We have $101$ tenorist, every two cooperated in exactly one concert, but there is no concert in which all participated. Prove that someone participated in at least $11$ concerts.
This is an old problem from Moscow math Olympiad. I did try to solve it but somehow I can't. Any thoughts?
Say we have $T_1,T_2,...,T_{101}$ tenorists and $A_1,A_2,...A_n$ concerts. So every pair $\{T_i,T_j\}$ is ''connected'' to exactly one concert. So we have:
$$ \sum {\deg(A_i)\choose 2} = \sum \deg(\{T_i,T_j\}) = {101\choose 2}\;\;\;\;\;(1)$$
We have to prove that the degree of some $T_j$ is at least $11$. Suppose there is no such $j$, then for every $T_j$ the degree is at most $10$ and we have: $$ \sum \deg(A_i) = \sum \deg(T_i) \leq 1010 \;\;\;\;\;(2)$$
I don't know what to do now.
Take an arbitrary tenorist, say $T_1$. He is involved in at most $10$ concerts, which combined pairs him once with each of the remaining $100$ tenorists. Hence, at least one of these concerts, say $A_k$, must involve at least $11$ tenorists.
Given concert $A_k$ involving $m\ge11$ tenorists, and a tenorist $T_p$ not in $A_k$, for each tenorist $T_i$ in $A_k$ there will be a unique concert which includes $T_i$ and $T_p$. This gives a list of $m$ concert, one for each $T_i\in A_k$, and these must all be distinct concert different from $A_k$ since no two members of $A_k$ may be in another concert.
So, this gives $m\ge11$ concert in which $T_p$ is a member.
Would the result also be true with a smaller number of tenorists? Unless I'm missing something, my proof should also work if there are $92$ tenorists.