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.

06/20/2026 UPDATE: Reader Marjan Milanovič writes:

You mention that the key component in the proof is that the individual watches spent most of the time in a desired state. Unfortunately, that is not enough for a correct proof — we would have to show that when the watch is in an undesired state, ‘the difference is small enough’. More formally, if p and d(t) are distances from the center of the table to the center of the watch and to the tip of the minute hand respectively, we need to show

But I see an elementary proof: if at a certain time the watches do not fulfil the condition than after 30 minutes they certainly do! It boils down to the triangle inequality.

(Thanks, Marjan.)