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Folding Papers and Planes: A Critical Difference
by Rosanna Nafziger
One of my favorite occupations during middle school was folding origami. As I
developed my folding precision, I could create more and more intricate works, until
eventually I was folding immense geometric fusions of many origami units and some
very acceptable imitations of spiders, winged horses, and three-headed dogs.
The mathematically puzzling piece is determining the limits and capabilities of
origami folding. The stipulations of origami are that we begin with a square piece of
paper and that a well-defined fold relies upon references—previously formed creases,
points, or folds—such that the fold can be replicated fairly accurately on another piece of
paper. If we consider the paper as a unit square with the bottom left corner at the origin,
what points could we mark with well-defined folds? We could certainly reach the point
1 1
 ,  , since it would be indicated by the intersection of the creases formed by folding
2 2
the paper in half vertically and horizontally. Following this type of operation, we can at
least approximate any point in the square as close as can be desired.
Theorem. Let r  R  and x, y   R 2 such that 0  x  1 and 0  y  1 . Then there
exists a series of folds of the unit square that define a point ( x1, y2 ) such that x  x1  r
and y  y1  r .
Proof. We will proceed by first proving that we can fold a vertical line within any given
distance of x, and a horizontal line within any given distance of y.
First, let me define a vertical folding algorithm. At the very outset, we determine
if x lies on either the left or right edge. If it does, we are done. In the first iteration, the
paper is folded in half vertically, aligning the vertical edges of the paper (which are the
two starting references), and then unfolded, leaving a crease we will call l1 . If the point
is on the crease, we are done. If the point is to the left of the crease, the paper is folded to
align l1 —the last crease folded—with the left edge—the reference immediately to the left
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of the point. If the point is to the right of the center crease, the paper is folded to align l1
with the right edge—the reference immediately to the right of the point. Again, the paper
is unfolded. If the point is on this second crease which we will call l2 , we are done. If
the point is to the left of l2 , we fold the paper to align l2 with the reference immediately
to its left, which will either be l1 or the left edge. If the point is to the right of l2 , we fold
the paper to align l2 with the reference immediately to its right, which will either be l1 or
the right edge. We continue this process of folding the nth crease between the previous
crease and one of the references for the previous crease such that the point lies between
the nth crease and the nearest reference for the previous crease. We continue until we
either reach the point or come within a certain given distance of it. We define a
horizontal folding algorithm analogously.
Now we want to prove that at the nth step of my algorithm, we have folded a
vertical line that passes within
1
of x. First, let d n be the closest distance from x to ln .
2n
We will proceed inductively. For the base case of n  1 , there are three possibilities. In
the case that x lies on l1 , d1  0 
1
1
. In the case that x is to the left of l1 , 0  x 
1
2
2
since x is neither on the left edge nor the center line and thus d1 
1
 x must be less than
2
1
1
. In the third case, x is to the right of l1 and  x  1 , so d1  1  x has to be less than
1
2
2
1
as well. Note that since l1 lies directly between its references, the distance between it
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and its references is
1
.
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Next, we assume that at the nth iteration of our algorithm, when n creases have
been folded,
1
1
 d n and the distance between ln and one of its references is n . Then
n
2
2
the n  1 th iteration will fold a crease exactly between the nth crease and one of the
references used to fold the nth crease, so the distance from ln 1 to either of its references
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will be
1
2 n 1
1 1
1
 n , or n 1 . Then, if the point lies on ln 1 , d n 1 is 0 and therefore less than
2 2
2
. If the point lies to either side of ln 1 , it must be between ln 1 and one of its
references, which implies that the horizontal distance between it and ln 1 can be no more
than the distance between ln 1 and its references, which is
1
2
n 1
, so d n 1 
1
2n 1
. Thus at
the nth iteration of our vertical folding algorithm, we have defined a vertical line that
passes within
1
of the point x, y  . The proof that a horizontal folding algorithm
2n
produces a horizontal line within
1
of x, y  is analogous.
2n
To demonstrate that we can produce two lines whose intersection yields a point
( x1, y1 ) such that x  x1  r and y  y1  r , we let k be an integer such that 2 k 
1
, so
r
1
 r .1 Then we perform both the horizontal and vertical algorithms k times for x, y 
2k
and let ( x1, y1 ) be the point where the horizontal l k ( lk h ) intersects the vertical l k ( lk v ).
Since the vertical distance between x, y  and any point on lk h is d k h , and the horizontal
distance between x, y  and any point on lk v is d k v and ( x1, y1 ) lies on both lk h and lk v ,
x  x1  dk v and y  y1  dk h . Therefore, since d k v < r and d k h < r, x  x1  r and
y  y1  r . ‫ٱ‬
Clearly, the algorithm I defined would not be physically possible with real-world
paper, since physical paper clearly is an imperfect medium and the physical act of folding
relies upon human manipulation, which is also imperfect. Not only that, but paper has
1
1
is a real number, there exists an integer N such that N  , and that the number
r
r
1
2 n  n for any natural number n. Thus for n  N , 2n  N  .
r
1
We know that since
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thickness, so folds lack perfect precision, and thus locations on the paper lack perfect
precision because they are determined by other folds or edges cut by humans.
Ideal paper, however, would also be unsatisfying. Consider the operation of
folding a plane in “half” about a line in the plane, defined as follows:
Definition. Let P be a plane in three-space, l a line on P, and Q a plane containing l. A
fold of P around l onto Q is a function f : P  Q such that the following conditions are
met:
(1) All points on l remain fixed.
(2) For any point p  Q not on l and l p on l such that l p is the closest point on l to
p, l p is also the closest point on l to f ( p) , and the distance from f ( p) to l p is the
distance from p to l p .
(3) For any q  ( x q , y q , z q ) in Q with l q  ( xl , yl , z l ) the point on l closest to q, if
q  l q  ( x q  xl , y q  y l , z q  z l ) is nonnegative, then for any other point s in Q
and l s , the closest point on l to s, s  l s is also nonnegative. If q  lq is not
positive, then s  l s is also not positive.
The second point provides the restriction that a fold is a reflection, and the third that we
are reflecting the plane P only to the half of another plane Q that is on one side of the
line. Note that this function is not onto. Observe that such an ideal fold seemingly can’t
be unfolded—that is, f
1
does not exist. I will prove not only that f is not one-to-one,
but that f takes pairs of points to the single points in Q. I will first need to introduce
another plane, however.
Theorem. Let P and Q be planes intersecting at a line l, and let f be a fold from P to Q
around l. Then for each point p in P not on l, the point l p on l closest to p, f ( p) and p all
lie on a plane perpendicular to l.
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Proof. Let R be the plane containing p, f ( p) , and l p . Suppose R is not perpendicular to
l. Then the line segment between l p and f ( p) is not perpendicular to l, because it lies in
R. Then l p cannot be the closest point on l to f ( p) , which is a contradiction. Thus R is
perpendicular to l.‫ٱ‬
Having introduced the plane R, I can now go back to my original goal of proving
that an ideal fold can’t be unfolded.
Theorem. Let P and Q be planes intersecting at a line l, and let f be a fold from P to Q
around l. Then for each point p in P not on l, there exists a point p' such
that f ( p)  f ( p' ) .
Proof. Let d be the shortest distance from p to l, and let p' be the reflection of p across l
in the plane P. Then l is a perpendicular bisector of pp' , so l p lies on pp' , as does the
point on l closest to p' , l p ' . Since l and pp' only intersect at one point, l p  l p ' , and pp'
contains pl p and l p ' p ' . Thus pl p  l p ' p'  pp' .
Let R be the plane containing p, l p , and f ( p) . Then R contains pl p , which is the
same as pp' , so p ' R . Similarly, the plane R ' containing p' , l p ' , and f ( p' ) contains
l p ' p ' , or pp' , so p  R ' . Knowing that R and R ' are perpendicular to l and both contain p
and p' is more than enough information to say that R and R ' are the same plane. Then R
contains f(p’), since f(p’) is in R’.
Since f(p) and f(p’) are in both R and Q, they lie on the line where R intersects
Q—a line in Q perpendicular to l. We know that the distance d between p and l p equals
the length of l p ' p ' , which is distance between l p and f ( p) , as well as between l p and
f ( p' ) . If f ( p) and f ( p' ) lie on the same line an equal distance away from l p and can’t
be on opposite sides of l p because it lies on l, they must be the same point. Thus for each
point p in P, a fold maps the reflection of p' to f ( p) .‫ٱ‬
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Thus we see that folding a plane isn’t a one-to-one function, so we can’t undo our
fold. In other words, a totally precise folding of completely flat paper, which seems like
ideal origami, is in fact completely impractical. The paper, having no thickness, appears
to absorb itself wherever it comes in contact with itself, leaving us nothing we can peel
apart and unfold! Clearly there should be another model of folding.
Oftentimes in origami, the most important result of a fold is not the new position
of the paper, but rather, the fact that a crease forms, marking the line of the fold. In
perfect paper lacking thickness, even if we could unfold a fold, a crease wouldn’t form,
because a crease results from the slight stretching and buckling of real-world paper when
it bends to meet itself and then is pressed. To see that paper without thickness wouldn’t
have to stretch and buckle, note that when real-world paper is folded, parts of the paper
relatively far from the crease are put under no stress. It’s only when we observe the
paper within half a millimeter or so of the folded edge that we can see the slight
wrinkling, stretching, and buckling. On the other hand, when we observe cardboard we
note that the material a centimeter away from the folded edge exhibits signs of buckling
and stretching. It appears that the buckling and stress associated with forming creases is
proportional to the thickness of the material being folded, from which we might infer that
paper without thickness would exhibit no buckling and stretching whatsoever, and thus,
not form a crease.
Once again, folding planes does not appear to model folding real-world paper
very well at all. The hunt for some ideal folding model founders. If no ideal paper and
no ideal precise fold exist that model real paper in any practical way, we can see that a
practical model of paper folding will require an axiomatic approach that assumes a real
paper can be folded accurately and still be unfolded. In fact, this assumption was
inherent in the first proof I gave.