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
d^{2} ≤ u^{2} + x^{2} + y^{2} + z^{2} < 2d^{2}.
Let the rectangle have side lengths 2a, 2b, 2a, 2b, with the vertices of the quadrilateral on and at distances p, q, r and s, respectively, from the midpoints of these sides, so that Pythagoras, applied to the four triangles which lie outside the quadrilateral, gives:
u^2 = (a  p)^2 + (b + q)^2
x^2 = (b  q)^2 + (a + r)^2
y^2 = (a  r)^2 + (b + s)^2
z^2 = (b  s)^2 + (a + p)^2
When these four equations are added, many terms cancel out, leaving:
u^2 + x^2 + y^2 + z^2 = 4(a^2 + b^2) + 2(p^2 + q^2 + r^2 + s^2)
The minimum value of this expression occurs when p=q=r=s=0 (i.e. when the quadrilateral becomes a diamond with vertices at the midpoints of the rectangle's sides).
Then: u^2 + x^2 + y^2 + z^2 = 4(a^2 + b^2) = d^2 since d^2 = (2a)^2 + (2b)^2 by Pythagoras.
The maximum value occurs when p^2, q^2, r^2 and s^2 all take their maximum values of a^2, b^2, a^2 and b^2 respectively (i.e. when pairs of vertices approach opposite corners of the rectangle and the quadrilateral reduces to a diagonal line).
Then: u^2 + x^2 + y^2 + z^2 = 4(a^2 + b^2) + 2(a^2 + b^2 + a^2 + b^2)
= 8(a^2 + b^2) = 2d^2.
So d^2 <= u^2 + x^2 + y^2 + z^2 < 2d^2
Edited, sadly, because all the indices and inequalty signs obtained from the extra font facility appeared as gobbledegook in the posted version. Why was that?
Edited on July 17, 2009, 9:37 pm
Edited on July 17, 2009, 9:47 pm

Posted by Harry
on 20090717 21:28:35 