Given a rectangle with diagonal length d. On each side pick an arbitrary point that is not a corner.
Let u, x, y, and z be the side lengths of the convex quadrilateral determined by these four points.
Prove that
d2 ≤ u2 + x2 + y2 + z2 < 2d2.
(In reply to
re(2): Solution - your reason to edit by Harry)
Yes, I've found that the special characters from the special-character button on the comment editor give bad results, as in Ó for a capital sigma. It looks good here in the comment box, but I'm confident that once this is posted, that won't look like a capital sigma.
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Posted by Charlie
on 2009-07-19 13:07:02 |