I Calibration of axis for Oragami folds

[ddddd28]

This problem came to existence as an attempt to justify a very common operation that an origami folder is often faced with.
Broadly speaking, when one folds a model, it is almost always necessary to divide the paper into some segments. The statement that has to be proven: given the proportion k/n, when k and n have no common factors except than 2^m, it is always possible to mark the segment 1/n, only by marking the mean of existing marks. (you can only divide a paper into two parts).

exampe: initial proportion: 5/11
we average 5 and 11 and get 8 and from here it's straightforward.
In other words, we need to prove that we always fall on 2^a mark.
I tried to do so, but each proportion has a unique sequence of convergence to 1, and it becomes messy.
Please, give me an advice of how I should prove it.
A reference to the reverse statement will be also appreciated, I think it is also easier.

 

[andrewkirk]

That's a very nice problem, because it is connected to something physical - origami - rather than just being abstract.

I can prove it but, if you don't mind, I'll first see if Greg would like to use it as a Challenge Problem.

Here's a hint to suggest how you might set about proving this. Although as you say there are different 'sequences of convergence' for different proportions, there may be an algorithm that works for all proportions. A proof would involve finding such an algorithm and proving that it always works. That algorithm would probably not give the shortest sequence of folds in each situation, but it would be guaranteed to get there in the end, which is all we want.

 

[Erland]

I don't understand. Which marks are "existing"? Which marks do we begin with?
Could you reformulate the problem in purely arithmetical terms, without talking about "marks"?

 

[I like Serena]

Suppose we start with 5 and 10.
Then we can only combine them to numbers of the form $2^p(2^q\cdot 5 + 2^r \cdot 10 + 0)$ can't we?
And this is always a multiple of 5, meaning we wouldn't be able to get to 1.

 

[mfb]

@I like Serena:

when k and n have no common factors except than 2^m

In addition, we can only go to the average of numbers, not to twice the numbers.

@Erland: See here