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        riddles >> putnam exam (pure math) >> Problem with a square
        
(Message started by: x2862 on Mar 24^th, 2010, 2:48am)

Title: Problem with a square
Post by x2862 on Mar 24^th, 2010, 2:48am

I'm pretty good at coordinate geometry but this problem a friend gave me is stumping me. any ideas? Should I be using the law of cosines?

A square with a side of length x has a interior point of p. the location of point p is shuch that the distance to 3 corners of the square is 500, 300, and 400 feet measured in a clockwise direction.what is the length of the side of the square?

Title: Re: Problem with a square
Post by towr on Mar 24^th, 2010, 4:38am

You can solve it algebraically. There are two ways to put the three distances in the square (see attachment). Putting the origin at the left lower corner, you can pick an x-coordinate for P, and calculate the rest from there by solving
x = sqrt(3²-a²) + sqrt(4²-a²)
x = a + sqrt(5²-(3²-a²))

or for the second case
x = sqrt(4²-b²) + sqrt(5²-b²)
x = b + sqrt(3²-(4²-b²))

It doesn't give particularly nice answers though.

Title: Re: Problem with a square
Post by Aryabhatta on Mar 24^th, 2010, 10:20am

Perhaps this is useful: http://www.cut-the-knot.org/proofs/swivel.shtml

Title: Re: Problem with a square
Post by towr on Mar 24^th, 2010, 11:00am

I don't see any way to use that.

Title: Re: Problem with a square
Post by Aryabhatta on Mar 24^th, 2010, 5:57pm

on 03/24/10 at 11:00:17, towr wrote:

I don't see any way to use that.

Sorry about that, I posted that link without trying it myself. In any case, it is kind of related to the problem and some people might find it interesting.