A meeting has R participants. All participants arrive and leave at different times, but it is always true that within any 3 of them there is always at least 1, who has met the other 2. Prove that then there is at least one person who met with all the other participants.
2026-05-05 05:49:04.1777960144
Meeting with R participants
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If you replace the phrase "has met" with "were both present at the same time", then the solution (for $R \geq 3$) is as follows:
Consider the first person to leave and the last person to arrive. If the first person left after the last person arrived, then there was a time when everyone was present at once. QED
If the first person left before the last person arrived, then consider any third person. By your criteria, of the group (first, last, other), one of them must have been present when each of the others was present, and since there was no overlap for first & last, that person must be other. Since everyone left after (or at the same time as) first left and everyone arrived before (or at the same time as) last arrived, this other person must have been present at the same time as each other person.
In fact, everyone except the first to leave and last to arrive must have had an overlap with every other person.
Which brings us to the case where $R = 2$. If the first person left before the last person arrived, the criteria are still satisfied since there are no groups of three people. On the other hand, if there were only two participants and the first one left before the second arrived, it can hardly be called a "meeting."