Square UVWX; A on UX, B on UV, C on VW, D on WX.

Area square UVWX = twice area quadrilateral ABCD.

Perimeter ABCD = 800. All side lengths are integers.

What's the area of square UVWX?

Have fun (Sleepy)

Printable View

- August 15th 2009, 09:37 PMWilmerSquare + Quad
Square UVWX; A on UX, B on UV, C on VW, D on WX.

Area square UVWX = twice area quadrilateral ABCD.

Perimeter ABCD = 800. All side lengths are integers.

What's the area of square UVWX?

Have fun (Sleepy) - August 15th 2009, 09:44 PMMatt Westwood
The intuitive and obvious-looking answer is that ABCD has to be the square formed when A, B, C, D are at the midpoints of the sides they are on. Anywhere else and the size of ABCD will be bigger than half UVWX.

Except now I note that this won't work because "All side lengths are integers", and that won't work because otherwise the sides of UVWX are which ain't. - August 16th 2009, 06:15 AMWilmer
- August 17th 2009, 12:27 AMOpalg
The problem would become a lot easier if we knew that AC was parallel to XW. Then each of the triangles ABC, ACD would have half the area of the corresponding rectangle (UVAC, ACWX). So the area of the quadrilateral ABCD would automatically be half the area of the square.

If in addition we also knew that XD = DW, then the triangles AXD and CDW would be congruent, so we would only have to deal with three pythagorean triples instead of four.

So let the side of the square be 2a, let AX = CW = h, and let UB = x. Then we want to find pythagorean triples (a,h,p), (2a–h,x,q) and (2a–h,2a–x,r) with 2p+q+r=800. But if you write out the condition 2p+q+r=800 in terms of a, h and x, there are too many square roots to indicate much prospect of progress. As far as I can see, the only way forward is to do some kind of a search. A computer program could check all possible values of h and x within some plausible range of a, say 120 ≤ a ≤ 150.

I don't have access to such computer facilities, so I tried another search method (Google). That quickly led me to a web site with a solution.

__Spoiler__:

But no explanation of a method was given, and I don't see any constructive way of finding this solution. A proof of uniqueness would be much harder still, because then one could no longer make the simplifying assumptions about AC being parallel to XW and D being the midpoint of XW. - August 17th 2009, 08:33 AMWilmer
Everything you're saying I agree with, Opal.

Of course, a computer program needs to be written to find solution.

I set this up simply for "ye who liketh programming"!

Was easy enough for me: already have a program that finds all

pythagorean triangles (limit optional, like short leg < 1000);

and search is quick since there ain't that many cases where the sum of

the hypotenuses of 4 of them does not exceed 800.

By the way, which site did you find this at? Barry Clarke's?

I also posted it there...but your "find" doen't seem to match the "style"

of the solution somebody posted there... - August 17th 2009, 12:17 PMOpalg
Yes, it was this page. I adapted the solution given there to fit in with the approach that I was taking.

- August 17th 2009, 08:01 PMWilmer
Thanks; yes, that was where I posted it...under Denis Borris