20070725, 12:49  #1 
Dec 2005
2^{2}·7^{2} Posts 
Problem 3 from IMO 2007
In a mathematical competition some competitors are friends. Friendship is always mutual. Call a group of competitors a clique if each two of them are friends. (In particular, any group of fewer than two competitiors is a clique.) The number of members of a clique is called its size.
Given that, in this competition, the largest size of a clique is even, prove that the competitors can be arranged into two rooms such that the largest size of a clique contained in one room is the same as the largest size of a clique contained in the other room. A small postscriptum: I don't have a solution yet 
20070725, 14:03  #2  
"Bob Silverman"
Nov 2003
North of Boston
1D58_{16} Posts 
Quote:


20070725, 14:08  #3 
Dec 2005
11000100_{2} Posts 
like in one of my algebra textbooks straight after a theorem:
the proof is long and tedious, is done by induction and is therefore left as an exercise... 
20070725, 22:09  #4 
Jun 2003
The Texas Hill Country
3^{2}·11^{2} Posts 
OK, I'm confused.
Assume that A, B and C are competitors, and that A is a friend of B, and also that B is a friend of C. Therefore we have the cliques {}, {A}, {B}, {C}, {A,B}, {B,C}. Is it true that we additionally have the cliques {A,C} and {A,B,C} iff A and C are also friends? Or are we to interpret "mutual" in some way to unconditionally infer the transitive relationship? Last fiddled with by Wacky on 20070725 at 22:14 
20070725, 23:34  #5 
Aug 2002
Ann Arbor, MI
433 Posts 
"Is it true that we additionally have the cliques {A,C} and {A,B,C} iff A and C are also friends?"
Yes. "Or are we to interpret "mutual" in some way to unconditionally infer the transitive relationship?" No, mutual (aka, symmetric) does not imply transitive. 
20070726, 07:09  #6 
Dec 2005
2^{2}·7^{2} Posts 
Mutual does (as I interpret the question) indeed not imply transitivity.
Note that the question becomes trivial if friendship is transitive 
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