Problem 25 · 2017 AMC 8
Hard
Geometry & Measurement
areaarea-decomposition
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Answer: B — 4√3 − 4π/3.
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Hint 1 of 2
The curvy region is awkward, but the two arcs are bites taken out of a clean shape. The trick with weird areas: find the simple figure they were carved from, then subtract the bites.
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Hint 2 of 2
Extend US and UT past S and T by the radius 2. The 60° at U and the equal sides make the completed figure an equilateral triangle of side 4 — and each arc is a 60° (one-sixth) sector of radius 2 scooped out of it.
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Approach: complete to a clean equilateral triangle, subtract the scooped sectors
- The region looks hard, so rebuild the simple shape behind it: extend US and UT past S and T by 2 each (the arc radius). With the 60° apex at U and two equal sides of 2 + 2 = 4, the angles force a 60-60-60 equilateral triangle of side 4. That reframing — carved shape = full triangle − bites — is the whole idea.
- Equilateral triangle area (side s): s2√34 = 42√34 = 4√3.
- The two bites are each a 60° sector (one-sixth of a circle, radius 2): sector area = πr2/6 = 4π/6 = 2π/3, so two of them total 4π/3.
- Region = 4√3 − 4π/3 (choice B).
- Sanity check: 4√3 ≈ 6.93 and 4π/3 ≈ 4.19, leaving ≈ 2.7 — a small positive area, right for this slim region. Why this transfers: for any region bounded by arcs, look for the straight-sided figure it was cut from and subtract (or add) the circular pieces.
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