Competition problem — Combinatorics

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Combinatorics Difficulty 5.6 AIME, harder Prove it JBMO

Problem:

Along a round table are arranged 11 cards with the names (all distinct) of the 11 members of the $16^{ ext{th}}$ JBMO Problem Selection Committee. The distances between each two consecutive cards are equal. Assume that in the first meeting of the Committee none of its 11 members sits in front of the card with his name. Is it possible to rotate the table by some angle so that at the end at least two members sit in front of the card with their names?

Solution

Solution:

Yes it is: Rotating the table by the angles $\frac{360^{\circ}}{11}, 2 \cdot \frac{360^{\circ}}{11}, 3 \cdot \frac{360^{\circ}}{11}, \ldots, 10 \cdot \frac{360^{\circ}}{11}$ , we obtain 10 new positions of the table. By the assumption, it is obvious that every one of the 11 members of the Committee will be seated in front of the card with his name in exactly one of these 10 positions. Then by the Pigeonhole Principle there should exist one among these 10 positions in which at least two of the $11(>10)$ members of the Committee will be placed in their positions, as claimed.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.