The old new thing: mathematics of paper folding
A few weeks ago I received the following quiz as one of the questions for an exam. The question is about paper folding, something that we all know from an early age, and seems simple but it seems that noone could answer it. I presented it in this article with my proposed solution in an attempt to show how a frustrating mathematics problem could be set from something so simple in our everyday life.
The Problem
Fold a sheet of A0 paper (841mm x 1189mm) in such a way that the longer side (1189mm) is divided into half its length, you will get a sheet of A1 paper (594mm x 841mm). Do the same for the resulting piece of paper and you’ll get a sheet of A2 paper (420mm x 594mm). Repeat the same twice and you will get the commonly used paper size, A4 (210mm x 297mm). This process could repeat over and over again to get a paper size of An after n times, as demonstrated in the following diagram:
n

Inner Folds

Outer Folds

Total

0

0

0

0

1

1

0

1

2

3

1

4

3

6

4

10

4

14

10

24

5

28

24

52

6

60

52

112

The first task is to find a general formula or algorithm to calculate the number of inner and outer folds after n times.
My Proposed Solution
As simple as it seems, there is no quick solution the problem. Since most people don’t have an A0 sheet of paper at home to try to fold so many times, they can only attempt with the common A4 paper and get frustrated after a short while when the paper becomes too thick to fold or the number of folds too big to count.
I proposed a simple solution below. My research also shows that there are several other approaches which may yield seemingly different formulas. Do not read it until you have attempted the question.:)
We first observe the total number of horizontal (from left edge to right edge) lines and vertical (from top edge to bottom edge) lines after every fold:
n

Horizontal

Vertical

1

1

0

2

1

1

3

3

1

4

3

3

5

7

3

6

7

7

This can be generalized into:
Now look at each horizontal and vertical line and calculate the number of segments on each line:
n

Segments
/hori. line 
Segments
/vert. line 
1

1

0

2

2

2

3

2

4

4

4

4

5

4

8

6

8

8

This can be generalized into (for n=2 and above)
The total number of folds after n steps is simply:
With some efforts we are able to derive the formula for the total number of folds:
Now that we have found out the total number of folds, let’s calculate the difference between the number of inner folds and outer folds after each step:
n

Inner Folds

Outer Folds

Diff.

1

1

0

1

2

3

1

2

3

6

4

2

4

14

10

4

5

28

24

4

6

64

56

8

Surprisingly enough, the difference is simply:
I leave this as an exercise for the reader to prove why. With the sum and the difference known, we can now calculate the number of inner and outer folds easily:
The first part of the problem is now solved. You can compare the results from our formulas with the example to make sure that the formulas are correct. Interestingly, the total number of folds after each step is equal to the number of outer folds in the next step.
Bonus Question
This is the next part of the problem:
“So far you have always been folding by the longer side of the resulting paper after every step (so that you could get an A1 paper from A0, A2 from A1, A3 from A2 and so on). Now you are told that this restriction is no longer necessary and you could fold the paper by either the shorter or the longer side. For example, fold an A0 paper (841mm x 1189mm) by the shorter side and you will get a piece of paper of size 420mm x 1189 mm.
How would the final formula for the number of inner and outer folds change? Derive a method to calculate the total number of inner and outer folds after a given set of fold steps, where one can fold either way in each step.”
Although I managed to write a program to solve this, I believe there is no need to describe it here as most readers by now would have understood the concept and would know how to approach this extension of the original problem.
Afterthoughts
The concept which the problem is based on is very simple and does not require any fancy maths knowledge – anyone with perhaps a secondary school education can understand the problem. However, among many whom I have asked, almost all would immediately attempt the question but most would just become frustrated and give up without ever finding the solution, or even the correct approach. I believe it requires a high level of concentration and logical thinking as the number of folds grows exponentially. Setting this as an exam question where students work under stress and time constraints would therefore be unreasonable.