I think this is from the 1976 Soviet mathematical olympiad:
Fifty accurate watches lie on a table. Prove that there exists a moment in time when the sum of the distances from the center of the table to the ends of the minute hands is greater than the sum of the distances from the center of the table to the centers of the watches.
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Imagine there’s a single watch on the table, directly north of the center, with 6 pointing due south and the minute hand at 3 o’clock. At that moment, the watch fulfills the condition described: The distance from the table’s center to the watch’s center is (slightly) less than the distance from the table’s center to the end of the minute hand. This shows that, as the minute hand turns, the watch spends most of its time in this state. And this remains true regardless of the watch’s position on the table or its orientation.
Add 49 more watches and the same must be true of them all, on average. So sometime during the hour the required state must be achieved.
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