1998 IMO Problems/Problem 2: Difference between revisions
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==Problem== | |||
In a competition, there are <i>a</i> contestants and <i>b</i> judges, where <i>b</i> ≥ 3 is an odd | In a competition, there are <i>a</i> contestants and <i>b</i> judges, where <i>b</i> ≥ 3 is an odd | ||
integer. Each judge rates each contestant as either “pass” or “fail”. Suppose <i>k</i> | integer. Each judge rates each contestant as either “pass” or “fail”. Suppose <i>k</i> | ||
is a number such that, for any two judges, their ratings coincide for at most <i>k</i> | is a number such that, for any two judges, their ratings coincide for at most <i>k</i> | ||
contestants. Prove that <i>k</i>/<i>a</i> ≥ (<i>b</i> − 1)/(2<i>b</i>). | contestants. Prove that <i>k</i>/<i>a</i> ≥ (<i>b</i> − 1)/(2<i>b</i>). | ||
==Solution== | |||
{{solution}} | |||
==See Also== | |||
{{IMO box|year=1998|num-b=1|num-a=3}} | |||
Revision as of 22:47, 18 November 2023
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
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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 | ||
