Do you mind posting the complete set of TST problems and your national olympiad ?

$k=2n-1$. For $k=2n$, there is an obvious counterexample ( two complete graphs with $n$ vertices). For $k=2n-1$, it is a known problem. In Russia, it was proposed by Vladimir Leonidovich Dolnikov to 239 olympiad in 2000-2001 season. As I know, it is even called "a Dolnikov lemma" in graph theoretical community. I wonder to know an Argentinian source.

although it is "known", it is no well-known at all, and is interesting to think about it (in particular, i like my solution, which unfortunately is veeery long). Anyway, thanks for the information. To Orl: I posted only this problem because is the only one that is "interesting" to this forum, we can say that the other five are trivial for you, and I'm sure you will find another source of the problems... Our national olympiad takes place in October, remember we are in the South...

RaMlaF wrote: To Orl: I posted only this problem because is the only one that is "interesting" to this forum, we can say that the other five are trivial for you, and I'm sure you will find another source of the problems... Our national olympiad takes place in October, remember we are in the South... Who says you are forced to post it in the olympiad section?

Me.... Because they are olympiad problems. Even if they are not that hard... This is the olympiad section, not the difficult section... Pierre.

pbornsztein wrote: Me.... Because they are olympiad problems. Even if they are not that hard... This is the olympiad section, not the difficult section... Pierre. OK. Though that should not prevent him from posting the other problems.

So, I wonder, will somebody post a solution? I found in a book something that is the "official" solution. It uses Hall's theorem, to show that for a minimal $n$ that doesn't satisfy the conditions, there exists a smaller $k$ that also doesn't satisfy the condition, so conclusion!

And what do you need else?

My solution is quite long, so i don't want to write it here... but I can tell you the basic idea: For each group $A$ of $2n-1$ boys consider $f(A)$ as the number of complete $n$ graphs among that group. Now, if there is a counterexample for some $n$ choose the one that has minimal $f(A)=k$, and using that for any $k-1$ complete $n$ graph from there there has to be non-empty intersection, one can do some things to show that either the configuration is not a counterexample, or there is a counterexample with $f(A)=k-1$...

it's used for our training also but i don't know the offical solution can anyone post it

~bump , Can someone write a complete solution?

Can someone plz post a complete solution to this? Am weeek in combi

BBUUMMMPPPP

2 solutions from Argentina

spacemaven wrote: i play minecraft lol nice same dude