applied geometry: foldings and unfoldings - tu wien · applied geometry: foldings and unfoldings...

Applied Geometry: Foldings and Unfoldings

Hellmuth Stachel

[email protected] — http://www.geometrie.tuwien.ac.at/stachel

ISTEC 2016, International Science and Technology Conference

July 13–15, Vienna University of Technology

Table of contents

1. Unfolding and folding

2. Flexible quad-meshes

3. Curved folding, Example 1


Funding source:

Partly supported by the Austrian Academy of Sciences

July 13: ISTEC 2016, Vienna University of Technology 1/57


unfold

fold

There are standard procedures provided for theconstruction of the unfolding (development,net) of polyhedra or developable surfaces.

The result is unique, apart from the placementof the different components, and it shows theintrinsic metric of the spatial structure.


E.g., the unfolding of the Oloid

Institut fuer GeometrieInstitut fuer Geometrie

Technische Universitaet WienTechnische Universitaet WienTechnische Universitaet Wien

Oloid

Oloid

Oloid

Oloid

1

2 3

3

Developable surfaces are ruled surfaces withvanishing Gaussian curvature


Institut fuer GeometrieInstitut fuer Geometrie

Technische Universitaet WienTechnische Universitaet WienTechnische Universitaet Wien

Oloid

Oloid

Oloid

Oloid

1

2 3

3

Oloid:

arc-length parametrization of theunfolding of the circles:

x(s) =2√3

9

[

arccos

√2 cos s√1 + cos s

−√

2(1− cos s)(1 + 2 cos s)(1 + cos s)

]

y(s) =

√3

9

[

ln2

1 + cos s+11 + 7 cos s

1 + cos s

]

.



unfold

fold

The inverse problem, i.e., the determination ofa folded structure from a given unfolding ismore complex. In the smooth case we obtaina continuum of bent poses.

In the polyhedral case the computation leads toa system of algebraic equations. Also here thecorresponding spatial object needs not be unique.



Only if the polyhedron bounds a convex solid then the result is unique, due to AleksandrDanilovich Alexandrov (1941).

In this case, for each vertex the sum of intrinsic angles for all adjacent surfaces is< 360◦ (= convex intrinsic metric).

Theorem: [Uniqueness Theorem]For any convex intrinsic metric there is a unique convex polyhedron.



If convexity is not required the unfoldingof a polyhedron needs not define itsspatial shape uniquely !

Definition 1: A polyhedron is calledglobally rigid if its intrinsic metricdefines its spatial form uniquely — up tomovements in space.

e.g., a tetrahedron

f1

f2

f4

f5

f 6

f1

f2

f4

f6

f7A flipping (or snapping) polyhedronadmits two sufficiently close realizations– by applying a slight force.



Definition 2: A polyhedron is called (continuously) flexible if there is a continuousfamily of mutually incongruent polyhedra sharing the intrinsic metric. Each member ofthis family is called a flexion.

f8

f 1

f5

f 6

f8

f 1

f6

f7

f 1

f2

Even a regular octahedron is flexible — after being re-assembled. The regular pose onthe left hand side is called locally rigid.


1. Unfolding and folding4

B

C

B′

C ′

A

A′

B

C

B′

C ′

A

A′Two particular examplesof flexible octahedrawhere two faces areomitted. Both have anaxial symmetry (types 1and 2)

Below: Nets of the twooctahedra.

B C

B′ C ′C ′ A

A′A′ BC

B′ C ′C ′ A

A′A′


1. Unfolding and folding8

N

A

A′

B

B′

C

C ′ t1

t2

P

S

According to R. Bricard (1897)there are three types of flexibleoctahedra (four-sided double-pyramids).

The first two have axis or plane ofsymmetry. Those of type 3 admittwo flat poses. In each such pose,the pairs (A,A′), (B,B′), and(C, C′) of opposite vertices areassociated points of a strophoid S.



NA

A′B

B′

C

C′

A′′

B′′

C′′

According to Bricard’s construc-tion, all bisectors must passthrough the midpoint N of theconcentric circles.

The two flat poses of a type-3flexible octahedron, when ABCremains fixed.



16

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This polyhedron called“Vierhorn” is locallyrigid, but can flipbetween its spatialshape and two flatrealizations in theplanes of symmetry(W. Wunderlich, C.Schwabe).

At the science exposition “Phänomena” 1984 in Zürich this polyhedron was exposedand falsely stated that this polyhedron is flexible.



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the “Vierhorn” and its unfolding

Wolfram MathWorld: A flexible polyhedron which flexes from one totally flat configuration to another, passing through

intermediate configurations of positive volume.



A polygonal mesh is a simply connected subsetof a polyhedral surface (sphere-like or withboundary) consisting of (not necessarily planar)polygons, edges and vertices in the Euclidean 3-space.

The edges are either internal when they areshared by two faces, or they belong to theboundary of the mesh.

When all polygons are quadrangles, then it iscalled a quadrilateral surface.



In modern architecture, most freeform surfaces are designed as polyhedral surfaces –like the Capital Gate in Abu Dhabi, built by the Austrian company Waagner Biro

(160 m high, 18◦ inclination)



V1

V2V3

V4

f0

f1

f2

f3

f4

a1

a2

a3

a4 Under which conditions is this 3 × 3mesh continuously flexible ?



V3

V4

f0

f1

f2

f3

I10

I20

I30

α1

β1

γ1

δ1

α2

β2 γ2

δ2

Transmission from f1 to f3via the quadrangle f2

I10I20

I30

A1B1

A2

B 2

ϕ1

ϕ2ϕ3

α1 β1

γ1

δ1

α2β2

γ2

δ2

Composition of two spherical four-barlinkages



A Kokotsakis mesh is continuously flexible⇐⇒the transmission from f1 to f3 can in twoways be decomposed into two sphericalfour-bar mechanisms, one via V1 and V2,the other via V4 and V3.

The internal edges can be arranged in two‘horizontal’ (blue) and two ‘vertical’ edgefolds (red).

V1

V2V3

V4

f0

f1

f2

f3

f4

I10

I20

I30

I40



I. Planar-symmetric type (Kokotsakis 1932):

The reflection in the plane of symmetry of V1 and V4maps each horizontal fold onto itself while the twovertical folds are exchanged.

V1

V2V3

V4

II. Translational type:

There is a translation V1 7→ V4 and V2 7→ V3 mappingthe three faces on the right hand side onto the tripleon the left hand side.

V1

V2

V3

V4

a4


2. Flexible quad-meshes6

Vi

αi

βiγi

δi

III: Isogonal type (Kokotsakis 1932):

A Kokotsakis mesh is flexible when at eachvertex Vi opposite angles are either equal orcomplementary, i.e.,

αi = βi , γi = δi or

αi = π − βi , γi = π − δi and (n = 4)

sinα1 ± sin γ1sin(α1 − γ1)

· sinα2 ± sin γ2sin(α2 − γ2)

=sinβ3 ± sin γ3sin(β3 − γ3)

· sinβ4 ± sin γ4sin(β4 − γ4)

A quad mesh where all 3 × 3 complexes are of this type is continuously flexible andcalled Voss surface (Kokotsakis, Graf, Sauer)



These are Voss surfaces:

Nadja Posselt:Synthese von zwangläufig beweglichen 9-gliedrigen VierecksflachenDiploma thesis, TU Dresden 2010



IIIa. Generalized isogonal type:

A. Kokotsakis (1932): At all vertices oppositeangles are congruent or complementary.

G. Nawratil (2010): At least at two of the fourpyramids opposite angles are congruent.

IV. Orthogonal type (Graf, Sauer 1931):

Here the horizontal folds are located in parallel (say:horizontal) planes, the vertical folds in vertical planes(T-flat).

V1 V2

V3V4



V. Line-symmetric type (H.S. 2009):

A line-reflection maps the pyramid at V1onto that of V4; another one exchanges thepyramids at V2 and V3.

This includes Kokotsakis’ example of aflexible tessellation.

V1

V2

V3

V4

f0

f1

f2

f3

f4

a1

a2

a3

a4

ρ1

ρ2



fold

ing

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Unfolded miura-ori;dashs are valley folds,

full lines are mountain folds

1) Miura-ori is a Japanese foldingtechnique (1970?) named afterProf. Koryo Miura, The University ofTokyo (military secret in Russia).

It is used for solar panels because itcan be unfolded into its rectangularshape by pulling on one corner only.

On the other hand it is used as kernelto stiffen sandwich structures.


2. Flexible quad-meshesreplacements

180◦

we start with twoparallelograms sharingone edge . . .


2. Flexible quad-meshesreplacements

2δε1

and rotate the right one againstthe left one through the angle2δ.

The lower sides span a plane ε1,the upper sides a plane ε2 parallelε1 .



ε1

ε2

By translations we generate a zig-zag strip of parallelogramsbetween the two parallel planes ε1 and ε2 .



ε1

ε2

By reflection in ε2 we generate a second zig-zag strip ofparallelograms sharing the border line in ε2 with the initial strip

— and we iterate . . .



Miura-ori, a Japanese folding technique




















Miura-ori, a Japanese folding techniqueMiura-ori, a Japanese folding technique








V

x

y

z

ϕϕ

ε2ψ

ψ

α

P1

P2

P3

P4

P∗3

P∗4

There is a hidden localsymmetry at eachvertex V :

The parallelogramsP1,P2 with angle αand the elogationsP∗3,P

∗3 of those with

angle 180◦ − α forma pyramid symmetricwith respect to thefixed planes.



Cf4

f3 f2

f1r

l

A. Kokotsakis, 1932Athens

2) Any arbitrary plane quadrangle isa tile for a regular tessellation of theplane. It is obtained from the initialquadrangle

• by iterated 180◦-rotations about themidpoints of the sides — or

• by iterated translations of centrallysymmetric hexagon.

For a convex f1 this polyhedral surfaceis continuously flexible.



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At each flexion all vertices are located on a right circular cylinder.




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Different flexions of a 9×6tessellation mesh (dashesindicate valley folds).

Under which conditions isthere a flexion where theright border zig-zag fitsexactly to the left border —apart from a vertical shift ?



f11

ρ4

l−1r

l−1rρ4

l−2r2

l−2r2ρ4

l−3r3

l−3r3ρ4

l r

l rρ4

r2

r2ρ4

l−1r3

l−1r3ρ4

l−2r4

l−2r4ρ4

l2r2

l2r2ρ4

l r3

l r3ρ4

r4

r4ρ4

l−1r5

l−1r5ρ4

l3r3

l2r4 l r

5r6

l−1ρ4 l

−2rρ4 l

−3r2ρ4

r

rρ4

l−1r2

l−1r2ρ4

l−2r3

l−2r3ρ4

l r2

l r2ρ4

r3

r3ρ4

l−1r4

l−1r4ρ4

l2r3

l2r3ρ4

l r4

l r4ρ4

r5

r5ρ4

This scheme of a 7×7 tesselation mesh indicateswhich product of helical motions l , r and 180◦-rotations ρ4 maps f11 onto f i j .

Theorem:

f i j=

li−j2 r

i+j2−1(f11)

for i + j ≡ 0 (mod 2)

li−j−12 r

i+j−32 ρ4(f11)

for i + j ≡ 1 (mod 2)

( r = ρ2 ◦ ρ1 and l = ρ4 ◦ ρ1)



f11

ρ4

l−1r

l−1rρ4

l−2r2

l−2r2ρ4

l−3r3

l−3r3ρ4

l r

l rρ4

r2

r2ρ4

l−1r3

l−1r3ρ4

l−2r4

l−2r4ρ4

l2r2

l2r2ρ4

l r3

l r3ρ4

r4

r4ρ4

l−1r5

l−1r5ρ4

l3r3

l2r4

l r5

r6

l−1ρ4 l

−2rρ4 l

−3r2ρ4

r

rρ4

l−1r2

l−1r2ρ4

l−2r3

l−2r3ρ4

l r2

l r2ρ4

r3

r3ρ4

l−1r4

l−1r4ρ4

l2r3

l2r3ρ4

l r4

l r4ρ4

r5

r5ρ4

Coaxial helical motions commute=⇒

Theorem:A flexion of a m × n tesselationmesh closes with a vertical shiftof k faces ⇐⇒there exist a, b ∈ Z with

lar b = d2π, k = −a − b.



Closing flexion of a 7× 9 tesselation mesh with l−6r = d2π, k = 5.

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f 91

f 81

f 71

f 61

f 51

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f 31 f 21

f 11

f 92

f 82 f 72

f 62 f 13

f 24

f 14

f 95

f25

f 96 f 8

6

f57 f37

f 27 f 17

How obtainable ?

• numerically:minimize a distance, or

• start with two coaxialhelical motions r, lsatisfying lar b = d2π.



The analogon of a Schwarz boot with trapezoids; a closing flexion of a 17 × 16tesselation mesh with l−10r 6 = d2π and k = 4.

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f 27 f 17

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f 1616

f31 f11

f32 f12

f39 f19f310

f110

f317 f117

f318

f118

Theorem:Each closing flexion of a m × n tesselation mesh is infinitesimally rigid.



A common way of producing small boxes is to push up appropriate planar cardbordforms Φ0 with prepared creases. Below the case of creases along circular arcs c0.

f

c0

Φ0

planar version with circular creases

Φ

c

corresponding box with planar creases



As proved by W. Wunderlich (1958), the spatial creases c are again planar andknown as meridians of surfaces of revolution with constant Gaussian curvature.

f

c0

Φ0

planar version with circular creases

Φ

c

corresponding box with planar creases



P

c

x

yc′

ρ1c′′

M1

M2ρ 1

ρ2

α

On surfaces of revolution the meridians andparallel circles are the principal curvature lines.Therefore, the signed principal curvatures are

κ1 = − y ′′

cosα, κ2 =

cosα

y.

The Gaussian curvature is definedas K = κ1κ2 . Hence,

K = const. ⇐⇒y ′′ +Ky = 0, x ′ =

√

1− y ′ 2.

provided that cosα 6= 0.



The general solution of y ′′ + Ky = 0with constant K 6= 0 is

for K > 0 :

y = a cos s√K + b sin s

√K ,

for K < 0 :

y = a cosh s√−K + b sinh s

√−K

with constants a, b ∈ R, and

x =∫

√

1− y ′ 2 ds.

we can restrict to six cases, up tosimilarities (Gauß, Minding).

Pseudosphere (tractroid)



a = 0.78 a = 1.00 a = 1.18

1. 2. 3.

xxx

yyy

b = 0.3

tractrix

a = 0.8

4.

5.

6.

M1

M2

xx

x

y

y

y

1

There are six types of meridians to distinguish at the surfaces of revolution withconstant Gaussian curvature K 6= 0.



x

yy0

z

a

P

T

c

β

Φ

ε

Let c0 satisfy y ′′0+Ky0 = 0 and bounda cylindrical patch Φ0 with generatorsorthogonal to the x-axis a0.

Theorem: If at a cylindrically bentpose Φ of Φ0 the boundary c lies ina plane ε, then it satisfies the samedifferential equation as c0.

Proof: y0(s) = y(s) cosβ withβ < π/2 being the (constant) angleof inclination of the cylinder.

The axis of c is the meet of ε and the plane of the orthogonal section a, which is thebent counterpart of the original axis a0 of c0.



Theorem:

Let Φ0 be a planar ‘ruled surface’ with a transversalcurve (crease) c0, which separates Φ0 into two patchesΦ10 and Φ20.

Suppose the generators of the ruling remain straight atthe bent pose Φ1, Φ2 with a curved edge c between.Then c must be a planar curve.

If all generators of Φ1 and Φ2 are extended to infinity,we obtain two torses, which are symmetric with respectto the plane of c .

Φ10

Φ20

c0α

1/κg

E.g., take a cone of revolution with a parabolic section c and reflect the part oppositeto the apex in the plane of c . In Origami this is called reflection operation.



Sketch of the Proof:

Let κ(s) and τ(s) denote the curvature and torsion of c . In terms of the angle γ1(s)between the osculating plane of c and the tangent plane of the torse Φ1, the geodesiccurvature of c w.r.t. Φ1 is

κg = κ cos γ1.

The geodesic curvature κg must be the same w.r.t. Φ2 =⇒ γ2 = −γ1.

The angle α between the tangent of c and the generator of Φ1 satisfies

cosα : sinα = (τ − γ ′1) : −κ sin γ1.The angle α must be the same w.r.t. Φ2 =⇒ (τ − γ ′1) = −(τ + γ ′1), hence τ = 0.



c0 P0

Q0

Unfolding and corresponding spatial form (photos: G. Glaeser)

The spatial form Φ is obtained by gluing together the semicircles with the straightsegments. How to model the resulting convex body ?



c

Φ

Unfolding and corresponding spatial form (photos: G. Glaeser)

The crucial point is here that the ruling is unknown.

M. Kilian, S. Flöry, Z. Chen, N.J. Mitra, A. Sheffer, H. Pottmann: Curved Folding.ACM Trans. Graphics 27/3 (2008), Proc. SIGGRAPH 2008.


4. Curved folding, Example 2y

A0

B0C0

D0

M0Φ0

c10

c ′10

c20

c ′20

r

A physical model shows:

• The spatial body with its developable boundary Φis convex and uniquely defined.

• The helix-like curve c = c1 ∪c2 is a proper edge of Φ; theresulting solid is the convex hullof c .

• The semicircular disks arebent to cones with apices Aand C. Hence, Φ is a C1-compound of two cones and atorse between.

• The body has an axis a ofsymmetry which connects themidpoint M with the remainingtransition point B = D on c .



A0

B0C0

B′0=D0

M0

E10

E ′10

E ′20

E20

Φ0

c10

c ′10

c20

c ′20

g0

r

Consequences:

• Because of the straight segments of c10, thedevelopable surface on the left hand side of c1belongs to the rectifying torse of c1.

• At A and C the surface Φcan be approximated by a rightcone with apex angle 60◦.

• The tangent tA to c1 at Ais a generator, the osculatingplane of c1 a tangent plane ofthis cone; the rectifying planepasses through the cone’s axis.

•When g0 meets both straightsides of c0, then g meets c1and c2 at points with paralleltangents =⇒ coincidingtangent indicatrices.



A0

B0C0

B′0=D0

M0

E10

E ′10

E ′20

E20

Φ0

c10

c ′10

c20

c ′20

gM

rψ

• The tangent at the point E2 ∈ c2 of transitionbetween the cone with apex A and the torse mustbe parallel to tA.

• The tangent at the analoguepoint E1 ∈ c1 is parallel to thefinal tangent tC of c2.

• The subcurves AE1 ⊂ c1and E2C ⊂ c2 have concidingtangent indicatrices.

At a first approximation thecone with apex A is specified asright cone with apex angle 60◦;c1 is a geodesic circle on thiscone.



A′

t ′A

B′

E ′1

c ′1

A′′

t ′′A

B′′E ′′1

c ′′1

A′′′

t ′′′A

B′′′

C ′′′

E ′′′1

E ′′′2

c ′′′1

c ′′′2

a′′′

A0

c0

Approximation 1:

c1 is an algebraic curve.

tA is parallel to the tangent atE2 ∈ c2. Analogously, tC is parallelto the tangent at E1 ∈ c1. Thisdefines the axis a of symmetry.

We notice a contradiction sincethe osculating plane of c1 at B isnot orthogonal to BC.



tE2=tA

tB

tC=tE1

‖ a

tE2=tA

tB

tC=tE1

‖ a

Left: Tangent indicatrices of c1 and c2 for the firstapproximation; no coinciding subcurves!

Approximation 2 is defi-ned by alined side viewsof the tangent indicatrices(right) =⇒

• the subcurve AE1 ⊂ c1 isa curve of constant slope.

• the central torse is acylinder,

• a translation maps AE1onto the subcurve E2C ⊂c2.


A′

B′

C ′

M ′

E ′1

E ′2

F ′1

F ′2

c ′1

c ′2

a′

A′′

B′′

C ′′

M ′′

E ′′1

E ′′2

F ′′1

F ′′2

c ′′1

c ′′2

A′′′

B′′′

C ′′′

M ′′′

E ′′′1

E ′′′2

F ′′′1

F ′′′2

c ′′′1

c ′′′2

a′′′

a′′′1

a′′′2

Approximation 2:

The product of the translation A 7→ E2and the half-rotation about a maps thesubcurve AE1 onto itself, but in reverseorder.

Therefore this portion AE1 has an axisa1 of symmetry passing through themidpoint F1.


Approximation 2 shows an excellentaccordance with the physical model.

. . . but there remains a contradiction.

A

B

C

E1

E2

F1

F2

c1

c20

a

a1

a2



A0

B0

M0 E10

E ′10

F10

F ′10

E ′20

Nψ

A

N

E1

F1

c1

a1

Due to the symmetry w.r.t. a1, the midpoint N of AE1 lies on a1. The distancesA0F10 and A0E10 are preserved, the triangle ANF1 is congruent to its counterpartA0N0F10 in the unfolding. But NF1 is not (exactly) orthogonal to the tangent of c1at F1.


Thank you for your attention!


References

[1] A.D. Alexandrov: Convex Polyhedra. Springer Monographs in Mathematics,Springer 2005, ISBN 9783540263401, (first Russian ed. 1950).

[2] A.I. Bobenko, I. Izmestiev: Alexandrov’s theorem, weighted Delaunay triangulations,and mixed volumes. Annales de l’Institut Fourier 58/(2), 447–505 (2008),arXiv:math.DG/0609447.

[3] R. Bricard: Mémoire sur la théorie de l’octaèdre articulé. J. math. pur. appl.,Liouville 3, 113–148 (1897).

[4] R. Bricard: Leçon de Cinématique, Tome II, de Cinématique appliqueé. Gauthier-Villars et Cie, Paris 1927.


[5] E.D. Demaine, J. O’Rourke: Geometric folding algorithms: linkages, origami,polyhedra. Cambridge University Press 2007.

[6] H. Dirnboeck, H. Stachel: The Development of the Oloid. J. Geom. Graph. 1,105–118 (1997).

[7] M. Kilian, S. Flöry, Z. Chen, N.J. Mitra, A. Sheffer, H. Pottmann: Curved Folding.ACM Trans. Graphics 27/3 (2008), Proc. SIGGRAPH 2008.

[8] A. Kokotsakis: Über bewegliche Polyeder. Math. Ann. 107, 627–647, 1932.

[9] G. Nawratil, H. Stachel: Composition of spherical four-bar-mechanisms. In D. Pislaet al. (eds.): New Trends in Mechanism Science, Springer 2010, pp. 99–106.

[10] G. Nawratil: Flexible octahedra in the projective extension of the Euclidean 3-space.J. Geom. 14/2, 147–169 (2010).


[11] G. Nawratil: Reducible compositions of spherical four-bar linkages with a sphericalcoupler component. Mech. Mach. Theory 46/5, 725–742 (2011).

[12] H. Stachel: What lies between the flexibility and rigidity of structures. SerbianArchitectural J. 3/2, 102–115 (2011).

[13] H. Stachel: On the Rigidity of Polygonal Meshes. South Bohemia Math. Letters19/1, 6–17 (2011).

[14] H. Stachel: A Flexible Planar Tessellation with a Flexion Tiling a Cylinder ofRevolution. J. Geometry Graphics 16/2, 153–170 (2012).

[15] H. Stachel: On the flexibility and symmetry of overconstrained mechanisms. Phil.Trans. R. Soc. A 372, num. 2008, 20120040 (2014).

[16] H. Stachel: Flexible Polyhedral Surfaces With Two Flat Poses. In Special Issue, B.Schulze (ed.): Rigidity and Symmetry, Symmetry 7(2), 774–787 (2015).


[17] H. Stachel: Two examples of curved foldings. Proceedings of the 17th Internat.Conf. on Geometry and Graphics, Beijing 2016 (to appear).

[18] W. Whiteley: Rigidity and scene analysis. In J.E. Goodman, J. O’Rourke(eds.): Handbook of Discrete and Computational Geometry, 2nd ed., Chapman& Hall/CRC, Boca Raton 2004, pp. 1327–1354.

[19] W. Wunderlich: Aufgabe 300. El. Math. 22, p. 89 (1957); author’s solution: El.Math. 23, 113–115 (1958).


applied geometry: foldings and unfoldings - tu wien · applied geometry: foldings and unfoldings...

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