1998 IMO Problems/Problem 2

Revision as of 13:55, 30 May 2024 by Nedsanger (talk | contribs) (Solution)

Problem

In a competition, there are a contestants and b judges, where b ≥ 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 k/a ≥ (b − 1)/(2b).

Solution

Let $c_i$ stand for the number of judges who pass the $i$th candidate. The number of pairs of judges who agree on the $i$th contestant is then given by

\begin{align*} {c_i \choose 2} + {{b - c_i} \choose 2} &= \frac{1}{2}\left(c_i(c_i - 1) + (b - c_i)(b - c_i - 1) \right) \\ &= \frac{1}{2}\left( c_i^2 + (b - c_i)^2 - b \right) \\ &\geq \frac{1}{2}\left( \frac{b^2}{2} - b \right) \\ &= \frac{b^2 - 2b}{4} \end{align*}

where the inequality follows from AM-QM. Since $b$ is odd, $b^2 - 2b$ is not divisible by $4$ and we can improve the inequality to

\[{c_i \choose 2} + {{b - c_i} \choose 2} \geq \frac{b^2 - 2b + 1}{4} = \left(\frac{b - 1}{2}\right)^2.\]

Letting $N$ stand for the number of instances where two judges agreed on a candidate, it follows that

\[N = \sum_{i = 1}^n {c_i \choose 2} + {{b - c_i} \choose 2} \geq a \cdot \left( \frac{b - 1}{2} \right)^2.\]

The given condition on $k$ implies that

\[N \leq k \cdot {b \choose 2} = \frac{kb(b - 1)}{2}.\]

Therefore

\[a \cdot \left( \frac{b - 1}{2} \right)^2 \leq \frac{kb(b - 1)}{2},\]

which simplifies to

\[\frac{k}{a} \geq {b - 1}{2b}.\]

See Also

1998 IMO (Problems) • Resources
Preceded by
Problem 1
1 2 3 4 5 6 Followed by
Problem 3
All IMO Problems and Solutions