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        riddles >> medium >> 50 Watches
        
(Message started by: Aryabhatta on Oct 12^th, 2007, 4:04pm)

Title: 50 Watches
Post by Aryabhatta on Oct 12^th, 2007, 4:04pm

There are 50 accurate watches lying on a circular table.

Show that at some point in time, the sum of distances from the center of the table to the ends of the minute hands is more than the sum of distances from the center of the table to the centres of the watches.

Title: Re: 50 Watches
Post by FiBsTeR on Oct 12^th, 2007, 4:38pm

Is it required that the clocks be oriented the same way? That is, if one clock's minute hand is facing north at a given time, are all of the minute hands pointed the same way? Even though the watches are all accurate, one might be upside-down, for example.

Title: Re: 50 Watches
Post by Aryabhatta on Oct 12^th, 2007, 4:50pm

For now, let's assume they are oriented the same way.

Later we might be able to prove it even for random orientation perhaps (Basically, i don't know :P)

Title: Re: 50 Watches
Post by FiBsTeR on Oct 12^th, 2007, 5:09pm

If you considered two watches both with their center at the center of the table, one watch turned 180^o with respect to the other, the sum to the ends of the minute hands will always be zero, equal to the sum of the distances to the centers.

Oops.

Title: Re: 50 Watches
Post by Aryabhatta on Oct 12^th, 2007, 5:14pm

on 10/12/07 at 17:09:08, FiBsTeR wrote:

No... distances are always non-negative. They can't sum to zero unless they are all zero.

Anyway, if they can be oriented anyway i think we can easily give counterexamples. So assume they are oriented the same way.

Title: Re: 50 Watches
Post by Grimbal on Oct 13^th, 2007, 1:33pm

Consider one watch. If you project the end of the minute hand (call it M) on the line joining the center of the watch with the center of the table, and consider the projection P of M on this line, then the distance of P to T: the center of the table, has a nice sine component that averages to zero. What remains is the distance from T to W: the center of the watch.
Now if you consider the actual end of the minute hand, MT is always larger than PT, because it is the hypotenuse of a triangle where PT is a small side. So, averaged over a day, MT is larger than PT, which equals WT.
Now consider 50 watches. 50 is 2 times 25, but the day has 24 hours. Over 2 days, you have 2 watches in excess. That is not neat, so you discretely pocket them.
Anyway, since for each watch the average distance MT exceeds the distance WT, it must also be true over the sum of all watches. On average, sum(MT) exceeds sum(WT). If sum(MT) never exceeded sum(WT), then the average of sum(MT) over a day cold not exceed the average of sum(W). Therefore, there must be a time interval over which the sum of distances of the ends of the minute hands to the center of the table exceeds the sum of the distances fo the centers of the watches to the table center.

Title: Re: 50 Watches
Post by Aryabhatta on Oct 14^th, 2007, 4:07pm

Well done, Grimbal. I had something similar...

I believe this problem is from All Soviet Union Maths Olympiad 1976.