Hi,Heres a nice problem.Let R be a rectangle which is subdivided into smaller rectangles. The smaller rectangles all have one property in common! One of its sides has integral length. Then show that the rectangle R also shares the property that one of its sides has integral length!By integral length i mean that the length belongs to the set 1,2,3,4,...

Funny, I was thinking for a while about posting the same problem. It has been one of my all time favourites - ever since I got it at a high school problem-solving contest (quite a while ago, alas).

Yes even i've been thinking of posting this for a while now. Its a very nice problem isn't it?

Yes, it is very nice indeed! May I ask you, how many different proofs do you know?

I know only one proof. I would definitely be very interested in hearing all the proofs you know! Excellent stuff!

The usual source is this paper. Requires subscription unfortunately

Very nice indeed!! Thanks very much!

Enjoy! Don't you think that his proof #3 ("Checkerboard") is based on an idea similar to the one used in your problem about the 6x6x6 cube? That is why I wanted to post this problem.

The 6x6x6 cube? Sorry which one was that again?

This one.

Oops sorry forgot about that one. Yes you're right! By the way were you involved with the IMO at all?

Well, I myself did not make the team. A close family member did though. Why?

The style of questions you ask made me think that perhaps you were involved with it. I really like all the questions and answers you've posted so far! Very elegant!

Well, the problem in a different thread is in fact from IMO 1979 (I just adjusted the numbers.) Thanks for the compliment though

Dear wannabequantie,I just came up with a proof. Let me descripe my idea here and give more details if necessary.Let's put our "big" rectangle to the xy plane. Let one of its side be on the x-axis and another on y-axis.(We call sides which are parellal to x-axis (resp. y-axis) x-side (resp. y-side).Suppose the argument is not true. From every point on a y-side of the "big" rectangle we draw a line whichis parallel to the x-axis. This line intersects with some small rectangles. At least one of them will have its x-side to be non-integral. And we pick the one which is to the very left. Its y-side must be integral (since its x-axis is not).We collect all these rectangles we picked from the process above. We can prove that if we project them down to they-axis they "don't" overlap (Actually they only overlap at their x-sides which becomes points after projection). We add up the legths and conclude the y-side of the "big" rectangle is integral. Contradiction!Well, I hope what I wrote is not too obscure. I will be happy to explain later if it is necessary.