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Quadrilateral in a rectangle (Posted on 2009-07-17) Difficulty: 2 of 5
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.

  Submitted by Bractals    
Rating: 4.0000 (1 votes)
Solution: (Hide)
Let m and n be the side lengths of the rectangle and a, b, c, and d
the distances, from the corners of the rectangle, of the points such that
   u2 = (m - a)2 + b2
   x2 = (n - b)2 + c2
   y2 = (m - c)2 + d2
   z2 = (n - d)2 + a2
Adding these together and rearranging we get
   u2 + x2 + y2 + z2 = m2 + n2 + 2(a - m/2)2
                             + 2(b - n/2)2
                             + 2(c - m/2)2
                             + 2(d - n/2)2
This and the following inequalities,
   0 ≤ 2(a - m/2)2 < m2/2
   0 ≤ 2(b - n/2)2 < n2/2
   0 ≤ 2(c - m/2)2 < m2/2
   0 ≤ 2(d - n/2)2 < n2/2
gives the desired result,
   d2 = m2 + n2 ≤ u2 + x2 + y2 + z2 < 2(m2 + n2) = 2d2

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re(4): Solution - your reason to editCharlie2009-07-19 13:08:13
re(3): Solution - your reason to editCharlie2009-07-19 13:07:02
re(2): Solution - your reason to editHarry2009-07-18 16:53:08
re: Solution - your reason to editbrianjn2009-07-17 22:59:13
SolutionSolutionHarry2009-07-17 21:28:35
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