Consider quadrilateral ABCD whose diagonals are perpendicular and meet at point E.

Minimize the perimeter of ABCD where AB, BC, CD, DA, EA, EB, EC, ED are all different integers.
(Or prove no such quadrilateral exists.)

After rereading this problem, I decided it would be better to consider this quadrilateral instead as four right triangles all joined at their right angles.

This brings to my mind one phrase: Pythagorean triples.  I'm no pythagorean expert, but I don't think there's anything to suggest a lack of a solution.  So let's start looking!

Here's a list of the most likely triples I'll use:
3,4,5
5,12,13
8,15,17
9,40,41

If I only consider EA, EB, EC, and ED, I only need to worry about the legs (and hopefully the hypotenuses will turn out distinct too).  If the length of the first leg is x, then the next leg can be 3x/4, 4x/3, 5x/12, 12x/5, etc.  And from that, we can determine the possibilities for the next leg, and the next.  However, it has to somehow circle around back to x, and each leg must be distinct.

Here are the ratios that can be used, derived from the above triples:
3:2*2
5:2*2*3
2*2*2:3*5
3*3:2*2*2*5

I shall reply to myself later with my results

Posted by Tristan on 2005-08-29 16:59:37