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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.

See The Solution Submitted by Bractals    
Rating: 4.0000 (1 votes)

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Solution Solution | Comment 1 of 5

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 mid-points 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 mid-points 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 2009-07-17 21:28:35

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