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

   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