A math problem I picked up while I was in the math lounge today:
Given a simple closed curve in the 2-dimensional plane, can one always find a square whose four vertices lie on the curve? (We found a nice proof for the case when the square is replaced by a rectangle. I don't know if one exists for the squares.)
First person to solve this problem gets... to be my hero. Or not :P
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I remember a bunch of us in PMC tried to solve this...we only got as far as showing it for rectangles, but I'm sure your proof is nicer. We essentially used a 2-D variant of IVT on the angle, when you sweep a line and its perpendicular companion out from a point, and while you vary the initial point around. I'm going to go out on a limb and say not many people can solve this one ;)
ReplyDeletehttp://quomodocumque.wordpress.com/2007/08/31/inscribed-squares-denne-speaks/
ReplyDeleteFirst person to solve it gets to publish her results in a journal!