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1989 USAMO Problems/Problem 2

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Problem

The 20 members of a local tennis club have scheduled exactly 14 two-person games among themselves, with each member playing in at least one game. Prove that within this schedule there must be a set of 6 games with 12 distinct players

Solution

Solution 1

Consider a graph with 20 vertices and 14 edges. The sum of the degrees of the vertices is 28; by the Pigeonhole Principle at least 12 vertices have degrees of 1 and at most 8 vertices have degrees greater than 1. If we keep deleting edges of vertices with degree greater than 1 (a maximum of 8 such edges), then we are left with at least 6 edges, and all of the vertices have degree either 0 or 1. These 6 edges represent the 6 games with 12 distinct players.

Solution 2

Let a slot be a place we can put a member in a game, so there are two slots per game, and 28 slots total. We begin by filling exactly 20 slots each with a distinct member since each member must play at least one game. Let there be m games with both slots filled and n games with only one slot filled, so 2m+n=20. Since there are only 14 games, m+n \leq 14 \Longrightarrow 2m+n \leq 14+m \Longleftrightarrow 20 \leq 14+m \Longrightarrow m \geq 6, so there must be at least 6 games with two distinct members each, and we must have our desired set of 6 games.

See also

1989 USAMO (Problems)
Preceded by
Problem 1
1 2 3 4 5 Followed by
Problem 3
All USAMO Problems and Solutions
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