Problem 4 · AMC 8 Stretch
Stretch
Geometry & Measurement
considering-extreme-casesvisual-representation
Two circles share the same center. A chord \(AB\) of the big circle just touches (is tangent to) the small circle. If \(AB = 8\), find the area of the ring (the shaded region between the two circles).
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Answer: \(16\pi\)
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Hint 1 of 3
Notice that the problem never tells you the sizes of the circles, yet it expects one answer. That is a big clue: the answer must NOT depend on the sizes. So try an extreme case.
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Hint 2 of 3
Shrink the small circle down until it is just a tiny dot at the center. The chord still has to touch it, so now the chord passes right through the center — which makes \(AB\) a diameter of the big circle.
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Hint 3 of 3
If \(AB=8\) is a diameter, the big radius is \(4\). With the small circle gone, the ring is now the whole big circle. Find its area.
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Approach: Considering an extreme case — shrink the inner circle to a point
- Because the problem gives no circle sizes but wants one answer, the answer cannot depend on the radii. So we are free to pick a convenient extreme case.
- Extreme case: shrink the inner circle to a single point at the center. The chord \(AB\) still must touch it, so \(AB\) now goes straight through the center, meaning \(AB\) is a diameter of the big circle. Since \(AB=8\), the big radius is \(4\).
- With the inner circle gone, the ring becomes the entire big circle: area \(= \pi r^2 = \pi (4)^2 = 16\pi\).
- Check with the Pythagorean theorem: the radius to the touch point is perpendicular to \(AB\) and cuts it in half, giving half-chord \(4\), so \(R^2 - r^2 = 4^2 = 16\), and the ring area \(\pi R^2 - \pi r^2 = 16\pi\) — the same answer.
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