Monday, April 14, 2008


While tutoring the other day I came across a puzzle I don't know how to do. This isn't exactly noteworthy, as I can't do most puzzles. But it seemed like an interesting enough puzzle, so I thought I'd give it broader attention:

50 watches are placed at random on a table. Show that at some time 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.

The problem didn't say anything about the sizes of the watches (like, if they were all the same). Also, I guess we should say that they are all facing up (does it matter?). I think the problem did mention that the watches weren't all necessarily all set to the same time, just that they were all running at the same, accurate, pace.

So, any thoughts?


Monkey Sri said...

Hi, I'm here via C'ville Blogs. I'm no mathematician, but I was intrigued by this puzzle - thought I'd give it a shot. Call the distance from the center of the table to the center of watch A as C(a), and to the end of the minute hand of watch A as M(a). If you draw a line between the center of the table and the center of watch A, then draw an arc across the watch face, the arc should cover the time when M(a) is less than or equal to C(a), right? And because C(a) is at the apex, that arc should cover less than 50% of the watch face - so more than 50% of the time, M(a) is greater than C(a). You could try to set watch B to a time where it could cancel out the effect of watch A: M(b) being < C(b) when M(a) > C(b), and vice versa, but the arc is not big enough. There will be some time when both watches' minute hands are outside the arc, and thus M(a+b) > C(a+b). I'm not sure how the fact that there are fifty watches comes into play, or if it matters at all. Does that make sense?

sumidiot said...

That seems to be about the idea. For convenience, I'll say a watch is "good" at some particular time if the minute hand is further than the center of the watch, from the center of the table. So, each watch is "good" more than half the time. That is, on average, each watch is "good". Adding up 50 watches that are on average "good" should make the average distance from the minute hands larger than the average (constant) distance from the centers of the watches. For the average of the minute hands distances to be bigger, there must be a time when the sum of their distances is bigger.

Or something like that. Averaging averages and such always makes me quesy, but that seems to be about right.