1998 IMO Problems/Problem 2
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 
stand for the number of judges who pass the 
th candidate. The number of pairs of judges who agree on the 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 
is odd, 
is not divisible by 
and we can improve the inequality to
![\[{c_i \choose 2} + {{b - c_i} \choose 2} \geq \frac{b^2 - 2b + 1}{4} = \left(\frac{b - 1}{2}\right)^2.\]](http://latex.artofproblemsolving.com/b/1/7/b171b18ac10a2c0f0a7bbd1eabf1410edeba864b.png)
Letting 
stand for the number of instances where two judges agreed on a candidate, it follows that
![\[N = \sum_{i = 1}^a {c_i \choose 2} + {{b - c_i} \choose 2} \geq a \cdot \left( \frac{b - 1}{2} \right)^2.\]](http://latex.artofproblemsolving.com/d/d/2/dd2364f4214f7010c2137216e277e6e54722e63e.png)
The given condition on 
implies that
![\[N \leq k \cdot {b \choose 2} = \frac{kb(b - 1)}{2}.\]](http://latex.artofproblemsolving.com/d/8/2/d822e4cde94139c63856f9ca61185a3157b3afb3.png)
Therefore
![\[a \cdot \left( \frac{b - 1}{2} \right)^2 \leq \frac{kb(b - 1)}{2},\]](http://latex.artofproblemsolving.com/c/6/6/c66c6465bc61dc52b9e264ec903870f9bd984f66.png)
which simplifies to
![\[\frac{k}{a} \geq \frac{b - 1}{2b}.\]](http://latex.artofproblemsolving.com/b/3/1/b3122f2a058fc60b564edde0125e319f117b2450.png)
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 | ||
