Britney Gallivan, who was at the time a junior in high school, solv...
In the larger perspective, the paper folding experiment connects to...
To understand better how the length is lost as we fold the paper le...
$$ L=\frac{\pi t}{8}(\sum_{i=1}^{n}2^{2i}+2\sum_{i=1}^{n}2^{i}) ...
Another way to see this is: $$ L=\sum_{i=1}^{n}\sum_{j=1}^{2^{i...
As we can see below the function $\frac{1}{6}(2^n+4)(2^n-1)$ grows ...
In 2012, students at St. Mark's School in Southborough, Massachuset...
If we don't consider the length lost in radii of earlier folds it m...
Alternate folding is special since folding in one direction doesn't...
At Right Angles | Vol. 4, No. 3, Nov 2015
9
Gaurish Korpal
Folding
Paper in Half
feature
Vinay Nair
Miles Please
When I was a small child I read the following claim about
paper folding: “It is impossible to fold any piece of paper in half
more than eight times no matter how big, small, thin or thick
the paper is” ([1], pp. 32-43). At that time I accepted the fact
after some experimentation. But last month, to my amazement,
I discovered that a grade 11 student in California, Britney C.
Gallivan, had mathematically disproved the above statement
in 2002 [1]. This article discusses the proof given by Gallivan
regarding “folding paper in half, n times”.
Say Crease!
Vol. 4, No. 3, November 2015 | At Right Angles 20
20 At Right Angles | Vol. 4, No. 3, November 2015
Gaurish Korpal
Folding
Paper in Half
When I was a small child I read the following claim about
paper folding: “It is impossible to fold any piece of paper in half
more than eight times no matter how big, small, thin or thick
the paper is” ([1], pp. 32-43). At that time I accepted the fact
after some experimentation. But last month, to my amazement,
I discovered that a grade 11student in California, Britney C.
Gallivan, had mathematically disproved the above statement
in 2002 [1]. This article discusses the proof given by Gallivan
regarding “folding paper in half, times”.
Introducing the Problem: Gallivan’s Rules
We �irst state the rules to be followed while folding a sheet of
paper in half, as enunciated by Britney Gallivan:
1. A single rectangular sheet of paper of any size and uniform
thickness can be used.
2. The fold has to be in the same direction each time. (Hence the
fold lines are all parallel to each other.)
3. The folding process must not tear the paper. (That is, it must
not introduce any discontinuities.)
4. When folded in half, the portions of the inner layers which
face one another must almost touch one another.
5. The average thickness or structure of material of paper must
remain unaffected by the folding process.
6. A fold is considered complete if portions of all layers lie in one
straight line (called folded section).
Keywords: Gallivan, paper folding, half fold, geometric progression, creep
section, radius section, series summation, quadratic equation
feature
Vol. 4, No. 3, November 2015 | At Right Angles 21
21 At Right Angles | Vol. 4, No. 3, November 2015
Following the rules, we claim that: The length of
the given sheet of paper decides the number of times
we can fold it in half. Thus if we have a sheet of
paper of given length, we can calculate the number
of times we can fold it theoretically (allowing a
reasonable amount of manpower and time).
HOW TO REACH THE SUN …ON A PIECE OF PAPER
A poem by Wes Magee
Take a sheet of paper and fold it,
and fold it again,
and again, and again.
By the 6th fold it will be 1-centimeter thick.
By the 11th fold it will be
32-centimeter thick,
and by the 15th fold - 5-meters.
At the 20th fold it measures 160-meters.
At the 24th fold - 2.5-kilometers,
and by fold 30 it is 160-kilometers high.
At the 35th fold it is 5000-kilometers.
At the 43rd fold it will reach the moon.
And by the fold 52
will stretch from here
to the sun!
Take a piece of paper.
Go on.
TRY IT!
Absolute folding limit
Geometric series. There is a short poem by Wes
Magee titled How to reach the sun …on a piece of
paper ([2], page 19), which illustrates the
geometric series involved in folding paper in half.
After folds (if possible), the width of the folded
paper will be approximately equal to the distance
between the sun and the earth!
Every time we fold the paper in half, we double
the number of layers involved. We have to fold

sheets of paper for the
th
fold. Thus for each
successive fold we need more and more energy.
Initially this was thought to be the reason for our
inability to fold a piece of paper more than times.
But, as stated earlier, the strength of the arm is
not the limiting factor for the number of folds.
Understanding folds. After each fold, some part
of the middle section of the previous layer
becomes a rounded edge. The radius of the
rounded portion is one half of the total thickness
of folded paper; see Figure 1.
Initially the radius is small as compared with the
length of the remaining part of sheet. As the folds
Figure 1. Paper folded in half 12 times illustrating the
decrease in folded section and increase in radius
section caused due to continued folding which leads to
fold losses [©Britney C. Gallivan]; source: [1], [4]
begin nearing their �inal thickness, the curved
portion becomes more prominent and begins
taking up a greater percentage of the volume of
the paper. The radius section is the part of the
paper ‘wasted’ in connecting the layers.
The section that projects past the folded section
on the side opposite the radius section is called
the creep (Figure 1). It is caused by the difference
in lengths of layers due to the rounded section of
the fold layers having different radii and
circumferences.
The limit to the number of folds is reached when a
fold has been completed but there is not enough
volume or length in the folded section of the
paper to �ill the entire volume needed for the
radius section of the next fold. Thus while making
folds there is loss of paper in the form of radius
section and creep section.
Limit formula. Since the radius and creep
sections are semicircular, the length to height
ratio of the paper being folded has to be greater
than to allow one more successful fold to occur.
If a folded section’s length is less than times the
height, the next fold cannot be completed.
Let be the thickness of a sheet of paper. On the
�irst fold, we lose a semicircle of radius , so the
length lost is (‘lost’ in the sense of ‘not
contributing to the length’). On the second fold,
we lose a semicircle of radius and another
semicircle of radius , so the length lost is
2
At Right Angles Vol. 4, No. 3, November 2015
Vol. 4, No. 3, November 2015 | At Right Angles 23 23 At Right Angles | Vol. 4, No. 3, November 2015
Figure 3. Britney Gallivan holding the first sheet of
paper ever to be folded twelve times [©Britney C.
Gallivan]; source: [1], [4]
thickness inch) would have been enough
to accomplish the task of folding paper in half
twelve times. For a nice writeup on this episode,
please see [5]. See also [6].
Exercise
After the above analysis, the following question
may arise: “How many times can we fold a sheet
of paper in half, by folding in alternate directions,
keeping all other rules the same?”
As expected, the answer is: The length and width
of the given sheet decide the number of times we
can fold the paper in half.
A good idea to proceed will be to start with a
square sheet of paper and calculate its limiting
width. Without considering the effects of
material lost in radii of earlier folds, we get a
crude bound for the width of a square sheet of
paper of thickness required for folding in half
times as:


I invite readers to derive the above formula.
Surely, the above formula does not give any
minimum limit. Using analysis seen in one
direction folding, we can derive a Limit Formula
for alternate folding. But in alternate folding, we
get separate equations for odd and even folds.
Note that odd folds accumulate losses in an odd
fold direction, and even folds accumulate losses in
an even fold direction. Also, each fold in an odd
direction doubles the amount of paper for the
next even direction, and vice versa.
References
1. Britney C. Gallivan, How to Fold Paper in Half Twelve Times, Historical Society of Pomona Valley Inc. (2002); see
http://pomonahistorical.org/12times.htm
2. Arvind Gupta, Hands on Ideas and Activities, Vigyan Prasar (An autonomous organization under the Department of
Science and Technology, Government of India), ISBN : 87-7480-118-9 (2005)
3. http://mathforum.org/library/drmath/view/60675.html
4. http://mathforum.org/mathimages/index.php/Bedsheet_Problem
5. http://www.abc.net.au/science/articles/2005/12/21/1523497.htm
6. Weisstein, Eric W. ”Folding. From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/Folding.html
4
At Right Angles Vol. 4, No. 3, November 2015
GAURISH KORPAL is an undergraduate student at National Institute of Science Education and Research (NISER),
Jatni; he is currently in his second year. He loves mathematics and is a regular blogger at gaurish4math.
wordpress.com. He may be contacted at korpal.gaurish@gmail.com.
Vol. 4, No. 3, November 2015 | At Right Angles 22
22 At Right Angles | Vol. 4, No. 3, November 2015
Figure 2. The folded portion and the circular portions
[Britney C. Gallivan]; source: [1], [4]
      
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paper required       
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 
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Realising the theoretical limit
 

      
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     
        
       
        
        
  
         
Vol. 4, No. 3, November 2015 | At Right Angles 23
23 At Right Angles | Vol. 4, No. 3, November 2015
Figure 3. Britney Gallivan holding the first sheet of
paper ever to be folded twelve times [©Britney C.
Gallivan]; source: [1], [4]
thickness inch) would have been enough
to accomplish the task of folding paper in half
twelve times. For a nice writeup on this episode,
please see [5]. See also [6].
Exercise
After the above analysis, the following question
may arise: “How many times can we fold a sheet
of paper in half, by folding in alternate directions,
keeping all other rules the same?”
As expected, the answer is: The length and width
of the given sheet decide the number of times we
can fold the paper in half.
A good idea to proceed will be to start with a
square sheet of paper and calculate its limiting
width. Without considering the effects of
material lost in radii of earlier folds, we get a
crude bound for the width of a square sheet of
paper of thickness required for folding in half
times as:


I invite readers to derive the above formula.
Surely, the above formula does not give any
minimum limit. Using analysis seen in one
direction folding, we can derive a Limit Formula
for alternate folding. But in alternate folding, we
get separate equations for odd and even folds.
Note that odd folds accumulate losses in an odd
fold direction, and even folds accumulate losses in
an even fold direction. Also, each fold in an odd
direction doubles the amount of paper for the
next even direction, and vice versa.
References
1. Britney C. Gallivan, How to Fold Paper in Half Twelve Times, Historical Society of Pomona Valley Inc. (2002); see
http://pomonahistorical.org/12times.htm
2. Arvind Gupta, Hands on Ideas and Activities, Vigyan Prasar (An autonomous organization under the Department of
Science and Technology, Government of India), ISBN : 87-7480-118-9 (2005)
3. http://mathforum.org/library/drmath/view/60675.html
4. http://mathforum.org/mathimages/index.php/Bedsheet_Problem
5. http://www.abc.net.au/science/articles/2005/12/21/1523497.htm
6. Weisstein, Eric W. ”Folding. From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/Folding.html
4
At Right Angles Vol. 4, No. 3, November 2015
GAURISH KORPAL is an undergraduate student at National Institute of Science Education and Research (NISER),
Jatni; he is currently in his second year. He loves mathematics and is a regular blogger at gaurish4math.
wordpress.com. He may be contacted at korpal.gaurish@gmail.com.

Discussion

$$ L=\frac{\pi t}{8}(\sum_{i=1}^{n}2^{2i}+2\sum_{i=1}^{n}2^{i}) $$ Using the sum formulas for geometric series $$ L=\frac{\pi t}{8}\left(\frac{2^2}{2+1}\frac{2^{2n}-1}{2-1}+4\frac{2^n-1}{2-1}\right)= \frac{\pi t}{8}\left(\frac{4}{3}(2^{2n}-1)+4(2^n-1)\right) $$ To understand better how the length is lost as we fold the paper let's take a look at how the paper looks like in the first 3 folds. ![](http://i.imgur.com/Qpnjlo2.png?1) For the first fold we see that the lost portion is the one in blue, which corresponds to a semicircle of radius $t$. The length lost is $\pi t$. For the second fold in addition to $\pi t$ we have the portions in red: $\pi t + 2\pi t$ . Finally for the third fold in addition to $\pi t$ and $\pi t + 2\pi t$ we have the portions in green: $\pi t + 2\pi t+ 3\pi t + 4\pi t$ . As we can see every additional fold adds $\sum^{2^{i-1}}_{i=1} j\pi t$ to the total length lost. Britney Gallivan, who was at the time a junior in high school, solved this well-known problem. She was asked by her teacher to fold a sheet of paper 12 times and as an incentive she would get extra credit. She failed multiples times. Later she succeeded after using a thin gold sheet and proved the assumption wrong. Gallivan was able to achieve 12 folds by folding a roll of thin toilet paper that stretched over three-fourths of a mile. It took seven hours in a shopping mall with her parents, but Gallivan was able to bust a myth as well as derive a formula relating the width, thickness of a paper and the number of folds achievable. The urban legend was disproved in 2001. In 2012, students at St. Mark's School in Southborough, Massachusetts visited MIT and beat Gallivan's record with 13 single-direction folds. They didn't actually use Gallivan's single-sheet method, instead choosing to layer the first 64 sheets (equivalent to six folds) on top of each other and then begin the folding. [![IMAGE ALT TEXT](https://slack-imgs.com/?c=1&o1=wi400.he300&url=https%3A%2F%2Fi.ytimg.com%2Fvi%2FvPFnIotfkXo%2Fhqdefault.jpg&width=400&height=300)](https://www.youtube.com/watch?v=vPFnIotfkXo "Paper folding record") Another way to see this is: $$ L=\sum_{i=1}^{n}\sum_{j=1}^{2^{i-1}}j\pi t $$ Now, $\sum_{j=1}^{2^{i-1}}j\pi t$ is a sum of an arithmetic series and so $$ \sum_{j=1}^{2^{i-1}}j\pi t= \frac{\pi t}{2}(2^{2(i-1)}+2^{i-1}) $$ As we can see below the function $\frac{1}{6}(2^n+4)(2^n-1)$ grows very rapidly with the number of folds. ![](http://i.imgur.com/zren5F1.png?1) If we don't consider the length lost in radii of earlier folds it means we just have to consider the length lost in the last fold. When we reach the final fold $n$, the side of the square must be at least equal to the length lost in the final fold which is $2^{n-1}\pi t$. Since the total area of the sheet $W^2$ (area = nb of sheets in penultimate step $\times$ area of square in penultimate step) is preserved $$ W^2=2^{n-1}\times (2^{n-1}\pi t)^2 \\ W=\pi t 2^{3/2(n-1)} $$ In the larger perspective, the paper folding experiment connects to exponential growth. This is the essence of exponential growth: very small amounts rapidly become astronomically large through simple doubling. Assuming it was possible to fold paper without restriction, the height of a piece of folded paper would double in thickness each time it was folded. Since one sheet of typical 20-pound paper has a thickness of about 0.1 millimeter, folding 50 times would produce a tower of height $1.13×10^{11}$ meters, and folding one more time would make the stack higher than the distance between the Earth and Sun. Alternate folding is special since folding in one direction doesn't influence the length lost in the other direction (the only impact is in terms of the amount of paper that needs to be folded in the next fold). To derive the formula for alternate folding we start by defining the length of the paper in the two directions as $L_{odd}$ and $L_{even}$. When $n=0$ it's simple: $L_{odd}=L_{even}=W$. After the first fold, as can be seen below: $L_{odd}=\frac{W-\pi t}{2}$ and $L_{even}=W$. After the second fold: $L_{odd}=\frac{W-\pi t}{2}$ and $L_{even}=\frac{W-\pi t -2 \pi t}{2}$. ![](http://i.imgur.com/5zaeukQ.jpg) It's not difficult to conclude that for an arbitrary fold count $n$ if $n$ is even $[L_{even}]_n=\frac{[L_{even}]_{n-1}-\sum^{2^{n-1}}_{j=1}j\pi t}{2} $ and $[L_{odd}]_n=[L_{odd}]_{n-1}$, if $n$ is odd $[L_{odd}]_n=\frac{[L_{odd}]_{n-1}-\sum^{2^{n-1}}_{j=1}j\pi t}{2} $ and $[L_{even}]_n=[L_{even}]_{n-1}$.