Problem 6 · 2026 AMC 8
Medium
Geometry & Measurement
area-decomposition
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Answer: E — 2/5.
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Hint 1 of 2
The reachable border is a frame around the field — an awkward shape. Don't measure it directly; what's the easy shape you'd subtract to leave only the frame?
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Hint 2 of 2
Find the unreachable inner rectangle instead (everything more than 1 m from every edge): a 1 m strip on each side shrinks each dimension by 2. Then border = whole − inner.
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Approach: complementary counting — subtract the easy inner rectangle
- The reachable strip is a frame, which is fiddly to measure head-on. Flip it: the unreachable part is the rectangle more than 1 m from every edge — a clean rectangle.
- A 1 m margin on each side trims 1 + 1 = 2 from each dimension: the inner rectangle is (10 − 2) × (8 − 2) = 8 × 6 = 48, out of the full 10 × 8 = 80.
- So the reachable frame is 80 − 48 = 32, giving the fraction 32/80 = 2/5.
- Why this transfers: a border/frame is almost always easiest as (whole rectangle) − (inner rectangle) — and remember a uniform margin shrinks each side by twice the margin, not once.
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