Challenge Math Challenge by Erland #1

73
4
33,236
8,950
Your approach was based on combining everything to get a contradiction. How does that work if you don't combine rectangles?
 
73
4
Your approach was based on combining everything to get a contradiction. How does that work if you don't combine rectangles?
"to get a contradiction"?
For the third lemma I had to make an assumption that the integer side is known. I claim that there is always an order of rectangles which satisfies the third lemma( both vertically and horizontally), and if not, there is the forth lemma. So far, I always succeeded to combine all the rectangles.
Give an example in which my rules are not enough.
 
33,236
8,950
"to get a contradiction"?
You don't necessarily have to write it like that, depends on the direction you take. Doesn't matter.
I claim that there is always an order of rectangles which satisfies the third lemma( both vertically and horizontally), and if not, there is the forth lemma.
Well, you claim that, but can you prove it? It doesn't work as you described in my example. It will work if I make a complete tiling out of it, but that is not the point.
 

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