TITLE: IMO 1998 - Combinatorics
QUESTION [7 upvotes]: I am trying this IMO Combinatorics problem $1998$ P2 that goes like this:

In a competition, there are $m$ contestants and $n$ judges, where $n \geq 3$ is an odd integer. Each judge rates each contestant as either “pass” or “fail”. Suppose $k$ is a number such that, for any two judges, their ratings coincide for at most $k$ contestants. Prove that $$\frac{k}{m}\geq \frac{n-1}{2n}$$

I am completely puzzled on how to start, could you give me any hints?

REPLY [2 votes]: Consider the number of combinations $(\{j_1, j_2\},c)$,  where $\{j_1, j_2\}$ is a pair of distinct judges, and $c$ is a contestant that they agree on.  You can arrive at this quantity two ways:

Sum over contestants, the number of pairs of judges who agree on them.

Sum over pairs of judges, the number of contestants they agree on.


Then the quantity being summed in 1. can be bounded below by an expression involving $n$ (remember $n$ is odd), whilst the quantity being summed in 2. may be bounded above by $k$.  Combining yields the desired inequality.