Problem 6 · AMC 8 Stretch
Core
Geometry & Measurement
considering-extreme-cases
A rope is wrapped tightly around the Earth's equator. Now you add just \(1\) extra meter of rope and spread the slack evenly so the rope floats at the same height all the way around. Could a mouse fit underneath it?
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Answer: Yes — the rope rises about \(\frac{1}{2\pi}\approx 0.159\) m \(\approx 16\) cm everywhere
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Hint 1 of 3
The rope and the Earth's surface are two circles with the same center. When a circle's circumference grows by some amount, how much does its radius grow?
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Hint 2 of 3
If the circumference grows by \(1\) meter, the radius grows by \(\frac{1}{2\pi}\) meters. (From \(C = 2\pi r\), a change in \(C\) of \(1\) means a change in \(r\) of \(\frac{1}{2\pi}\).) Surprisingly, this gap is the same no matter how big the planet is!
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Hint 3 of 3
Work out \(\frac{1}{2\pi}\) in meters, then convert to centimeters and decide if a mouse fits.
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Approach: Considering an extreme case — the gap is independent of the planet's size
- The rope circle and the Earth circle share a center, so the rope's height above the ground is the difference in their radii.
- When a circumference grows by \(1\) meter, the radius grows by \(\Delta r = \frac{\Delta C}{2\pi} = \frac{1}{2\pi} \approx 0.159\) m \(\approx 16\) cm.
- This gap does not depend on the Earth's size at all (the same \(1\) meter of slack on a basketball would lift the rope by the same \(16\) cm).
- A gap of about \(16\) centimeters all the way around is easily enough room for a mouse, so yes — a mouse can fit.
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