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 Post subject: A question regarding rectangle ratios.Posted: Tue May 14, 2013 5:20 pm

Joined: Sun Nov 23, 2008 2:18 am
You have a set of n real numbers(x1, x2, ... xn) that represent the ratio between the length and width of a rectangle such that:
Removing the largest square from a rectangle with ratio x1 renders a rectangle with ratio x2.
Removing the largest square from a rectangle with ratio x2 renders a rectangle with ratio x3.
...
Removing the largest square from a rectangle with ratio xn renders a rectangle with ratio x1.

It is well known that the golden ratio phi satisfies these conditions for the case n = 1, and that the sqrt(2) and the sqrt(2)+1 satisfy these conditions for the case n =2.

Upon finding a page that refers to a rectangle with a ratio of sqrt(3) as a Bronze rectangle, I fed the numbers into Google calculator and discovered that sqrt(3), 1/(sqrt(3)-1) and sqrt(3)+1 satisfy the above conditions for the case n=3.

I suspect that these solutions are unique, but have no idea of how to determine this. I also wonder if solutions exist for n>=4.

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 Post subject: Re: A question regarding rectangle ratios.Posted: Wed May 15, 2013 2:09 am

Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
It has been years since I've done any math related to this but my gut reaction is that you should be able to setup and solve a recurrence relation. You'll probably end up with a polynomial and all of the roots of the polynomial would be solutions.

A question like this is probably better suited at MathOverflow.

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 Post subject: Re: A question regarding rectangle ratios.Posted: Wed May 15, 2013 3:40 am

Joined: Wed Mar 15, 2000 9:11 pm
Location: Delft, the Netherlands
Instead of cutting off squares, it is more insightful to add squares. This way you see that you actually have choice - you can add it to the long side or to the short side. This also shows that the golden ratio example is fundamentally different from the other two examples you gave.

If you always add the square to the long side (which means that you alternate adding it left/right or top/bottom), then you will get an aspect ratio of phi regardless of how many steps you specify as your goal. This is the only solution.

If you add a square to the short side exactly once, and one or more times to the long side, then you get your other two examples. These cases have the solution sqrt( F(n+1)/F(n-1) ) where F(n) is the Fibonacci sequence.

If you add one square to the long side, and the other n-1 squares to the short side, then you'll get a different answer. Suppose the rectangle is 1 by x, with x>1 as the long side. The added squares will all be in a row to create a 1+nx by x rectangle with the former as the long side. This gives the equation:
1/x = x/(1+nx)
Simplifying gives
x^2=nx+1
which has the positive solution
x = (n+sqrt(n^2+4))/2

For n=1 (only long side) it is phi, and for n=2 (one long one short) it is 1+sqrt(2), both of which we have already seen. The next two values are (3+sqrt(13))/2 and 2+sqrt(5).

For any sequence of long/short choices (that includes at least one long) you can set up an equation which will be quadratic, and turn out to have a unique positive solution.

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