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

Title: Problem with a square
Post by x2862 on Mar 24th, 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 24th, 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 24th, 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 24th, 2010, 11:00am
I don't see any way to use that.

Title: Re: Problem with a square
Post by Aryabhatta on Mar 24th, 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.



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