Difference between revisions of "1998 IMO Problems/Problem 2"
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− | where the inequality follows from AM-QM. Since <math>b</math> is odd, <math>b^2 - 2b</math> is not divisible by <math>4</math> and we can | + | where the inequality follows from AM-QM. Since <math>b</math> is odd, <math>b^2 - 2b</math> is not divisible by <math>4</math> and we can strengthen the inequality to |
<cmath>{c_i \choose 2} + {{b - c_i} \choose 2} \geq \frac{b^2 - 2b + 1}{4} = \left(\frac{b - 1}{2}\right)^2.</cmath> | <cmath>{c_i \choose 2} + {{b - c_i} \choose 2} \geq \frac{b^2 - 2b + 1}{4} = \left(\frac{b - 1}{2}\right)^2.</cmath> |
Revision as of 14:28, 30 May 2024
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
where the inequality follows from AM-QM. Since is odd, is not divisible by and we can strengthen the inequality to
Letting stand for the number of instances where two judges agreed on a candidate, it follows that
The given condition on implies that
Therefore
which simplifies to
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 |