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 Quadrilateral in a rectangle (Posted on 2009-07-17)
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 ```

 Subject Author Date re(4): Solution - your reason to edit Charlie 2009-07-19 13:08:13 re(3): Solution - your reason to edit Charlie 2009-07-19 13:07:02 re(2): Solution - your reason to edit Harry 2009-07-18 16:53:08 re: Solution - your reason to edit brianjn 2009-07-17 22:59:13 Solution Harry 2009-07-17 21:28:35

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