Geometry problem

[almostbusto]

you are right borisp. i guess if you can prove that there exists a path from one end of R to the other, then my proof holds. i am not sure there necessarily is such a path however.

I suspected it might have problems. mostly because i rarely figure out proofs where you have to be clever on my first try.

[blah_blah]

I have seen this problem before (it certainly is well known enough to be considered part of mathematical folklore).

One of the nicer ways is to integrate exp(2 \pi i (x+y)) over R (which is of course the same as integrating exp(2\pi i (x+y)) over the union of all of the rectangles ...)

there is another solution involving checkerboards but I am an analyst by heart and by training and so I stick to my most familiar tools [img]/images/graemlins/smile.gif[/img]

[borisp]

[ QUOTE ]
I have seen this problem before (it certainly is well known enough to be considered part of mathematical folklore).

One of the nicer ways is to integrate exp(2 \pi i (x+y)) over R (which is of course the same as integrating exp(2\pi i (x+y)) over the union of all of the rectangles ...)

there is another solution involving checkerboards but I am an analyst by heart and by training and so I stick to my most familiar tools [img]/images/graemlins/smile.gif[/img]

[/ QUOTE ]
Shut up!!...no offense [img]/images/graemlins/tongue.gif[/img]

[blah_blah]

since the majority of 'recreational' mathematicians are not capable of integrating a complex function over a curve in R^2, I think that I haven't spoiled your problem [img]/images/graemlins/smile.gif[/img]

[almostbusto]

[ QUOTE ]
Ok, now I subtract points from myself for agreeing to a fake solution. When you choose to traverse along an integer path, you are not necessarily guaranteed to land at a "corner." It is necessary to be at a corner before you are guaranteed to have an integer path along which to travel.

Try again.

[/ QUOTE ] wait a minute, yes you are guaranteed to land at the corner point with at least one other R_i, but i am not sure i can prove that the path necessarily hits the other side.


can you visualize a corner of an R_i that doesn't share that corner with another R_i? maybe you could demonstrate with a pic because i can't see it right now.

[pzhon]

[ QUOTE ]
[ QUOTE ]
Ok, now I subtract points from myself for agreeing to a fake solution. When you choose to traverse along an integer path, you are not necessarily guaranteed to land at a "corner." It is necessary to be at a corner before you are guaranteed to have an integer path along which to travel.

Try again.

[/ QUOTE ] wait a minute, yes you are guaranteed to land at the corner point with at least one other R_i, but i am not sure i can prove that the path necessarily hits the other side.


[/ QUOTE ]
This is part of one of the relatively well-known proofs for this problem, which was discussed in Stan Wagon's award-winning paper, "Fourteen Proofs Of a Result About Tiling a Rectangle" American Mathematical Monthly, Aug/Sep 1987. There is something left to prove. In the spirit of this thread, though, I won't post the rest of the proof in the next few days.

[Freyalise]

Bah I've seen the checkerboard solution now and spoiled it.
Very nice problem though which I had not seen before.

[thylacine]

[ QUOTE ]
I have seen this problem before (it certainly is well known enough to be considered part of mathematical folklore).

One of the nicer ways is to integrate exp(2 \pi i (x+y)) over R (which is of course the same as integrating exp(2\pi i (x+y)) over the union of all of the rectangles ...)

there is another solution involving checkerboards but I am an analyst by heart and by training and so I stick to my most familiar tools [img]/images/graemlins/smile.gif[/img]

[/ QUOTE ]

I thought there was a rule against problems where the easiest solution involves calculus. Right? Does this mean there is an even easier solution than this?

[Freyalise]

[ QUOTE ]
[ QUOTE ]
I have seen this problem before (it certainly is well known enough to be considered part of mathematical folklore).

One of the nicer ways is to integrate exp(2 \pi i (x+y)) over R (which is of course the same as integrating exp(2\pi i (x+y)) over the union of all of the rectangles ...)

there is another solution involving checkerboards but I am an analyst by heart and by training and so I stick to my most familiar tools [img]/images/graemlins/smile.gif[/img]

[/ QUOTE ]


I thought there was a rule against problems where the easiest solution involves calculus. Right? Does this mean there is an even easier solution than this?

[/ QUOTE ]

There is an extremely simple solution, one that a child could understand.

I didn't see it myself so of course I won't spoil the problem by posting it.