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arXiv:1807.09557v1 [math.HO] 25 Jul 2018
Construction of a regular hendecagon by two-fold
origami
Jorge C. Lucero∗
July 26,2018
Abstract
The regular hendecagon is the polygon with the smallest number of sides that
cannot be constructed by single-fold operations of origami on a square sheet of
paper. This article shows its construction by using an operation that requires two
simultaneous folds.
1Introduction
Single-fold origami refers to geometric constructions on a sheet of paper by performing
a sequence of single folds, one at a time [1]. Each folding operation achieves a minimal
set of specific incidences (alignments) between given points and lines by folding along
a straight line, and there is a total of eight possible operations [2]. The set of single-
fold operations allows for the geometric solution of arbitrary cubic equations [3,4].
As a consequence, the operations may be applied to construct regular polygons with
a number of sides nof the form n=2r3sp1p2. . . pk, where r,s,kare nonnegative
integers and p1,p2,...,pkare distinct Pierpont primes of the form 2m3n+1, where
m,nare nonnegative integers [5]. For example, previous articles in Crux Mathemati-
corum have shown the construction of the regular heptagon [6] and nonagon [7]. Let
us note that this family of regular polygons is the same that can be constructed by
straightedge, compass and angle trisector.
Number 11 is the smallest integer not of the above form (the next are 22,23,25,
29,31,. . . ); therefore, the hendecagon is the polygon with the smallest number of sides
that can not be constructed by single-fold origami. In fact, its construction requires
the solution of a quintic equation [5], which can not be obtained by single folds. It
has been shown that any polynomial equation of degree nwith real solutions may
be geometrically solved by performing n−2simultaneous folds [1]. Hence, a quintic
equation could be solved by performing three simultaneous folds. However, a recent
paper presented an algorithm for solving arbitrary quintic equations with only two
simultaneous folds [8]. In this article, the algorithm will be applied to solve the quintic
equation associated to the regular hendecagon, and full folding instructions for its
geometric construction will be given.
∗Dept. Computer Science, University of Brasília, Brazil. E-mail: lucero@unb.br
1
J. C. Lucero: Regular hendecagon by two-fold origami
x
y
2π/11
1
z0
z1
z2
z3
z4
z5
z6
z7
z8z9
z10
Figure 1:The regular hendecagon.
Let us note that another problem related to a quintic equation, the quintisection of
an arbitrary angle, has also been solved by two-fold origami [9].
2Quintic equation for the regular hendecagon and its solution
2.1Equation
A similar approach to that used for the heptagon [6] is followed.
Consider an hendecagon in the complex plane, inscribed in a circle of unitary
radius (Fig. 1). Its vertices are solutions of the equation
z11 −1=0(1)
One solution is z=z0=1, and the others are solutions of
z11 −1
z−1=z10 +z9+z8+···+z+1=0(2)
Assuming z6=0and dividing both sides by z5produces
z5+z4+z3+···+1
z3+1
z4+1
z5=0(3)
Next, let t=z+z=2Re z. Since |z|=|z|=1, then z=z−1and t=z+z−1.
Expressing Eq. (3) in terms of t, the equation reduces to
t5+t4−4t3−3t2+3t +1=0(4)
Solutions of Eq. (4) have the form tk=2Re zk=2cos(2kπ/11), where k=
1,2,...,5and zkare vertices as indicated in Fig. 1. Note that, due to symmetry on
the x-axis, vertices except z0appear as complex conjugate pairs with a common real
part.
2
J. C. Lucero: Regular hendecagon by two-fold origami
2.2Solution
Any quintic equation may be solved by applying the following two-fold operation (see
Fig. 2): given two points Pand Qand three lines ℓ,m,n, simultaneously fold along a
line γto place Ponto m, and along a line δto place Qonto nand to align ℓand γ[8].
ℓδγ
n
Q
m
P
Figure 2:A two-fold operation. Red lines γand δare the fold lines.
The coordinates of points Pand Qand equations of lines ℓ,m, and nare computed
from the coefficients of the quintic equation to solve. In the case of Eq. (4), we have
P(−5
2,−3),Q(0,1),ℓ:x=0,m:x= −3
2, and n:y= −1(Fig. 3). Their calculation is
omitted here for brevity; complete details of the algorithm may be found in Ref. [8].
x
y
Q
ℓ
m
n
Q′
R
S
P
P′
T
δ
γ
Figure 3:Geometric solution of Eq. (4), for t=2cos(2π/11).
Let us demonstrate that the folds solve Eq. (4). Folding along line δreflects point
Qonto Q′∈n. Assume that point Q′is located at (2t,−1), where tis a parameter.
Then, the slope of segment QQ′is −1/t, and its midpoint Ris at (t,0). The fold line δ
is perpendicular to QQ′and passes through R; therefore, it has an equation
y=t(x−t)(5)
Folding along line γreflects point Pon P′∈m. Assume that point P′is located
at (−3
2,2s), where sis another parameter. Then, the slope of segment PP ′is 2s +3,
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J. C. Lucero: Regular hendecagon by two-fold origami
and its midpoint Tis at (−2,s−3
2). The fold line γis perpendicular to PP ′and passes
through T; therefore, it has an equation
y= − x+2
2s +3+s−3
2
= − x
2s +3+2s2−13
2
2s +3(6)
Now, the same fold along δreflects line ℓover γ. Let Sbe the point of intersection
of δand ℓ. The y-intercept may be obtained by letting x=0in Eq. (5), which produces
y= −t2. Then, the slope of segment SQ′is (t2−1)/(2t). Line γmust pass through
both Q′and S, and therefore it has an equation
y=t2−1
2t x−t2(7)
Since Eqs. (6) and (7) describe the same line, then their respective coefficients must be
equal:
−1
2s +3=t2−1
2t (8)
2s2−13
2
2s +3= −t2(9)
Solving Eq. (8) for sand replacing into Eq. (9), we obtain Eq. (4). Therefore, the x-
intercept of δ(i.e., tat point R) is a solution of Eq. (4). Note that the equation has five
possible solutions, and Fig. 3shows the case of t=2cos(2π/11)≈1.6825 . . ..
3Folding instructions
The following diagrams present full instructions for folding the regular hendecagon
on a square sheet of paper.
Steps (1) to (7) produce points Pand Qand lines ℓ,m,nof Fig. 3. The center
of the paper is assumed to have coordinates (0,−1), and each side has length 8. In
step (1), the vertical and horizontal folds define lines ℓand n, respectively. In step
(3), the intersection of the fold line with the vertical crease is point Q. In step (5), the
small crease at the bottom will mark the position of line m, after folding the paper
backwards in the next step. Finally, step (7) defines point P.
Next, steps (8) and (9) produce the fold lines γand δ, respectively, of Fig. 3. As a
result, point Q′in step (11) is at a distance of 4cos(2π/11)from the center of the paper.
In the same figure, point Ais adopted as one vertex (z0in Fig. 1) of an hendecagon
with radius of 4units. The fold in step (11) produces a vertical line through Q′, and
steps (12) and (13) rotate a 4-unit length (from the paper center to point A) so as to
find the next vertex of the hendecagon (point Din step 14). The remainder steps, (14)
to (20) are used to find the other vertices and folding the sides.
4
J. C. Lucero: Regular hendecagon by two-fold origami
(1) Fold and unfold verti-
cally and horizontally.
(2) Make a small crease at the
midpoint.
(3) Fold backwards the top
edge to the horizontal center
line.
(4) Fold and unfold.
m
(5) Make a small crease at the
bottom edge, and call this
crease m.
(6) Fold backwards to align
the vertical crease made in
step (4) with the small crease
at the right.
P
(7) Make a small crease at the
left edge, and call this point
P.
P
m
(8) Fold backwards to place
Ponto crease mmade in step
(5), so that...
Q
Q′
S
P
m
(9) ...a fold through point S
places Q′onto Q.
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J. C. Lucero: Regular hendecagon by two-fold origami
(10) Unfold.
Q′
A
(11) Fold along a vertical line
through point Q′to place
point Aonto the horizontal
crease and unfold.
A
(12) Fold through the center
of the paper to place point A
onto the vertical crease made
in step (11), and unfold.
B
B′
C
C′
(13) Fold backwards to place
point Bonto point B′, and
point Conto point C′.
D
(14) Fold through the center
of the paper and point D, at
the upper intersection of the
crease made in step (11) with
the right edge.
(15) Fold backwards the
lower layer using the edges
of the upper layer as guide-
lines, and next unfold the
upper layer.
E
(16) Fold through the center
of the paper and point E, at
the lower intersection of the
crease made in step (11) with
the right edge.
(17) Fold backwards the
lower layer using the edges
of the upper layer as guide-
lines, and next unfold the
upper layer.
(18) Repeat the fold in step
(14).
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J. C. Lucero: Regular hendecagon by two-fold origami
(19) Fold backwards the
lower layer using the edges
of the upper layer as guide-
lines, and next unfold the
upper layer.
(20) Finished hendecagon.
References
[1] R. C. Alperin and R. J. Lang. One-, two-, and multi-fold origami axioms, in R. J.
Lang, editor, Origami 4- Fourth International Meeting of Origami Science, Mathematics
and Education, A. K. Peters, pp. 371–393,2006.
[2] J. C. Lucero. On the elementary single-fold operations of origami: reflections and
incidence constraints on the plane, Forum Geometricorum,17:207–221,2017.
[3] R. C. Alperin. A mathematical theory of origami constructions and numbers, New
York J. Math.,6:119–133,2000.
[4] R. Geretschläger. Euclidean constructions and the geometry of origami, Math.
Mag.,68:357–371,1995.
[5] A. M. Gleason. Angle trisection, the heptagon, and the triskaidecagon, Amer. Math.
Monthly,95:185–194,1988.
[6] R. Geretschläger. Folding the regular heptagon, Crux Mathematicorum,23:81–88,
1997.
[7] R. Geretschläger. Folding the regular nonagon, Crux Mathematicorum,23:210–217,
1997.
[8] Y. Nishimura. Solving quintic equations by two-fold origami, Forum Mathematicum,
27:1379–1387,2015.
[9] R. J. Lang, Angle quintisection, 2004. Available at http://www.langorigami.com/
article/angle-quintisection.
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