## Saturday, November 6, 2010

### Fox 313

Bleaug submitted the following fox, saying: "a problem which can be expressed in geometric terms and that is elegantly solved using pure algebraic arguments, still giving deep insight."
General case of this problem can be found in literature (with at least 14 proofs :)
But let's try this smaller version here.

1. You can do this by contradiction. First, you just choose one of the sides of the inner rectangles to be an integer. And from there, you deduce what sides have to be an integer, and which do not.

The key to the proof, is that an integer plus a non-integer is never an integer, however, a non-integer plus another integer may be an integer. You will get a contradiction with the square in the center.

If I could get photoshop to work right now, I'd show it by example. Do you know any good Geometry program I could use?

2. Minor correction:
"a non-integer plus another integer may be an integer" must read:
"a non-integer plus another NON-integer may be an integer"

Contradiction in this case would work.

For geometry software, search:
CaRMetal
http://en.wikipedia.org/wiki/CaRMetal

GeoGebra:
http://www.geogebra.org/cms/

Among others...

3. Ah yes, thank you. That was a typo. And thanks for the software! I will check them out!

4. since we speak of software, don't forget Inkscape (http://inkscape.org/?lang=en). More like a vector graphics tool but can be very helpful if you look at the svg (xml) under the hood.

bleaug