Problem 15 · AMC 8 Stretch
Stretch
Geometry & Measurement
Counting & Probability
pigeonholevisual-representation
Seven points are placed inside a circle of radius 1. Show that 2 of the points are less than distance 1 apart.
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Answer: two points less than distance 1 apart
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Hint 1 of 4
Use the center to slice the disk into equal wedges (sectors) where any two points are close.
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Hint 2 of 4
Draw 6 radii to cut the disk into 6 equal wedges. What is the angle at the center of each wedge?
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Hint 3 of 4
Each wedge has a \(60^\circ\) angle. The two longest sides of a wedge are radii of length 1, and they meet at \(60^\circ\) — so the wedge is never wider than 1.
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Approach: Pigeonhole — 7 points into 6 sixty-degree wedges
- Draw 6 radii from the center, splitting the disk into 6 equal wedges, each with a \(60^\circ\) angle. These 6 wedges are the boxes.
- Drop the 7 points into the 6 wedges (the center may be counted in any one). Since \(7 > 6\), some wedge holds at least 2 points.
- In one wedge, the two straight sides are radii of length 1 meeting at \(60^\circ\); the greatest distance between two points there is exactly 1 (the radii tips form an equilateral triangle of side 1).
- So two points sharing a wedge are less than distance 1 apart.
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