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A tennis tournament is played between two teams. Each member of a team plays with one or more members of the other team, so that i) Two members of the same team have exactly one opponent in common. ii) No two members of a team facing together all members of the other team.

Prove that each player must play the same number of matches.

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    $\begingroup$ Why does this have the projective space tag? $\endgroup$ – Jorge Fernández Hidalgo Mar 27 '13 at 19:05
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    $\begingroup$ Are the games singles or doubles?? That is are the games 1vs1 or 2vs2 $\endgroup$ – Jorge Fernández Hidalgo Mar 27 '13 at 19:06
  • $\begingroup$ Because this is an exercise that comes immediately after the definition of projective plane $\endgroup$ – leticia Mar 27 '13 at 19:11
  • $\begingroup$ Probably you are meant to use the duality of points and lines in the projective plane. But the second condition (what you are calling ii) doesn't really make sense as it is written. $\endgroup$ – user641 Mar 27 '13 at 19:31
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    $\begingroup$ @Steve: For a player $t$ let $O(t)$ be the set of opponents faced by $t$. Condition (ii) says that if $t_1$ and $t_2$ are distinct members of one team, then $O(t_1)\cup O(t_2)$ is not the entire opposing team. $\endgroup$ – Brian M. Scott Mar 27 '13 at 20:24
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Call the members of one team points and the members of the other team lines. If a point plays a line, say that the point is incident with the line and vice versa. Condition (i) says that for any two points, there is exactly one line incident with both points, and for any two lines, there is exactly one point incident with both lines. Condition (ii) says that for any two points there is a line that is not incident with either point, and for any two lines there is a point that is not incident with either line. You can check that you now have a finite projective plane and use whatever theorems you already know about projective planes. The relevant ones can also be found here.

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