Geom Dedicata (2013) 166:1–14DOI 10.1007/s10711-012-9781-6

ORIGINAL PAPER

Constructing the Cubus simus and the Dodecaedronsimum via paper folding

Urs Hartl · Klaudia Kwickert

Received: 31 January 2012 / Accepted: 27 August 2012 / Published online: 11 September 2012© Springer Science+Business Media B.V. 2012

Abstract The archimedean solids Cubus simus (snub cube) and Dodecaedron simum (snubdodecahedron) cannot be constructed by ruler and compass. We explain that for general rea-sons their vertices can be constructed via paper folding on the faces of a cube, respectivelydodecahedron, and we present explicit folding constructions. The construction of the Cubussimus is particularly elegant. We also review and prove the construction rules of the otherArchimedean solids.

Keywords Archimedean solids · Snub cube · Snub dodecahedron ·Construction · Paper folding

Mathematics Subject Classification (2000) 51M15 · 51M20

1 Construction of platonic and archimedean solids

The archimedean solids were first described by Archimedes in a now-lost work and werelater rediscovered a few times. They are convex polyhedra with regular polygons as faces andmany symmetries, such that their symmetry group acts transitively on their vertices [1,3].They can be built by placing their vertices on the faces of a suitable platonic solid, a tetra-hedron, a cube, an octahedron, a dodecahedron, or an icosahedron. In this article we discussthe constructibility of the archimedean solids by ruler and compass, or paper folding.

All the platonic solids can be constructed by ruler and compass. Such constructions arealready explained by Euclid in Book XIII, Propositions 13–17, e. g. [4]. Of course the con-structions have to be carried out in space. But in addition to the usual constructions with rulerand compass within a plane one only needs the following two extra constructions: to raise aperpendicular to a given plane through a point in this plane, and to construct a plane containingthree (non-collinear) points.

U. Hartl (B) · K. KwickertMathematisches Institut, Universität Münster, Einsteinstr. 62, 48149 Münster, GermanyURL: www.math.uni-muenster.de/u/urs.hartl/

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2 Geom Dedicata (2013) 166:1–14

The construction of the tetrahedron, the cube, and the octahedron in this manner is straightforward. To construct the dodecahedron and the icosahedron, one first needs to construct theregular pentagon or the golden ratio � = 1+√

52 ≈ 1.618. Euclid even describes a construction

of the tetrahedron together with the center of its circumsphere (Book XIII, Proposition 13).Starting from the platonic solids, the archimedean solids can be obtained by constructing

their vertices on the faces of the platonic solids. On the face of the tetrahedron the verticesof the truncated tetrahedron are constructed by trisecting the edges.

On the face of the cube the vertices of the truncated cube, the cuboctahedron, the rhom-bicuboctahedron, and the truncated cuboctahedron are constructed according to Fig. 1.

On the face of the octahedron the vertices of the truncated octahedron, the cuboctahe-dron, the rhombicuboctahedron, and the truncated cuboctahedron are constructed accordingto Fig. 2.

On the face of the icosahedron the vertices of the truncated icosahedron, the icosidodeca-hedron, the rhombicosidodecahedron, and the truncated icosidodecahedron are constructedaccording to Fig. 3.

Finally on the face of the dodecahedron the vertices of the truncated dodecahedron, theicosidodecahedron, the rhombicosidodecahedron, and the truncated icosidodecahedron areconstructed according to Fig. 4. These construction rules are well known and easily verified(also see our Appendix).

Fig. 1 Vertices of archimedean solids on the face of a cube

Fig. 2 Vertices of archimedean solids on the face of an octahedron

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Geom Dedicata (2013) 166:1–14 3

Fig. 3 Vertices of archimedean solids on the face of an icosahedron

Fig. 4 Vertices of archimedean solids on the face of a dodecahedron

All these subdivisions of faces can be carried out with ruler and compass. In contrast thetwo remaining archimedean solids, Cubus simus (also called snub cube) and Dodecaedronsimum (also called snub dodecahedron) cannot be constructed with ruler and compass only.This was proved by Weissbach and Martini [6]. They actually showed that the ratio e : dbetween their edge length e and the diameter d of their circumsphere cannot be constructedby ruler and compass. Since one obtains this diameter and the edge length from the vertices,there can be no construction of the Cubus simus and the Dodecaedron simum with ruler andcompass.

From an algebraic point of view the reason is that the inverse ratio d : e satisfies the

polynomial equation( d

e

)6 − 10( d

e

)4 + 22( d

e

)2 − 14 = 0; see [6, p. 126]. This equa-tion is irreducible over Q by Eisenstein’s criterion (using the prime 2) [2, Chapter 11,Theorem 4.5], and hence d

e generates a field extension Q( de ) of degree 6 over Q. On

the other hand, with ruler and compass one can construct only ratios (or points withcoordinates) lying in field extensions K of Q with K ⊂ R whose degree is a powerof 2. More precisely, K must contain subfields K = Kn ⊃ Kn−1 ⊃ · · · ⊃ K1 ⊃K0 = Q such that Ki has degree 2 over Ki−1 for all i ; see [2, Chapter 13, Theo-rem 4.9].

Now there are various enhancements of construction techniques, allowing all construc-tions for which the degree of Ki over Ki−1 is either 2 or 3. We want to focus on one of theseenhancements namely paper folding, see Martin [5, Chapter 10], in which one allows thefollowing type of construction. Let two points P and Q and two lines g and h be drawn on apiece of paper but forbid the case where P lies on g, Q lies on h, and g and h are parallel orequal. We want to fold the paper creating one crease such that P is placed onto g and at thesame time Q is placed onto h. It turns out that there are at most three possibilities to achievethis and any crease obtained in this way will be a line constructible by paper folding; see[5, p. 149f]. Mathematically the folding can be seen as a reflection at the crease. Noticethat in the forbidden case, where P already lies on g, Q already lies on h, and g and hare parallel or equal, every fold along a crease orthogonal to g and h will achieve the goal.

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4 Geom Dedicata (2013) 166:1–14

Since one does not want that all the lines orthogonal to a given line are constructible, thiscase is forbidden. All ruler and compass constructions can be carried out also by paper fold-ing. Moreover, two of the classical Greek construction problems which are impossible byruler and compass, the Delian problem to double the cube’s volume, and the trisection ofangles, can be solved by paper folding; see Martin [5, Lemmas 10.12 and 10.13 and Exercise10.14].

In the next two sections we will describe paper folding methods to construct the verticesof the Cubus simus on the faces of a cube, and of the Dodecaedron simum on the faces of adodecahedron. The construction of the Cubus simus is quite elegant, whereas our construc-tion of the Dodecaedron simum is not. We are uncertain whether there is a nicer constructionfor the latter.

2 The Cubus simus constructed out of a cube

The Cubus simus (or snub cube) is constructed by placing smaller squares on the faces ofthe cube which are slightly rotated either to the left or to the right. Then the vertices of thesmaller rotated squares are connected such that each vertex belongs to one square and fourtriangles; see Fig. 5.

We normalize the edge length of the cube to 1. Let x, y be the distances of a vertex A1

from the edges of the cube (i.e. the lengths of the perpendiculars from A1 onto these edges)such that x ≥ y. By reasons of symmetry the distances x, y must be the same for all the

Fig. 5 The vertices of the Cubus simus on the faces of a cube (modified from [7, p. 3])

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Geom Dedicata (2013) 166:1–14 5

vertices of the Cubus simus. From Pythagoras theorem we compute

|A1 A2|2 = (x − y)2 + (1 − x − y)2

|A1 B1|2 = (x − y)2 + x2 + y2

|A2 B1|2 = (1 − 2x)2 + y2 + y2.

Since the triangle �A1 A2 B1 should be equilateral we obtain the two equations

|A1 A2|2 = |A1 B1|2 and |A1 A2|2 = |A2 B1|2.The first implies (x − y)2 + x2 + y2 = (x − y)2 + (1 − x − y)2 = (x − y)2 + 1 + x2 +y2 − 2x − 2y + 2xy, whence

1 − 2x − 2y + 2xy = 0. (2.1)

The second implies

x2 − 2xy + y2 + 1 + x2 + y2 − 2x − 2y + 2xy = 1 − 4x + 4x2 + 2y2,

whence 2x2 + 2y2 + 1 − 2x − 2y = 1 − 4x + 4x2 + 2y2, and −2x2 + 2x = 2y, and

y = x − x2. (2.2)

We substitute (2.2) into (2.1) to obtain

2x3 − 4x2 + 4x − 1 = 0. (2.3)

By Cardano’s formula for the solutions of degree three equations we find

x = 2

3+ 1

6·(

3√

−26 + 6 · 2√

33 − 3√

26 + 6 · 2√

33

)

≈ 0.352201129

y ≈ 0.228155494.

From Eq. (2.3) one obtains another proof for the fact that the vertices of the Cubus simuscannot be constructed with ruler and compass on the face of the cube. Namely (2.3) is irre-ducible in Q[x]. Indeed, if it were reducible then it would have a solution x ∈ Q which wewrite as a reduced fraction x = p

q with relatively prime integers p, q . Multiplying with q3

yields

2p3 − 4qp2 + 4q2 p − q3 = 0

and one sees that q must be divisible by 2, say q = 2r . Substituting this and dividing by 2gives

p3 − 4r p2 + 8r2 p − 4r3 = 0.

It follows that also p is divisible by 2. But this contradicts that p and q are relatively prime.Thus (2.3) is irreducible and Q(x) is a field extension of degree 3 over Q. This proves that xcannot be constructed by ruler and compass.

However it can be constructed by paper folding. Begin with a square face of the cube anddivide one of its edges into four equal segments. Let P be one endpoint of this edge and letQ be the first division point. Let g be the opposite edge of the square. Construct h as theperpendicular of the last division point onto g as in Fig. 6. Then fold the square such that Pcomes to lie on g and Q comes to lie on h. In Fig. 6 the crease f is marked as a bold line

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6 Geom Dedicata (2013) 166:1–14

Fig. 6 Constructing the vertices of the Cubus simus by paper folding

and P and Q come to lie on P ′, respectively Q′. We claim that the distance of P ′ from thenearest corner of the cube’s face is x and the distance of Q′ from g is y.

Before we prove the claim, notice that in this construction no ruler and compass are needed.The division of the edge into quarters and the construction of the line h can be achieved bythree times folding in halves. Finally A1 is obtained as the image of Q′ under the foldingwhich places P ′ onto the line h and the line g onto itself.

To prove the claim we fix coordinates on the square such that P = (00

), Q = (1/4

0

), and

the lines are given as g ={(0

1

) + s · (10

) : s ∈ R

}and h =

{(3/40

) + t · (01

) : t ∈ R

}. For

the fold f we set f ={(0

b

) + u · (1c

) : u ∈ R

}with unknown b, c. Folding along f can be

viewed as the reflection at f which in our coordinates is given as the map

R f :(

v1

v2

)−→

(0

b

)+ 1

1+c2

(1 − c2 2c

2c c2 − 1

)·((

v1

v2

)−

(0

b

)).

This reflection sends P and Q to

P ′ = R f (P) = R f

((0

0

))=

(0

b

)+ 1

1+c2

( −2bc

−bc2 + b

)and

Q′ = R f (Q) = R f

((1/4

0

))=

(0

b

)+ 1

1+c2

(−2bc + 1/4 − c2/4

−bc2 + b + c/2

).

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Geom Dedicata (2013) 166:1–14 7

Our requirements that P ′ lies on g and Q′ lies on h mean P ′ = (1−d1

)and Q′ = (3/4

1−e

)for

some d, e. This translates into

b + 11+c2 (−bc2 + b) = 1 and

11+c2 (−2bc + 1/4 − c2/4) = 3/4 ,

whence

2b = 1 + c2 and (2.4)

−2bc = c2 + 1/2. (2.5)

Plugging (2.4) into (2.5) we obtain c3 + c2 + c + 1/2 = 0. Writing this as an equation forc +1 yields (c +1)3 −2(c +1)2 +2(c +1)−1/2 = 0, which is the same as our old equation(2.3) for x . Since the derivative of (2.3) is 6x2 − 8x + 4 = 2x2 + 4(x − 1)2 which is alwayspositive for real numbers x , we conclude that (2.3) has exactly one real solution. Thereforec + 1 = x and c = x − 1 = y

x . Using 1 + c2 = x2 − 2x + 2 = 12x we can now compute the

first coordinate of P ′ = (1−d1

)as follows.

d = 1 − −2bc

1 + c2 = 1 + c2 − (c2 + 1/2)

1 + c2 = 1

2(1 + c2)= x .

Also we compute the second coordinate of Q′ = (3/41−e

)as follows.

e=1−b− −bc2 + b + c/2

1 + c2 = 1+c2−b−bc2 + bc2−b−c/2

1 + c2 = −c

2(1+c2)= x · y

x= y.

Altogether this proves our claim that the distance of P ′ from the nearest corner(1

1

)of the

cube’s face is x and the distance of Q′ from g is y. ��

3 The Dodecaedron simum constructed out of a dodecahedron

The Dodecaedron simum (or snub dodecahedron) is constructed by placing smaller penta-gons on the faces of a dodecahedron which are slightly rotated either to the left or to theright. Then the vertices of the smaller rotated pentagons are connected such that each vertexbelongs to one pentagon and four triangles; see Fig. 7.

We normalize the edge length of the dodecahedron to 1. Then the vectors−→O P,

−−→O Q, and−→

O R all have length 1. Since all three angles at O have the same size �P O Q = �P O R =�QO R = 108◦ we compute for the scalar product

〈−→O P,

−−→O Q〉 = 〈−→

O P,−→O R〉 = 〈−−→O Q,

−→O R〉 = cos 108◦ = − 1

2�, (3.6)

where again � = 1+√5

2 ≈ 1.618 is the golden ratio.Join the vertex A1 of the Dodecaedron simum by line segments with the edges [O P] and

[O R] of the dodecahedron to form a parallelogram with the vertex O (the dotted lines labeledx and y in Fig. 7). Let x, y be the lengths of these segments such that x ≥ y. By reasons ofsymmetry the distances x, y must be the same for all the vertices of the Dodecaedron simum.

Since the “diagonal”−→P S of the dodecahedron’s face has length � and is parallel to

−→O R we

have−→O S = −→

O P + � · −→O R and

−→RS = −→

O S − −→O R = −→

O P + (� − 1) · −→O R.

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8 Geom Dedicata (2013) 166:1–14

Fig. 7 The vertices of the Dodecaedron simum on the faces of a dodecahedron

Then the vertices of the Dodecaedron simum have the following coordinates

−−→O A1 = x · −→

O P + y · −→O R ,

−−→O A2 = −→

O R − x · −→O R + y · −→

RS = y · −→O P + (1 − x − y + y�) · −→

O R ,

−−→O B1 = x · −→

O R + y · −−→O Q.

Using the values for the scalar product from (3.6) we compute the distances

|−−−→A1 A2|2 = ∣∣(y − x) · −→

O P + (1 − x − 2y + y�) · −→O R

∣∣2

= (3 − �)x2 + (3 − �)y2 + (4 − 3�)xy − (3 − �)x − (3 − �)y + 1 ,

|−−−→A1 B1|2 = ∣∣−x · −→

O P + (x − y) · −→O R + y · −−→

O Q∣∣2

= (1 + �)(x2 + y2 − xy) ,

|−−−→B1 A2|2 = ∣∣y · −→

O P + (1 − 2x − y + y�) · −→O R − y · −−→

O Q∣∣2

= 4x2 + 3y2 + 4(1 − �)xy − 4x + 2(� − 1)y + 1.

Since the triangle �A1 A2 B1 should be equilateral we obtain the two equations

|A1 A2|2 = |A2 B1|2 and |A1 B1|2 = |A2 B1|2

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Geom Dedicata (2013) 166:1–14 9

which give us

y2 = −�x2 + �x − �y + xy and (3.7)

(� − 2)y2 = (3 − �)x2 − 4x + 1 + (5 − 3�)xy + 2(� − 1)y (3.8)

We substitute (3.7) into (3.8) to obtain

y = (4 − 2�)x2 + (� − 5)x + 1

(4� − 7)x + 1 − �. (3.9)

Substituting this back into (3.8) yields((25� − 41)x3 + (30 − 17�)x2 − 5x + �

) · x = 0.

In the case x = 0, which implies y = −�, the point A1 does not lie on the face of thedodecahedron. Therefore we must have x �= 0 and

(25� − 41)x3 + (30 − 17�)x2 − 5x + � = 0 (3.10)

By Cardano’s formula for the solutions of degree three equations we find

x = −3√

2802410√

5 + 6281714 + 186√

2522364702 + 1128044106√

5

186

+ 2066/31 + 2986√

5/93

3√

2802410√

5 + 6281714 + 186√

2522364702 + 1128044106√

5+ 163

186+ 53

√5

186

≈ 0.3944605378 and

y ≈ 0.2604339436.

Similarly to the Cubus simus, Eq. (3.10) implies that the vertices of the Dodecaedron si-mum cannot be constructed with ruler and compass on the face of the dodecahedron. Namely(3.10) is irreducible over the field Q(�). This can be seen in many ways. For example byreducing (3.10), which lies in Z[�][x] modulo 2 to (�̄ + 1)x3 + �̄x2 + x + �̄ ∈ F2(�̄)[x].By plugging in all four elements of F2(�̄) = F4, one easily checks that the latter equation hasno solution in F2(�̄), hence is irreducible over F2(�̄). Therefore also (3.10) is irreducibleover Q(�) by [2, Chapter 11, Theorems 3.9 and 4.2] and Q(�)(x) is a field extension ofdegree 3 over Q(�). This proves that x cannot be constructed by ruler and compass but thatit can be constructed by paper folding.

Unfortunately we could not find a folding construction similarly elegant to our construc-tion of the Cubus simus in Sect. 2. The problem is actually that there are infinitely manyfolding constructions and one wants to find the most elegant among them. In all construc-tions we found, the number 31 plays an important role. We cannot explain this phenomenon

but we observe that multiplying (3.10) with its conjugate (in which � = 1+√5

2 is replaced

by 1 − � = 1−√5

2 ) and dividing by 31 yields

x6 − 16331 x5 + 386

31 x4 − 30631 x3 + 89

31 x2 − 531 x − 1

31 = 0.

This is irreducible over Q, because its reduction modulo 2 is irreducible over F2, and henceequals the minimal polynomial of x over Q. Therefore the algebraic number x ∈ Q is integralat all primes different from 31 but is not integral at 31.

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10 Geom Dedicata (2013) 166:1–14

Fig. 8 Constructing the vertices of the Dodecaedron simum by paper folding

The nicest construction we found proceeds as follows. It creates a crease through thefollowing two points X and Y ; see the magnifying glass in Fig. 8. The point X lies on thesegment [O P] at distance x from O . The point Y lies on the segment [O S] at distance yfrom O . This crease is created by placing (any two of) the points G1, G2, G3, G4, G5 ontotheir respective line g1, g2, g3, g4, g5 as in Fig. 8.

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Geom Dedicata (2013) 166:1–14 11

In this figure the crease is the line f and reflection at f maps Gi to the point G ′i on the

line gi . The points G1, . . . , G5 can be found as follows

−−→OG1 =

(3731� + 15

31

)· −→

O P + ( 2231� + 24

31

) · −→O R ,

−−→OG2 = ( 42

31� + 1231

) · −→O P + ( 29

31� + 2631

) · −→O R ,

−−→OG3 = ( 47

31� + 931

) · −→O P + ( 17

31� + 2731

) · −→O R ,

−−→OG4 = ( 46

31� + 2231

) · −→O P + ( 27

31� + 2131

) · −→O R ,

−−→OG5 = ( 49

31� + 1431

) · −→O P + ( 26

31� + 3431

) · −→O R.

The line g1 is the perpendicular to the line P R at distance 1231� − 1

31 from P .The line g2 is the perpendicular to the line P O at distance 33

62� + 1831 from O .

The line g3 is the perpendicular to the line S P at distance 1162� + 37

31 from P .The line g4 is the perpendicular to the line SO at distance 39

31� + 2031 from O .

The line g5 is the perpendicular to the line T O at distance 3831� + 35

62 from O .Note that the points G1, . . . , G5, and hence also the points G ′

1, . . . , G ′5 form regular penta-

gons. Also note that all the lines g1, . . . , g5 have one point in common which is not equal toG ′

5 but very near so that the difference cannot be seen in Fig. 8.Once the crease f has been constructed the point A1 is obtained by one more folding

along the dashed line � (in the magnifying glass in Fig. 8) which places the point X onto O .This folding also places the point Y onto A1.

We found this (and many other) folding constructions using the computer algebra systemMaple. We leave it to the interested reader to verify our construction, or even better, to findmore elegant constructions for the Dodecaedron simum. We are happy to send our Maplecode on request.

Appendix: Verifying the construction rules from figs. 2, 3, and 4

First note that the construction rules for the truncated tetrahedron, the truncated cube,the truncated octahedron, the truncated icosahedron, and the truncated dodecahedron,as well as for the cuboctahedron, and the icosidodecahedron are well known; seefor example [1]. Likewise the rules to construct the vertices of the rhombicuboctahe-dron and the truncated cuboctahedron on the face of a cube are straight forward toprove.

To verify the rule for the vertices of the rhombicuboctahedron on the face of an octahedronand the vertices of the rhombicosidodecahedron on the face of an icosahedron consider atriangle whose edge is divided into three parts with ratios a : b : a. Note that |A′′ A′| =|AA′| as sides of the small rhombus. So we obtain for the ratio |A′′ B′′|

|AA′| = |A′ B′|−2|A′′ A′||AA′| =

|C A′|−2|AA′||AA′| = a+b

a − 2 = b−aa .

If the triangle is the face of an octahedron and A′′ and B ′′ are the vertices of a rhom-bicuboctahedron, they form part of an octagon. In order that this octagon is regular weneed that |A′′ B′′|

|A′′ A′| = √2 (compare the picture for the truncated cube in Fig. 1). This implies

b =(√

2 + 1)

· a as desired.

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12 Geom Dedicata (2013) 166:1–14

If the triangle is the face of an icosahedron and A′′ and B ′′ are the vertices of a rhombi-cosidodecahedron, they form part of a decagon. In order that this decagon is regular we needthat |A′′ B ′′| = � · |A′′ A′| (compare the picture for the truncated dodecahedron in Fig. 4).This implies b = (� + 1) · a as desired.

To construct the truncated cuboctahedron on the octahedron and the truncated icosido-decahedron on the icosahedron consider a triangle whose edge is divided into five partswith ratios a : b : c : b : a. We compute |CC ′|

|C P| = 2a2a+2b+c , as well as |C R|

|C P| = a+b2a+2b+c ,

and |RQ||C P| = b+c

2a+2b+c . This yields |C ′ R| = |C R| − |CC ′| = b−a2a+2b+c · |C P|. Therefore

the hexagon in the middle is regular if and only if |RQ| = 2|C ′ R|, that is if and only ifc = b − 2a.

Furthermore, we compute |U V | = |C ′U | = |C ′ R| · |AC ||C P| = b−a

2a+2b+c · |AC | and |T U | =a

2a+2b+c · |AB|.

If the triangle is the face of an octahedron and U and V are the vertices of a truncatedcuboctahedron, they form part of an octagon. In order that this octagon is regular we needthat |U V | = √

2 · |T U | (compare the picture for the truncated cube in Fig. 1). This impliesb = (

√2 + 1) · a and c = (

√2 − 1) · a as desired.

If the triangle is the face of an icosahedron and U and V are the vertices of a truncatedicosidodecahedron, they form part of a decagon. In order that this decagon is regular we needthat |U V | = � · |T U | (compare the picture for the truncated dodecahedron in Fig. 4). Thisimplies b = (� + 1) · a and c = (� − 1) · a as desired.

To construct the vertices of a rhombicosidodecahedron on the face of a dodecahedrondivide the edges of a dodecahedron in three parts with ratios 1 :1 :1.

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Geom Dedicata (2013) 166:1–14 13

We compute |P B| = � · |AB| and |A′ B ′| = 1/3+��

· |AB| = (1 + 13�

) · 3|B B ′| =(2 + �)|B B ′|. Since |B ′ B ′′| = |B B ′| as sides of a rhombus, |A′′ B ′′| = |A′ B ′| − 2|B ′ B ′′| =� · |B B ′|. Moreover, |Q R| = �|B R| = 2� · |B B ′|. The point B ′′ together with twomore vertices of the rhombicosidodecahedron forms an equilateral triangle of edge length12 |Q R| = �|B B ′| = |A′′ B ′′|. So all edges of the rhombicosidodecahedron have the samelength and this proves that our construction is correct.

Finally look at the face of a dodecahedron and divide its edges into five parts with lengthsa, b, c, b, a to construct the vertices of the truncated icosidodecahedron.

This time |P B| = � · |AB| and |A′ B ′| = a+�·|AB|�·|AB| · |AB| = |AB| + a

�= 2a + 2b + c +

a�−a = a + 2b + c +a�. Also |B ′ B ′′| = b + c as sides of an isosceles triangle. Therefore|A′′ B ′′| = |A′ B ′| − 2|B ′ B ′′| = a − c + a�. On the other hand |QS| = � · (a + b + c)and |B ′′Q| = a

a+b+c · |QS| = a� imply |B ′′ R| = |QS| − 2|B ′′Q| = � · (b + c − a). Thepoints B ′′ and R together with four other vertices of the truncated icosidodecahedron forma hexagon. It is regular if and only if |B ′′ R| = |B ′′Q| (compare the picture for the truncatedicosahedron in Fig. 3), that is, if and only if b + c = 2a. In addition, the decagon containing

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A′′, B ′′ and R is regular if and only if |A′′ B ′′| = |B ′′ R| = |B ′′Q|, which is equivalent toa = c. Thus we find a = c = b.

References

1. Adam, P., Wyss, A.: Platonische und archimedische Körper. Ihre Sternformen und polaren Gebilde. HauptVerlag, Bern (1984)

2. Artin, M.: Algebra. Prentice Hall Inc., Englewood Cliffs, NJ (1991)3. Cromwell, P.: Polyhedra. Cambridge University Press, Cambridge (1997)4. Densmore, D. (ed.): Euclid: Elements, All Thirteen Books Complete in One Volume, the Heath, T.L. (trans).

Green Lion Press, Santa Fe, NM (2002)5. Martin, G.: Geometric Constructions. UTM, Springer, New York (1998)6. Weissbach, B., Martini, H.: On the chiral Archimedean solids. Beiträge Algebra Geom. 43(1), 121–

133 (2002)7. Wyss, A.: Die Sonderlinge. Die Archimedischen Körper Cubus simus und Dodecaedron simum. Das Rätsel

ihrer Herkunft. Haupt Verlag, Bern (1986) (Supplement in [1])

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