Difference between revisions of "1998 IMO Problems/Problem 2"

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==Problem==
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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>).
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==Solution==
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{{solution}}
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==See Also==
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{{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