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x2862
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 Problem with a square   « on: Mar 24th, 2010, 2:48am » Quote Modify

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?
 « Last Edit: Mar 24th, 2010, 3:03am by x2862 » IP Logged
towr
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 Re: Problem with a square   rect3312.png « Reply #1 on: Mar 24th, 2010, 4:38am » Quote Modify

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.
 « Last Edit: Mar 24th, 2010, 4:46am by towr » IP Logged

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Aryabhatta
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 Re: Problem with a square   « Reply #2 on: Mar 24th, 2010, 10:20am » Quote Modify

Perhaps this is useful: http://www.cut-the-knot.org/proofs/swivel.shtml
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towr
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 Re: Problem with a square   « Reply #3 on: Mar 24th, 2010, 11:00am » Quote Modify

I don't see any way to use that.
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Aryabhatta
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 Re: Problem with a square   « Reply #4 on: Mar 24th, 2010, 5:57pm » Quote Modify

on Mar 24th, 2010, 11:00am, 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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