.5π: θ360(5π) = 1.5π → θ360 = 0.3 → θ = 108°.
- This transfers: when a problem says two regions are equal, immediately rewrite it as "each = half the whole" — one equation instead of two, and the constant parts often vanish into the half.
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