1998 IMO Problems/Problem 2

Problem

In a competition, there are $a$ contestants and $b$ judges, where $b\ge3$ 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 $\frac{k}{a}\ge\frac{b-1}{2b}$ .

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

$\begin{aligned} (\binom{c_{i}}{2}) + (\binom{b - c_{i}}{2}) & = \frac{1}{2} (c_{i} (c_{i} - 1) + (b - c_{i}) (b - c_{i} - 1)) \\ = \frac{1}{2} (c_{i}^{2} + (b - c_{i})^{2} - b) \\ \geq \frac{1}{2} (\frac{b^{2}}{2} - b) \\ = \frac{b^{2} - 2 b}{4} \end{aligned}$

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

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

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

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

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

Page

Toolbox

Search

1998 IMO Problems/Problem 2

Problem

Solution

See Also