Algebra and geometry by paper-folding
Benedetto Scimemi
The game, the art and the science of paper-folding
Throughout the ages and all over the world, children have enjoyed folding paper
in order to make hats, vessels and various flying objects. For centuries, in the
Far East, delicate and variously coloured papers have been used to produce
refined objects that are either useful or just beautiful (the first ones to appear in
Europe were probably the Chinese fans, lanterns and umbrellas). In Japan,
paper-folding is a genuine art, Origami, which requires ingenuity and aesthetical
taste. The pedagogical importance of paper-folding has also been largely
recognised by educators, as this activity requires both manual and mental
control, while at the same time leaving room to imagination and creativity. And
one must not undervalue the fact that all can be made with very inexpensive,
harmless and common materials.
Less attention has perhaps been paid to the scientific aspects of paper-folding,
which are crucial for all such applications: why, for instance, are creases
perfectly straight? Note that this does not depend on our will: it is so even if the
paper happens to fall and slip under our shoe. Why can a piece of paper
perfectly wrap, without creases, the main body or the neck of a bottle, but not
both of them? Without giving a physical explanation, we shall take advantage of
these features to provide an experimental approach to some problems of plane
geometry and algebra. More precisely, we shall describe how paper-folding,
without any other device (except possibly a pencil, as a track reinforcement) can
yield geometrical constructions, to be compared with those one gets by the
traditional ruler-and-compass methods. These constructions will be introduced
as a game whose rules correspond to a list of "permitted" folding procedures.
What is surprising is that this game turns out to be more powerful than the
traditional method: besides achieving all the traditional constructions, it also
solves problems, such as the trisection of an angle and the duplication of a cube,
which cannot be solved by ruler and compass. Even if the reader does not have
enough patience to follow the details of these last aspects, we hope that some
teachers will be encouraged by this article to use paper-folding in introducing or
reviewing part of their program (e.g. symmetries); we can guarantee that the
class will show unusual interest and enjoyment.
Some historical background
The idea of using paper-folding for teaching geometry is not a new one. At the
beginning of this century an Indian teacher of mathematics who had used this
approach to geometry in a class where compasses had never been seen wrote
a little book [8] which enjoyed some international success. Many years later the
idea was reconsidered by some American high-school teachers, who wrote
various articles and booklets to popularise the subject. In 1960 Martin Gardner in
Scientific American [1] gave a good review and a large bibliography on the
subject. In none of these books and articles, however, was paper-folding shown
to go beyond ruler-and-compass constructions.
In 1980, in the Japanese edition of Scientific American , an elegant way to trisect
the angle by paper-folding was described [2]. This article aroused the curiosity of
a Japanese physicist, H.Huzita, who requested my help in order to understand
which mathematics was behind it. Short later we could retrieve some papers
written on the subject in the 1930's by Margherita Piazolla Beloch, a teacher of
mathematics at the University of Ferrara. In her little-known works [6], [7], written
in Italian, the algebraic aspects are not fully developed and the classical
problems are not solved, but her arguments more than suffice to explain why by
her approach traditional constructions could be surpassed. I reconsidered the
whole subject and wrote a paper which was submitted to The American
Mathematical Monthly in 1988. Similar papers were later published in the
proceedings of some meetings on mathematical games [3] but these had little
more than private circulation.
First experiments and practical instructions.
The reader is invited to obtain some standard sheets of thin white paper and to
devote some time to practice manipulation, in order to produce various creases.
We shall first observe two facts:
a) As we said, creases are perfectly straight: thus any crease represents a
straight line on the plane of the sheet. Therefore producing a crease is
equivalent to drawing a line and in fact, from now on, creases and lines will be
identified. If the line's track must be made more evident, one may run a pencil
down the crease.
b) Each crease t produces a (partial) overlapping of the sheet on itself:
thus each point P of the plane is superimposed on another point P'. Since in
folding the paper is not stretched or expanded, P' is the mirror image of P with
respect to line t (fig.1). We shall say that t takes (or carries) P onto P'. If a point
P has been marked on the plane, the crease t produces a new point P'. In this
way any crease realizes a line symmetry (reflection) of the plane.
In order to identify and mark a new point as P', we shall take advantage of a
special feature of thin paper, i.e. its translucency: putting a light source behind
the paper, or possibly sticking the paper against a windowpane, we first mark P'
on the back side of the folded paper and then replicate it, by translucency, on the
main side (alternatively, one could use a fresh ink pen so that a point's ink track
is transferred to the folded sheet with which it comes in contact ).
Creases that are useful for geometry.
So far, our creases have had no special purposes. Now we want to produce
particular creases, that meet specific geometrical requirements. Here is a list of
folding operations (or procedures ) which can be easily performed in practice, as
the reader will discover, with great precision. Each procedure solves a
geometrical problem. The first examples deal with very familiar problems:
1) Given a point P, find creases that take P into itself: P=P'. Equivalently,
we shall say that such creases fix P. Clearly, there are an infinite number of
such creases, i.e. all the lines passing through P.
2) Given a line r, find creases that take r into itself: r'=r. These are (r
itself and) all the lines that are normal to r.
3) Given two (distinct) points P and Q, find a crease that fixes both : P'=P
and Q’=Q. This is the line passing through P and Q.
4) Given two (distinct) points P and Q, find a crease that interchanges the
two points: P'=Q and Q'=P. This is the perpendicular bisector of the segment
PQ.
5) Given a point P and a line r (not containing P), find a crease that fixes
both: P'=P and r'=r. This is the line that is normal to r and that passes through
P.
6) Given two (different) lines, r and s, find creases that interchange the
two lines: r'=s and s'=r. Here, two distinct creases both solve the problem, i.e.
the two bisectors of the angle having sides r and s (angle bisection).
Let us stop for a moment to note that, even if we were to limit ourselves to these
constructions, paper-folding would already be a powerful device. First of all, proc.
3) clearly replaces the ruler: in traditional ruler-and-compass constructions,
centimetres or other distances are not marked on the ruler (in fact, marks on the
ruler must not be used), and this procedure allows one only to draw a line
through two given points. Also, if we compare procedures 4), 5), and 6) with the
traditional constructions, we notice that the latter require repeated and combined
applications of ruler and compass, whereas just one crease yields the result.
Incidentally, by making two creases (i.e. applying 5) twice), we also find the line
through a given point and parallel to a given line.
Let us now continue and introduce new reasonable ways of making creases:
7) Given a point F and a line d (not containing F), find creases that carry
point F onto line d : F'∈ d . We find infinitely many creases, i.e. all the lines that
are tangents to the parabola (F, d ), having F as focus and d as directrix.
Fig. 2 suggests a proof of this characterisation, based on the familiar property of
the parabola as a locus: its points have equal distances from the
focus F and the directrix d. Having at one's disposal a high number of tangent
creases, one can try to sketch the parabola (which itself obviously cannot be
drawn by ruler and compass).
8) Given two points C and F and a line d (not containing F), find creases
that fix the first point and carry the second point onto the given line: C'=C. If C
lies outside the parabola (F,d) there are two such creases: the two tangents from
point C. If C lies inside, no tangent can pass th rough it and the problem has no
solutions.
This procedure has also a different and useful interpretation: since the
distance |F'C| equals the distance |FC|, the point F' is an intersection of line d
with the circle that has centre C and that passes through F. In fig. 3 the circle
has two such intersections F', F" ; if C is inside the parabola, the distance |FC|
(i.e. the circle radius) is smaller than the distance of F from the line d, and
therefore the circle and the line have no common point.
Notice that - as before - we have not drawn the circle (this is the only operation
where a compass is really indispensable) but we have found its intersections with
a line, and this is what is needed in traditional ruler-and-compass constructions.
At this point we can see why paper-folding can fully
replace ruler and compass. But we want to go further.
9) Given two (distinct) points R, S and two (distinct) lines r, s, find
creases that carry the first point onto the first line and, simultaneously, the
second point onto the second line: R'∈r and S'∈ s. These are the common
tangents to two parabolas. Depending on the relative positions of the two
parabolas (R, r), (S, s), one finds either three such creases (as in fig.5), or two or
just one such crease (infinitely many tangents are only found in the exceptional
case that r, s are parallel and the two points lie on them). Each solution creates
an isosceles trapezoid RR'SS' or RR'S'S (fig.4). In order to find the paper
creases, the following practical suggestion helps: taking advantage of
translucency, first superimpose R on r, then let R run along r and meanwhile pay
attention to the motion of the point S, waiting until it comes to lie on line s.
For those who are familiar with analytical geometry, the existence of either one or
three solutions suggests that an algebraic problem of degree 3 is involved
somehow. In order to show this, we choose a coordinate system such that S
(0,-1); and s : y = 1 (as in fig.6. This can be done without loss of generality).
Then a line (a crease) t reflects (carries) S onto S' (c,1) on the line s if and
only if t passes through the midpoint (c/2, 0) of segment SS' and is normal to it.
Let t reflect the second point R(p,q) onto R'(x,y). Then we must have
2/c = (q-y)/(p-x) and -c/2 = [ (q+y)/2 - 0] / [(p+x)/2 - c/2 ]
the former equation meaning that RR' and SS' are parallel with slope 2/c, the
latter that the midpoint of RR' lies on c, which is normal to SS' and has slope -c/2
. By eliminating c one gets
( p-x ). [(p+x)(q-y) - 2(p-x)] + (q-y)2(q+y) = 0
This algebraic curve, which may be called the paper-folding cubic, is the locus of
R' when S' runs along s. We now introduce the second given line r: y = ax+b
and require that R' belong to r (procedure 9). This implies intersecting the
paper-folding cubic with line r . We eliminate y = ax+b by substitution into the
previous algebraic equation and end up with an algebraic equation of order 3 in
x, whose coefficients are functions of the parameters p, q, a, and b. Depending
on their values (i.e. on the relative positions of S, R, r, and s) this yields either
one or three real solutions for x (and for y=ax+b), i.e. one or three points R',
each leading to a crease. Of course, special situations may lead to some
coincidence (multiple roots).
The rules of the game
Assume that on our sheet a set of points and lines are given (i.e., marked), that
we shall call the initial set . If we do not want the game to stop immediately, we
should assume that the initial set contains at least three non collinear points.
The game consists of producing new points and lines by applying, one after the
other, a number of permitted procedures, as described below. As an example,
starting with two initial points (stage 0) a new line is
obtained by joining them (stage 1). Then a new point is obtained (stage 2) as
intersection of this line with another line that was in the initial set, and so on.
In general, each step (stage n) enriches the former set of points and lines. We
shall say that a new point or a new line is obtained by permitted procedures
either from the initial set or from pre-existing points and lines, i.e. objects
obtained in previous stages (0, 1, .., n-1).
List of the permitted procedures
A new point must be either
i) the intersection of two (distinct) pre-existing lines
or
ii) the point where a pre-existing line carries a pre-existing point.
A new line must be either
j) the crease through two pre-existing points (proc.3)
or
jj) a crease that carries two pre-existing points onto two pre-existing lines
(proc.9, exceptional cases being excluded).
Notice that we have excluded all procedures that admit of infinitely many
solutions (proc. 1, 2, 7). Among the other procedures, one might wonder why we
have not listed 4, 5, 6, 8. As a matter of fact, they turn out to be permitted, as
particular cases of 9). In fact, if R ∈ s and S ∈ r , then the three solutions are
the perpendicular bisector of the segment RS and the two bisectors of the angle
rs. If R ∈ r and S ∉ s , then two of the solutions are just those of 8). Finally, if
R ∈ r and S ∈ s (and r and s are not parallel) one finds the line normal to r
through S, the line normal to s through R and once more the line through R and
S.
Now that we have specified our game's rules, we can look at specific
goals and see if we can achieve them.
Classical problems and their solution.
We shall mainly treat problems of algebraic degree 3, as our treatment of
problems of lower order would not substantially differ from what can be done with
ruler and compass (some problems of degree 2 will be considered within the
construction of the regular polygons, see e) below).
a) Trisection of an angle. If in the initial set, two lines r and v form an angle α,
can we play with our rules until we see two lines appear on our sheet, that form
an angle α/3?
We shall mainly treat problems of algebraic degree 3, as our treatment of
problems of lower order would not substantially differ from what can be done with
ruler and compass (some problems of degree 2 will be considered within the
construction of the regular polygons, see e) below).
b) Duplication of a cube. If we are given two points O and S, can we apply our
game rules until two points appear, whose distance is OS 3√2 ?
c) Solving an algebraic equation of degree 3. If p0 , p1 , and p2 are three real
numbers and we can find, in the initial set, three segments of length |p0|, |p1|,|p2|,
can we play until a segment of length |v| appears, where v is a root of the
polynomial x3+p2x2 +p1x+p0 ?
d) Characterizing all constructible points and lines by algebraic conditions on
their co-ordinates and coefficients.. If the initial set consists of points and lines
whose coordinates and coefficients belong to a certain field, can we find
algebraic conditions with respect to that field for the coordinates and coefficients
of the new points and lines that we get by our paper-folding procedures?
e) Constructing a regular polygon with n sides (e.g. the heptagon). If two
points O and A1, are in the initial set, can we play according to our rules until
new points A2, A3, A4, ..., An appear, so that they form a regular n-gon A1A2 ...An
that has O as center?
Here again, it will be convenient to give our problems an analytical interpretation
and to choose a suitable coordinate system. Since our initial set must contain
three non-collinear points O, X, and Z , we can easily produce two orthogonal
creases that will serve as the axes of a coordinate system (proc. 5) and 8)) with
O=(0,0) and X=(1,0) . Then it is easy to construct the set of lines x=h and y=k
(where h and k are arbitrary integers) as well as their intersections, i.e. the
integer points. From these, by traditional constructions, one can also produce all
rational points. Let us now solve the classical problems.
a) Trisection of an angle.
We first construct a line s, normal to line r and passing through the intersection
O of lines r and v (figure 7). On line v we produce (using some element of the
initial set) two points R and S, which are symmetrical with respect to O. We now
apply procedure 9 to find a crease t that carries
R into R'∈ r,
S into S'∈ s
and O into O'
We want to show that the angle β = ∠O'OR' is one third of the angle α =
∠SOR'. In fact ∠S'OR' is a straight angle by construction, and hence the same
holds for ∠SO'R , which is obtained from the former angle by reflection across
crease t. It follows that points S, O' and R lie on a circle with centre O , and
therefore |OO’|=|OR|. Therefore triangle ROO’ is isosceles, as
∠ORO'=∠RO'O=∠O'OR'=β, where the symmetry was used again. The external
angle ∠SOO' equals 2β and therefore we have α = ∠SOR' = ∠SOO'+∠O'OR' =
2β + β = 3β as we wanted. Thus line OO' trisects α.
This is a slight variation of the construction that was published in [2]. In this case,
three creases solve the problem. The reader will easily carry them out, andwill
see why three solutions should be expected.
b) Duplication of a cube.
We first produce a net of mutually orthogonal creases as in figure 8. Then we
choose the points O=(0,0), S=(0,-1) and R(-2,0) and the lines s: y=1 and r: x=2.
If we now apply again our procedure 9, we find only one crease t . Let V(v,0)
denote the intersection of line t with the x-axis and U(0,u) the intersection of t
with the y-axis. Then triangles SOV, VOU and UOR are similar and therefore we
have v/1=u/v=2/u , i.e. v3=2. But then we have VO=3√2 SO and the problem is
solved: a cube with side |VO| has twice the volume of a cube with side |SO|.
This problem is a special case of problem c), namely that of the polynomial x3-2 .
Any other real number d can play the role of 2. It suffices to take point R as (d,0), line r as x=d and leave the rest unchanged. One finds V(3√d, 0).
Thus we have shown that paper-folding can produce all cubic roots.
c) Real roots of a real polynomial of degree 3.
In the same way we can prove that paper-folding solves any algebraic equation
of degree 3. All we need to do is properly choose points R and S and lines r
and s , as a function of the coefficients of the polynomial. Then we can again
apply procedure 9 and exploit the similarity of three triangles. Here are the
details of the construction (the basic idea is in [7]).
Given the real polynomial x3+p2x2+p1x+p0 , choose S(0,-1); R(p0-p2,-p1); s: y=
1; and r : x= -p0-p2. Find a crease t carrying S into S' ∈ s and R into R' ∈ r
.
Let V(v,0) and U(-p2, u) denote the points where the crease t meets the
lines y = 0 and x= - p2 . Then V is the midpoint of SS' and U is the midpoint
of RR'. The similarity of triangles SOV, V(-p2,0)U and U(-p2, -p1)R yields the
equations
-v /1 = -u /(p2+v) = p0 /(p1+u)
By eliminating u one finds indeed v3 + p2v2 + p1v + p0 = 0 so that the point V and therefore the root v - has been constructed by paper-folding.
Equations of degree 4 can be reduced - by algebraic means - to those of
lower order (see, e.g. [4]) and therefore, in theory, they too could be solved by
paper-folding. However, this procedure is somewhat artificial and cumbersome.
These remarks are already in [6] and [7]. Equations of order greater than 4
cannot in general be solved by this method.
d) Algebraic characterization of constructible points and lines.
The arguments we have just developed can be formalized in order to prove what
follows. A point P (or a line p ) can be constructed by paper-folding from a set
A if and only if there exists a chain of fields K0 ≤ K1 ≤ ... ≤ Ks-1 ≤ Ks such that
K0 is generated by the coordinates (coefficients) of the elements of A, Ks
contains the coordinates of P (coefficients of p) and for all i =0, .. , s-1 the
intermediate extension Ki+1 | Ki has degree [Ki+1 : Ki ] ≤ 3 .
Here is a sketch of the proof. Suppose point P is constructible. Then P is
obtained from the initial set A by successive applications of permitted
procedures. At the i-th stage of the construction we consider field Ki which
contains the coordinates and coefficients of the geometrical objects that have
appeared up to that stage. Now the i+1st step produces new objects whose
coordinates and coefficients either are obtained by rational operations on
elements of Ki (this is the case, e.g., if we intersect two pre-existing lines or
produce a new line through two pre-existing points) or are roots of a polynomial
of degree ≤ 3 with coefficients in Ki (if we apply rule jj), equivalent to procedure 9
). In any case, they belong to a field Ki+1 that is an extension of Ki of degree ≤ 3.
This covers the necessary condition. As for sufficiency, we know that each
extension Ki+1 | Ki of degree ≤ 3 is obtained by adding to Ki a root of a
polynomial of degree ≤ 3. Since such a root, as we have seen in c), can be
produced by paper-folding, a chain of fields K0 ≤ K1 ≤ ≤ ...Ks-1 ≤ Ks implies a
corresponding set of procedures yielding the final object P (or p).
e1) Construction of a regular pentagon.
It is well-known that this problem is solvable by ruler and compass, so that our
paper folding procedure 8) should certainly suffice. We shall construct a
pentagon whose vertices are the points (2cos(2kπ/5), 2sin(2kπ/5)) for
k=1,2,3,4,5. We first consider the complex root of unity z = exp(2πi/5) = cos
(2π/5) + i sin (2π/5) and easily deduce from 0 = z5-1 = (z-1)(z4+z3+z2+z+1) that
the real number z + z-1 = 2 cos (2π/5) is a root of the polynomial x2+x-1. In
order to apply the general method for polynomials of degree 3, we artificially
introduce the extra-root x=0 and deal with the polynomial x3+x2-x. Now we
look for creases that carry point S(0,-1) onto line s: y=1 and point R(-1,1) onto
line r : x=-1. We find three such creases: two of them meet the x-axis in the
points V1(v1,0) = (2 cos(2π/5), 0) = ((-1+√5)/2, 0) and V2(v2,0) = (2 cos 4π/5, 0)
= ((-1-√5)/2, 0)). The third crease is the x-axis and yields the trivial root.
Incidentally, we notice that the similarity of the three triangles now reads v/1 = 1/
(1+v), i.e. the well-known definition for the golden section v . Thus we have
constructed both v1 = 2 cos(2π/5) = 2 cos(8π/5) and v2 = 2 cos(4π/5) = 2 cos
(6π/5). The next step consists in intersecting the lines y=v1 and y=v2 with the
circle whose centre is O(0,0) and radius 2. To do this we once more apply
procedure 8 and look for creases fixing O(0,0) and carrying C(2,0) onto the lines
x=v1 and x=v2. For each case there are two solutions, so that in the end we find
for C' the four points C2, C3, C4, C5 that (together with C1=(2,0)) are the vertices
of a regular pentagon.
e2) Construction of the regular heptagon. This does involve a problem of
degree 3. As in c1) we wish to produce the complex roots of unity zk = exp
(2πik/7) for k=1,2,...6. All these are roots of z7-1 = (z-1). (z6+z5+z4+z3+z2+z+1)
and therefore are roots of the second factor. By an easy calculation we find that
the real numbers vk = zk + zk-1 = 2 cos (2πk /7), for k=1,2,3 are the roots of the
polynomial x3+ x2-2x-1. We now look for creases carrying S(0,-1) into some
point S' of the line s: y=1 and R(-2,2) into R' on the line r: x= 0. We find
three such creases, which meet the x-axis in the three points V1(v1,0)=(2 cos
(2π/7), 0), V2(v2,0)=(2 cos(4π/7), 0) andV3(v, 0)=(2 cos(6π/7), 0) (see fig. 9).
Thus we have constructed cos(2πk/7) for k= 1, 2, .., 6. We then go on as in
e1): we apply procedure 8 to find creases fixing O and carrying C1(2,0) onto line
x = vk. For each k=1,2,3 , we find two creases and six vertices, as we wanted
(see fig. 10).
e3) Construction of the regular n-gon.
According to a theorem of Gauss (see e.g. [4]), a regular n-gon is constructible
by ruler-and-compass if and only if n = 2h p1p2...ps where the pi are different
primes of the form p=1+2k, e.g. 3, 5, 17, 257... (the exponent k must also be a
power of 2). A similar result can be proved for paper-folding: here it is possible to
construct regular n-gons if and only if n has the form n = 1 + 2h3kq1q2...qm, and
the different primes qj have the form q=1+2u+13v. This implies that constructing
the regular polygon with 7, 9, 13, 19... sides is impossible with ruler-andcompass but becomes possible by paper-folding (n=11 is the smallest impossible
case). Obviously the construction becomes more and more difficult for increasing
values of n. For q=1+2u+13v one needs u+v +1 creases: u+1 of them are of type
8) and of degree 2, while v are of type 9) and of degree 3. For example, we
need two creases for n=7 and three creases for n=13. As for n=9, the best way
is first to construct an angle 2π/3 (this is easy by ruler and compass) and then
trisect it to produce the angle 2π/9.
We sketch out some ideas that explain the form of the number n for
which a regular n-gon is constructible by paper-folding. If the construction of an
n1n2-gon is possible, then, of course, a n1-gon is constructible as well, by
jumping over a certain number of vertices. Since bisections and trisections of
angles are possible, this explains the factors 2h and 3k in the formula for n. Thus
we need merely consider possible prime-powers qs, with q>3. As in the cases
we have already studied, the problem reduces to constructing v = cos(2π/qs). By
arguments standardly used for ruler-and-compass constructions [4], one finds
that the degree of the field extension is [Q(v):Q]=qs-1(q-1)/2 , so that the
preceding characterization of constructible points gives qs-1(q-1)/2 = 2u3w i.e.
s=1, q=1+2u3w and we have found the required form. The last step of the proof
consists in proving that q=1+2u3w is also a sufficient condition for the existence
of a chain of intermediate fields Q= K0 ≤ K1 ≤... Ku+w = Q(v) where [Ki+1 : Ki ]=2
for u values of i and [Ki+1 : Ki ]=3 for w values of i. This again is a standard
argument, but requires some knowledge of Galois theory. In general there exist
several chains of intermediate fields, all of them leading to different paper-folding
contructions. For example, for n=7, in the chain Q= K0 < K1 < K2 = Q(cos(2π/7),
sin(2π/7)) we have two choices for the intermediate field K1 : either K1 = Q(cos
2π/7) or K1 = Q(√7) , the former corresponding to the construction we gave
above, the latter to first constructing the number √7 (by traditional procedures)
then finding the roots of the polynomial x3+√7x2 -√7.
Unlike the Fermat primes 3, 5, 17, 257 ... whose list of known members
stops rather soon, there seems to be an abundance of primes of the form q=1+
2u3w. Here is the list of such primes with at most 5 digits
3, 5, 7, 13, 17, 19, 37, 73, 97, 109, 163, 193, 257, 433, 487, 577, 769, 1153,
1297, 1459, 2593, 2917, 3457, 3889, 10369, 12289, 17497, 18433, 39367,
52489, 65537. (Fermat primes are in bold). Their properties and distribution may
well merit independent investigation. Research on this subject was recently made
by Kirfel and Rødseth; their paper [5] appears in this volume.
Conclusion
As we said in the beginning, from some of these exercises one can derive ways
to make the teaching of geometry more attractive: since the definitions and
theorems become experiments, this approach may be successful when one
deals with students who find the traditional approach to geometry too abstract.
However, from an educational standpoint, it is worth emphasizing another aspect
: we have solved new problems because we have welcomed new rules
(substantially, the procedure 9)) into our permitted set. Nobody has obliged us to
do so (however acceptable the new rules may seem); and nobody can prevent
us from inventing a new game where new procedures are permitted, and where
new problems can be solved. This situation, by analogy, suggests a more
general remark: many parts of mathematics have developed merely because
their contributors were quite reasonably treating our discipline not as an
absolutely fixed science, but rather as a game whose rules can be changed:
traditional geometry, as an example, is the game whose rules - the postulates were codified by Euclid; if we drop some of the rules or invent new ones, we find
a new geometry, with new problems and new solutions!
Literature
[1]
M.Gardner, Origami, reprinted as cap.16 inMore mathematical puzzles
and diversions, Penguin books, 1961.
[2]
K. Husimi, Origami no kagaku (Science of Origami), Saiensu (Japanese
translation of the Scientific American), Huroku (Appendix) Oct. 1980, p.8.
[3]
H. Huzita and B.Scimemi, The algebra of paper-folding (Origami),
Proceed. of the meeting of Origami Science and Technology, Ferrara 1989.
[4]
N. Jacobson, Basic Algebra I, Freeman and Co., S. Francisco, 1974.
[5]
C.Kirfel and Ø.Rødseth, On the primality of 2h.3n+1 in this volume.
[6]
M. Piazolla Beloch, Sul metodo del piegamento della carta per la
risoluzione dei problemi geometrici, Periodico di Matematiche, Serie IV, 41, p.
104.
[7]
M. Piazolla Beloch, Sulla risoluzione dei problemi di 3. o e 4.o grado col
metodo del ripiegamento della carta, Scritti matematici offerti a Luigi Berzolari,
1936.
[8]
T.S. Row, Geometric Exercises in Paper Folding, Chicago: The Open
Court Pub.Co., 1905, reprint by Dover Pub.Inc., New York, 1966.
This article, except for a few additions (p.10-12), is the translation from Italian of Algebra e
Geometria piegando la carta published in Matematica: gioco ed apprendimento, a cura di
B.d'Amore, ed. Apeiron, Bologna,1990. A version in Norwegian appeared in NORMAT, 46
(1998), p.170-182. A substantially equivalent treatment of this subject can be found in the book
Geometric Constructions by George.E.Martin (Springer,1998) where the full chapter 10 is
dedicated to paper-folding.