Problem 11 · AMC 8 Stretch
Stretch
Logic & Word Problems
Geometry & Measurement
proof-by-contradictionpigeonhole
Six round disks lie in the plane. No disk contains the center of any other disk. Prove that the six disks cannot all share a single common point.
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Answer: Proven impossible: a shared point forces two centers within 60°, so one disk would contain another's center
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Hint 1 of 4
Argue by contradiction: pretend all six disks DO share a common point \(P\). Draw a segment from \(P\) out to each of the six centers.
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Hint 2 of 4
The six segments spread out around \(P\), and the angles around a point add up to \(360^\circ\). With six angles sharing \(360^\circ\), at least one angle is \(60^\circ\) or smaller (pigeonhole: they can't all be bigger than \(60\)).
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Hint 3 of 4
Take the two centers making that small angle. Call their distances from \(P\) \(a\) and \(b\), with \(a\) the smaller. In that skinny triangle, the side across from the small (\(\le 60^\circ\)) angle is not the longest side.
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Approach: Proof by contradiction with pigeonhole on the angles around a point
- Suppose, for contradiction, that all six disks share a common point \(P\). Draw the six segments from \(P\) to the centers \(O_1,\dots,O_6\).
- The angles around \(P\) total \(360^\circ\). If all six were bigger than \(60^\circ\) they'd exceed \(360^\circ\), so by pigeonhole at least one angle is \(\le 60^\circ\). Call its two centers \(O_a\) and \(O_b\), at distances \(a\le b\) from \(P\).
- In triangle \(PO_aO_b\) the angle at \(P\) is \(\le 60^\circ\), so it isn't the largest angle, and the side opposite it, \(O_aO_b\), isn't the longest side; hence \(O_aO_b\le b\).
- Since \(P\) is inside disk \(b\), \(b\) is at most that disk's radius, so \(O_aO_b\le b\le(\text{radius of disk }b)\). That puts center \(O_a\) inside disk \(b\) — one disk contains another's center, contradicting the rule. So the six disks cannot all share a common point.
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