perplexus.info :: Just Math : Math. competition,

Math. competition, (Posted on 2012-03-21)

In a mathematical competition, in which 6 problems were posed to the participants, every two of these problems were solved by more than 2/5 of the contestants.
Moreover, no contestant solved all the 6 problems.
Show that there are at least 2 contestants who solved exactly 5 problems each.

source: IMO 2005

No Solution Yet Submitted by Ady TZIDON

Rating: 3.7500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)

The error of my ways

| Comment 2 of 103 |

Oh, I see where I went wrong.

If there are 7 contestants, each of the 15 pairs needs to be solved by 3 of them, for a minimum total of 45 solved solved pairs. If each student solves 4 problems, then there are only 7*6 = 42 solved pairs, so at least one student must solve 5 problems to bring the total up to 46 solved pairs.

I am not yet clear why 2 contestants must solve 5, but I guess it is because of overlaps. I'll pursue it further this evening.

Posted by Steve Herman on 2012-03-21 11:12:53

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