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Quadrilateral fun (Posted on 2004-01-10) Difficulty: 4 of 5
Begin with an arbitrary, convex quadrilateral. Next, draw squares outwardly on the sides of the quadrilateral, and join the centers of opposite squares.

You might find the the two resulting lines are equal in length and intersect at precisely 90 degrees.

Prove (or disprove) the notion, that this is always true.

See The Solution Submitted by SilverKnight    
Rating: 4.0000 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Small problem | Comment 5 of 6 |
(In reply to re: Small problem by TomM)

For what it's worth, I was looking for a "simpler" solution that didn't involve complex numbers. (When I say "simpler", I mean using real numbers only.) What Federico submitted is a valid solution, but certainly, not everyone in Flooble's audience is familiar with imaginary numbers and rotating complex vectors by multiplication of -i.

That being said, here's a nice link to an applet that helps explain this, and perhaps someone may choose to submit a different (perhaps Cartesian) solution.
  Posted by SilverKnight on 2004-01-12 00:32:39

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