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Apologies in advance, I really cannot think of an intelligent or easy way to explain this.

You start out with a rectangle. Then you draw a straight line out of a right angled corner at 45 degrees until you hit a side of the rectangle at which point you reflect the line at 90 degrees and continue the line. Repeat this process until the line goes into a second corner.

Sorry if that image is hard to imagine.

If you add the length and width of the rectangle (providing they are both integers and have no common factors) that is the number of points of contact the lines have with the outline of the rectangle: (the walls and the 2 corners)

If the lengths aren't integers then you have to make then integers before you add them eg. 0.2cm and 0.3cm = 2cm and 3cm. Or if they have factors then factorise them eg. 6cm and 4cm = 3cm and 2cm

QUESTION If anyone understands all of that, can anyone explain if and why that is true for every rectangle? Is it just a coincidence or is there maths behind it that I can't find? And can it work if one of the lengths was an irrational number like pi?

Jim
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user75617
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  • Am I correct that the question can be reformulated as follows: "Suppose we have a rectangle with sides length $m$ and $n$, where $m$ and $n$ are relatively prime. We now draw lines inside this rectangle: we start with a line at 45 degrees from one corner and every time a side of the rectangle is hit, the line is reflected inward. Eventually we hit another corner of the rectangle. This appears to be the case after $m+n$ reflections (including the corners where we start and and). Why is this the case? Does this still hold if $m,n$ are not coprime? Or even if their proportion is irrational?" – HSN May 02 '13 at 23:11
  • If so, I'm very interested in the relatively-prime case already. It seems very plausible by just drawing a few examples, yet I don't see how to prove it yet. – HSN May 02 '13 at 23:12
  • Thanks for the comment that's a far better way of phrasing it, but I think it would hit another corner after m+n-2 reflections since you are excluding the corners come up with a lot of questions similar to this. The idea started from me just drawing rectangles and I just haven't had the patience to look into them very far. – user75617 May 02 '13 at 23:22
  • Well, I never said it would have to be the same corner, did I? The solution by Ross seems to be right, though. – HSN May 02 '13 at 23:26
  • I don't think it could ever be the same corner because ithe lines would have to go back on itself which isn't possible if it always reflects at 90 degrees – user75617 May 02 '13 at 23:31
  • This subject is often referred to as 'rectangular billiard dynamics'. Billiard dynamics in other shapes is a hot topic at the moment. – Dan Rust May 02 '13 at 23:35
  • I had no idea that this had a name. I've been trying to research some similar things to this and now that I know what it's called that does help a lot. – user75617 May 03 '13 at 00:09

1 Answers1

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Imagine tiling the plane with your rectangles. Then instead of reflecting, you just keep on going through them. You are asking where you hit the image of one of the corners. If your rectangles are $m$ high and $n$ wide with $m,n$ coprime, you will go through $n$ rectangles upward and $m$ rectangles leftward until you hit point $(mn,mn)$

If the ratio of lengths is irrational, you will never hit another corner. The same tiling argument shows it.

Ross Millikan
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