128 reputation
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location Naarden, Netherlands
age 28
visits member for 1 year, 7 months
seen Nov 4 '13 at 12:25

Dec
29
comment A small geometry puzzle out of curiosity
Thanks :) Your "tiling" suggestion really set me off to the right path, I didn't think of that myself, so I accepted your answer.
Dec
29
comment A small geometry puzzle out of curiosity
@CalvinLin Haha I like puzzling but officially I'm just asking for an answer here of course ;) It's not my homework or something, I really just thought of this out of curiosity. And I think I have it in my head how it should work, but how to formally write it down.. I'm not enough mathematician for that anymore I guess ;) The answer that @ barto is giving seems quite right btw. :)
Dec
29
comment A small geometry puzzle out of curiosity
Hm. How do you call that then.. For me $3\sqrt{2}$ and $4\sqrt{2}$ have a "common divisor", being $\sqrt{2}$. And if I do this one on paper, it also works out, you end up in another corner. But the rectangle of 1 by $\pi$ will never work out, just like the rectangle of 3 by $\sqrt{2}$, right? Because you can't divide both sides into an integer number of equally sized parts... (whatever that may be called)
Dec
29
comment A small geometry puzzle out of curiosity
Ah, tiling them, that's a good one. But that means that, for this to work, the two sides should have a common divisor, right? And that would mean that it would work for any rectangle with sides from $\mathbb Q$ but not for example for rectangles size 1 by $\pi$ ?
Dec
29
comment A small geometry puzzle out of curiosity
Nope, as I said, I'm looking for a proof that it's true for all rectangles, also with non-integer sizes. Or for a counter-example of course... But to analyze this in my head or with some paper, I use integer sides and for the few cases I did for my self, it always seems to work.