# A flexible polyhedron without self-intersections in Euclidean 3-space, all
of whose dihedral angles change during a flex

V.A. Alexandrov  Sobolev Institute of Mathematics, Koptyug ave., 4,
Novosibirsk, 630090, Russia and Department of Physics, Novosibirsk State
University, Pirogov str., 2, Novosibirsk, 630090, Russia
ORCID iD: 0000-0002-6622-8214<EMAIL_ADDRESS>and  E.P. Volokitin  Sobolev
Institute of Mathematics, Koptyug ave., 4, Novosibirsk, 630090, Russia and
Department of Physics, Novosibirsk State University, Pirogov str., 2,
Novosibirsk, 630090, Russia
ORCID iD: 0000-0002-2646-7800<EMAIL_ADDRESS>
###### Abstract.

We construct a sphere-homeomorphic flexible self-intersection free polyhedron
in Euclidean 3-space such that all its dihedral angles change during some flex
of this polyhedron. The constructed polyhedron has 26 vertices, 72 edges and
48 faces. To study its properties, we use symbolic computations in the Wolfram
Mathematica software system.

Key words: Euclidean 3-space, flexible polyhedron, dihedral angle, small
diagonal of a polyhedron, segment-triangle intersection algorithm.

UDK: 514.1

MSC: 52C25, 65D18, 68U05

##  1\. Introduction

In this article, a compact polyhedral surface in Euclidean 3-space
$\mathbb{R}^{3}$ with or without boundary, all of whose faces are triangles,
and which a priori can have self-intersections of any type is called a
polytope.

A polyhedron $P$ is called flexible if its spatial shape can be changed
continuously (i.e., without jumps) only by changing its dihedral angles, i.e.,
if $P$ can be included in a continuous family of polyhedra
$\\{P_{t}\\}_{t\in[\alpha,\beta)}$ such that $P=P_{\alpha}$ and, for any
$t\in(\alpha,\beta)$, $P_{\alpha}$ and $P_{t}$ are combinatorially equivalent,
their corresponding faces are congruent, while $P_{\alpha}$ and $P_{t}$
themselves are not congruent. We call such a family
$\\{P_{t}\\}_{t\in[\alpha,\beta)}$ a flex of $P$, and we call $t$ the
parameter of the flex.

The first examples of flexible polyhedra without boundary in $\mathbb{R}^{3}$
were constructed by R. Bricard in 1897 in [1]. All of them have self-
intersections and are combinatorially equivalent to a regular octahedron
(which is why they are called flexible octahedra). Moreover, in [1] a
classification of all flexible octahedra in $\mathbb{R}^{3}$ is given.
Nowadays, flexible octahedra are commonly called Bricard octahedra. One can
read more about them, e.g., in [1, 2, 3, 4, 5].

The theory of flexible polyhedra began to flourish after R. Connelly
constructed a flexible polyhedron in $\mathbb{R}^{3}$ in 1977 in [6] that is
homeomorphic to a sphere and has no self-intersections. As the theory of
flexible polyhedra developed, they turned out to have many remarkable
properties; for example, for every orientable flexible polyhedron without
boundary in $\mathbb{R}^{3}$, its integral mean curvature [7], volume [8, 9,
10, 11], and every Dehn invariant [12] are preserved diring the flex.

At the same time, there are still many interesting open questions in the
theory of flexible polyhedra. These include the following problem posed by
I.Kh. Sabitov:

######  Problem 1.

Is there a flexible polyhedron in $\mathbb{R}^{3}$, without boundary and
without self-intersections, for which all dihedral angles change during the
flex?

Its formulation can be found, e.g., in [13, p. 182] and [11, Problem 1.3].

The absence of self-intersections is fundamentally important in Problem 1.
Indeed, it is easy to see (see Lemma 1 below) that every dihedral angle of any
Bricard octahedron changes during the flex. But all Bricard octahedra have
self-intersections.

Theorem 1 gives a positive answer to Problem 1 and is the main result of this
article.

######  Theorem 1.

In Euclidean 3-space $\mathbb{R}^{3}$, there exists a polyhedron $\mathscr{P}$
with the following properties:

  1. (i)

$\mathscr{P}$ has only triangular faces, has no self-intersections and is
homeomorphic to the sphere $\mathbb{S}^{2}$,

  2. (ii)

there exists a flex of $\mathscr{P}$ such that none of its dihedral angles
remains constant.

As we were finishing this article, A.A. Gaifullin sent us the graduate work of
O.A. Zaslavsky [14], defended under his supervision in 2019 at the Chair of
Higher Geometry and Topology of the Mathematics and Mechanics Department at
Lomonosov Moscow State University. It turned out that the polyhedron
$\mathscr{P}$ constructed by us from the formulation of Theorem 1 had already
been constructed in [14]; moreover, it was constructed specifically to answer
Problem 1. It also turned out that O.A. Zaslavsky argues for the absence of
self-intersections in $\mathscr{P}$ differently than we do, and does not
substantiate the fact that all dihedral angles change. The results of O.A.
Zaslavsky’s graduate work [14] have not yet been published in a refereed
journal. Therefore, we decided to publish our proof of Theorem 1.

The plan of the article is as follows. In § 2 we clarify the terminology and
recall the construction and properties of the Bricard’s octahedron of the
first type. In § 3 we recall the information we need about the construction
and properties of the Steffen’s flexible polyhedron. In § 4 we modify the
Steffen’s polyhedron so that it can be used to construct the polyhedron
$\mathscr{P}$ of Theorem 1. In § 5 we propose an algorithm for solving the
problem of whether a given polyhedron has self-intersections; by applying its
computer implementation to the modified Steffen’s polyhedron we verify that it
has no self-intersections. In § 6 we explicitly construct a polyhedron
$\mathscr{P}$ of Theorem 1 and, using the algorithm of § 5 and the Mathematica
software system [15], verify that it has no self-intersections. In § 7 we
construct a special 1-parameter flex of $\mathscr{P}$ in which none of
dihedral angles of $\mathscr{P}$ remains constant; in this section, we again
make use of Mathematica. Finally, in § 8 we gather the results of previous
sections to obtain a proof of Theorem 1 and formulate open problems related to
Problem 1.

##  2\. Clarification of terminology and Bricard octahedron of type 1

Since the article will feature polyhedra with self-intersections, it makes
sense to clarify the terminology associated with this now.

Let $M$ be an abstract two-dimensional manifold glued from a finite number of
Euclidean triangles $\Delta_{k}$, $k=1,\dots,n$. It is possible that $M$ has a
non-empty boundary. Let $f:M\to\mathbb{R}^{3}$ be a continuous mapping whose
restriction to each $\Delta_{k}$ is a linear isometric embedding. Then we call
$f(M)$ a polyhedron or polyhedral surface in $\mathbb{R}^{3}$. If
$\delta\subset M$ coincides with one of $\Delta_{k}$’s, or with its side or
vertex, then we call $f(\delta)$ a face, edge, or, respectively, a vertex of
the polyhedron $f(M)$. A diagonal of a polyhedron is a straight line segment
that connects two of its vertices, but is not an edge. A diagonal is called
small if its ends belong to adjacent faces.

We say that $f(M)$ has no self-intersections, if $f:M\to\mathbb{R}^{3}$ is
injective. We call $x\in f(M)$ a point of self-intersection of $f(M)$, if its
complete preimage $f^{-1}(x)\subset M$ consists of more than one point.

An octahedron is a polyhedron $f(M)$ (convex or non-convex, self-intersecting
or not) such that the abstract manifold $M$ is combinatorially equivalent to
the regular convex octahedron shown in Fig. 1. Unless otherwise stated, we use
notations of Fig. 1 for the vertices of an arbitrary octahedron.

Figure 1. Designations of the vertices of a regular tetrahedron.

For our purposes it will be sufficient to recall the construction of only the
Bricard octahedron of type 1. For the constructions of Bricard octahedra of
types 2 and 3, we refer the reader to the articles [1, 2, 3, 4, 5] and the
literature cited therein.

Consider a disk-homeomorphic polyhedron $\mathscr{D}$ in $\mathbb{R}^{3}$
consisting of four triangles $A_{1}B_{1}C_{2}$, $B_{1}A_{2}C_{2}$,
$A_{2}B_{2}C_{2}$, and $B_{2}A_{1}C_{2}$. Its boundary is the closed spatial
broken line $A_{1}B_{1}A_{2}B_{2}$, from which we require that the lengths of
its opposite sides are equal to each other, i.e., we require that
$|A_{1}B_{1}|=|A_{2}B_{2}|$ and $|B_{1}A_{2}|=|B_{2}A_{1}|$ (see the left part
of Fig. 2).

Figure 2. Polyhedron $\mathscr{D}$ and Bricard octahedron of type 1
$\mathscr{B}$.

Let $N_{A}$ denote the midpoint of the segment $A_{1}A_{2}$ and $N_{B}$ denote
the midpoint of the segment $B_{1}B_{2}$ (see the central part of Fig. 2). If
$N_{A}\neq N_{B}$, then we denote by $\lambda$ the line passing through
$N_{A}$ and $N_{B}$ (see the central part of Fig. 2). If $N_{A}=N_{B}$, then
we denote by $\lambda$ the line passing through the point $N_{A}=N_{B}$
perpendicular to the plane containing the segments $A_{1}A_{2}$ and
$B_{1}B_{2}$.

First of all, we note that the quadrilateral $A_{1}B_{1}A_{2}B_{2}$ is mapped
onto itself under the rotation of the entire space $\mathbb{R}^{3}$ around the
line $\lambda$ by $180^{\circ}$. This is obvious if $N_{A}=N_{B}$. If
$N_{A}\neq N_{B}$, then the equality of triangles $A_{1}B_{1}B_{2}$ and
$A_{2}B_{1}B_{2}$ implies $|A_{1}N_{B}|=|A_{2}N_{B}|$ (see the central part of
Fig. 2). Therefore, the triangle $A_{1}A_{2}N_{B}$ is isosceles. This means
that its median $N_{A}N_{B}$ is also its height, i.e. the line
$\lambda=N_{A}N_{B}$ is perpendicular to the line $A_{1}A_{2}$. Therefore,
when the entire space is rotated around $\lambda$ by $180^{\circ}$, the points
$A_{1}$ and $A_{2}$ exchange places. Similarly, starting from the triangles
$A_{1}A_{2}B_{1}$ and $A_{1}A_{2}B_{2}$ we conclude that the line
$\lambda=N_{A}N_{B}$ is perpendicular to the line $B_{1}B_{2}$, and therefore
the points $B_{1}$ and $B_{2}$ also exchange places when rotating around
$\lambda$ by $180^{\circ}$. Thus, we have proved that under such a rotation
the quadrilateral $A_{1}B_{1}A_{2}B_{2}$ is mapped into itself.

Now we glue the polyhedron $\mathscr{D}$ and its image under the rotation of
$\mathbb{R}^{3}$ around $\lambda$ by $180^{\circ}$ along the sides of the
quadrilateral $A_{1}B_{1}A_{2}B_{2}$ (see the right part of Fig. 2). The
resulting polyhedron is a Bricard octahedron of type 1. We denote it by
$\mathscr{B}$. The image of the point $C_{2}$ under the rotation of
$\mathbb{R}^{3}$ around $\lambda$ by $180^{\circ}$ is denoted by $C_{1}$.

It follows directly from the above construction that $\mathscr{B}$ is
combinatorially equivalent to an octahedron, has self-intersections and allows
a one-parameter flex (recall that, by definition, the Euclidean distance
between at least some two vertices of the polyhedron is not preserved during
the flex).

To prove Theorem 1 we need one well-known property of the Bricard octahedra
(of all types, not just type 1). We formulate and prove it in the form of
Lemma 1.

######  Lemma 1.

Let a Bricard octahedron be located in $\mathbb{R}^{3}$ in such a way that
none of its dihedral angles is equal to 0 or $\pi$. Then the value of each of
its dihedral angles does not remain constant during the flex.

###### Proof.

Assume the converse, i.e., assume that one of the dihedral angles remains
constant. Without loss of generality, we can assume that this is the angle at
the edge $A_{1}B_{1}$ (recall that we use the same notation for the vertices
as in Fig. 1). Then the length of the diagonal $C_{1}C_{2}$ is constant during
the flex. Hence the lengths of all edges of the tetrahedra
$A_{1}B_{1}C_{1}C_{2}$ and $A_{1}B_{2}C_{1}C_{2}$ remain constant. Thus all
dihedral angles of these tetrahedra remain constant. Therefore, the dihedral
angle at the edge $A_{1}B_{2}$ of the Bricard octahedron is constant (indeed,
it is equal to the dihedral angle of $A_{1}B_{2}C_{1}C_{2}$ at $A_{1}B_{2}$).

Observe that the triangle $A_{1}C_{1}C_{2}$ is non-degenerate in the sense
that its vertices do not lie on the same line (otherwise the dihedral angle at
the edge $A_{1}B_{1}$ of the Bricard octahedron would be equal to 0 or $\pi$,
which contradicts the conditions of Lemma 1). Consequently, the tetrahedra
$A_{1}B_{1}C_{1}C_{2}$ and $A_{1}B_{2}C_{1}C_{2}$ are adjacent to each other
along the nondegenerate face $A_{1}C_{1}C_{2}$. This means that the dihedral
angles at the edges $A_{1}C_{1}$ and $A_{1}C_{2}$ of the Bricard octahedron
are either the sum or the difference of the dihedral angles at the same edges
in the tetrahedra $A_{1}B_{1}C_{1}C_{2}$ and $A_{1}B_{2}C_{1}C_{2}$, and
therefore are also constant during the flex.

Thus, we have proven that if the dihedral angle at the edge $A_{1}B_{1}$ of
the Bricard octahedron is constant during the flex, then the dihedral angles
at its edges $A_{1}B_{2}$, $A_{1}C_{1}$ and $A_{1}C_{2}$ are also constant. In
other words, we have proven that if, under the conditions of Lemma 1, a vertex
is incident to an edge, the dihedral angle at which remains constant, then the
dihedral angles of all edges incident to this vertex remain constant. It
immediately follows that the dihedral angles remain constant for all edges of
the Bricard octahedron, and therefore the lengths of all its diagonals remain
constant. The latter, however, contradicts our definition of the flex. This
contradiction proves Lemma 1. ∎

##  3\. Steffen polyhedron $\mathscr{S}$

As is known, the flexible Steffen polyhedron, denote it by $\mathscr{S}$, is
obtained by gluing together a certain tetrahedron, denote it by $\mathscr{T}$,
and two copies of the same Bricard octahedron of type 1, denote it by
$\mathscr{B}$. Gluing is carried out along congruent faces. In § 3 we resemble
this well-known construction. It can also be found, for example, in [3] and
the literature mentioned there. Additional insight into the flex of
$\mathscr{S}$ can be obtained via the computer animation [16].

Let the tetrahedron $\mathscr{T}=T_{1}T_{2}T_{3}T_{4}$ (see the left side of
Fig. 3) have the following edge lengths: $|T_{1}T_{4}|=17$,
$|T_{1}T_{2}|=|T_{1}T_{3}|=|T_{2}T_{4}|=|T_{3}T_{4}|=12$, and
$|T_{2}T_{3}|=11$. $\mathscr{T}$ will not change its spatial shape during the
flex of $\mathscr{S}$. Therefore, throughout §§ 3–7 we assume that the points
$T_{j}$ ($j=1,\dots,4$) occupy a fixed position in space.

Figure 3. Gluing the Steffen polyhedron $\mathscr{S}$.

Let $\mathscr{B}=A_{1}A_{2}B_{1}B_{2}C_{1}C_{2}$ be a Bricard octahedron of
type 1 (see the upper right part of Fig. 3), whose edges have the following
lengths: $|A_{1}C_{1}|=|B_{1}C_{2}|=|A_{2}C_{2}|=|B_{2}C_{1}|=12$,
$|A_{2}C_{1}|=|B_{1}C_{1}|=|A_{1}C_{2}|=|B_{2}C_{2}|=10$,
$|A_{1}B_{1}|=|A_{2}B_{2}|=5$, and $|A_{1}B_{2}|=|A_{2}B_{1}|=11$.

Let us move $\mathscr{B}$ in $\mathbb{R}^{3}$ by means of an orientation-
preserving motion so that the following pairs of points coincide: $T_{1}$ and
$C_{1}$, $T_{2}$ and $B_{2}$, and $T_{3}$ and $A_{1}$. Note that here we do
not care whether $\mathscr{T}$ intersects $\mathscr{B}$, after the above
mentioned motion, somewhere outside the faces $T_{1}T_{2}T_{3}$ and
$A_{1}B_{2}C_{1}$. In this case, we say that we have glued $\mathscr{T}$ and
$\mathscr{B}$ along $T_{1}T_{2}T_{3}$ and $A_{1}B_{2}C_{1}$. Since we want the
result of gluing two polyhedra to be a polyhedron again, here and below we
mean that both faces along which the gluing was made are deleted. We denote
the result of gluing $\mathscr{T}$ and $\mathscr{B}$ by
$\mathscr{T}\sqcup\mathscr{B}$.

Now consider another copy
$\overline{\mathscr{B}}=\overline{A}_{1}\overline{A}_{2}\overline{B}_{1}\overline{B}_{2}\overline{C}_{1}\overline{C}_{2}$
of the Bricard octahedron of type 1 which is obtained from $\mathscr{B}$ by an
orientation-preserving motion of $\mathbb{R}^{3}$. Thus, each edge of
$\overline{\mathscr{B}}$ has the same length as the corresponding edge of
$\mathscr{B}$. For example,
$|\overline{A}_{1}\overline{C}_{1}|=|A_{1}C_{1}|=12$.

Glue $\mathscr{T}\sqcup\mathscr{B}$ and $\overline{\mathscr{B}}$ along the
faces $T_{2}T_{3}T_{4}$ and
$\overline{A}_{1}\overline{B}_{2}\overline{C}_{1}$, i.e., move
$\overline{\mathscr{B}}$ in $\mathbb{R}^{3}$ by means of an orientation-
preserving motion so that the following pairs of vertices coincide: $T_{2}$
and $\overline{B}_{2}$, $T_{3}$ and $\overline{A}_{1}$, as well as $T_{4}$ and
$\overline{C}_{1}$ (see the lower right part of Fig. 3). The polyhedron
obtained as a result of such gluing is denoted by
$(\mathscr{T}\sqcup\mathscr{B})\sqcup\overline{\mathscr{B}}$. Obviously, it is
homeomorphic to the disk.

Since $\mathscr{B}$ and $\overline{\mathscr{B}}$ are Bricard octahedra, the
positions of the vertices $C_{2}$ and $\overline{C}_{2}$ of
$(\mathscr{T}\sqcup\mathscr{B})\sqcup\overline{\mathscr{B}}$ are not
determined uniquely by the above gluing $\mathscr{T}$ and $\mathscr{B}$ and
the subsequent gluing $\mathscr{T}\sqcup\mathscr{B}$ and
$\overline{\mathscr{B}}$. In fact, after such gluing, $C_{2}$ and
$\overline{C}_{2}$ can lie at any points of the circle $\gamma$ defined by the
following conditions: $\gamma$ is located in a plane perpendicular to the
segment $T_{2}T_{3}$; the center of $\gamma$ is the midpoint of $T_{2}T_{3}$;
the radius of $\gamma$ is equal to
$\sqrt{|A_{1}C_{2}|^{2}-|A_{1}B_{2}|^{2}/4}=\sqrt{10^{2}-11^{2}/4}=(3/2)\sqrt{31}\approx
8.35$. Thus, by choosing some position of $C_{2}$ on $\gamma$, we can (without
changing the position of $\mathscr{T}$ in space) bend the Bricard octahedron
$\overline{\mathscr{B}}$ so that $\overline{C}_{2}$ coincides with $C_{2}$. In
this position we glue the polyhedron
$(\mathscr{T}\sqcup\mathscr{B})\sqcup\overline{\mathscr{B}}$ with itself along
the faces $A_{1}B_{2}C_{2}$ and
$\overline{A}_{1}\overline{B}_{2}\overline{C}_{2}$. Denote the resulting
polyhedron by
$((\mathscr{T}\sqcup\mathscr{B})\sqcup\overline{\mathscr{B}})_{\overline{C}_{2}=C_{2}}$.
Observe that, in order
$((\mathscr{T}\sqcup\mathscr{B})\sqcup\overline{\mathscr{B}})_{\overline{C}_{2}=C_{2}}$
is a polyhedron, in the course of gluing we remove not only the internal
points of the glued faces $A_{1}B_{2}C_{2}$ and
$\overline{A}_{1}\overline{B}_{2}\overline{C}_{2}$ (this was already explained
above), but also the internal points of the segment
$A_{1}B_{2}=\overline{A}_{1}\overline{B}_{2}$ (which, therefore, is not an
edge of
$((\mathscr{T}\sqcup\mathscr{B})\sqcup\overline{\mathscr{B}})_{\overline{C}_{2}=C_{2}})$).

It follows directly from the construction that
$((\mathscr{T}\sqcup\mathscr{B})\sqcup\overline{\mathscr{B}})_{\overline{C}_{2}=C_{2}}$
form a continuous family of combinatorially equivalent polyhedra with
congruent corresponding faces, and that the polyhedra corresponding to
different positions of the point $\overline{C}_{2}=C_{2}$ on $\gamma$ are not
congruent to each other. Thus, this family is a flex of any of the polyhedra
included in it, or, equivalently, any polyhedron in this family is flexible.
Let us select one polyhedron in this family as follows.

Let us denote by $X$ the midpoint of the segment $T_{1}T_{4}$, by $Y$ the
midpoint of the segment $T_{2}T_{3}$, and by $Z$ the intersection point of the
ray $\stackrel{{\scriptstyle\longrightarrow}}{{XY}}$ with the circle $\gamma$.
If the point $\overline{C}_{2}=C_{2}$ coincides with $Z$, then we call
$((\mathscr{T}\sqcup\mathscr{B})\sqcup\overline{\mathscr{B}})_{\overline{C}_{2}=C_{2}}$
the Steffen polyhedron and denote it by $\mathscr{S}$. We consider it a
generally known fact that $\mathscr{S}$ has no self-intersections. Note,
however, that the technology we develope in § 5 for checking the absence of
self-intersections in a polyhedron allows us to strictly prove that
$\mathscr{S}$, as well as any polyhedron
$((\mathscr{T}\sqcup\mathscr{B})\sqcup\overline{\mathscr{B}})_{\overline{C}_{2}=C_{2}}$,
sufficiently close to $\mathscr{S}$, indeed has no self-intersections.
Although this will not be needed in this article, we can even specify that
$((\mathscr{T}\sqcup\mathscr{B})\sqcup\overline{\mathscr{B}})_{\overline{C}_{2}=C_{2}}$
has no self-intersections provided the angle between the segments $XZ$ and
$XC_{2}$ is less than $7.5^{\circ}$.

Using the following agreements, we assign a permanent designation to each
vertex $V$ of $\mathscr{S}$ and always use it in §§ 5–7:

$\bullet$ if, before gluing, $V$ belonged to only one of the polyhedra
$\mathscr{T}$, $\mathscr{B}$ or $\overline{\mathscr{B}}$, then we reserve for
it the designation that it had on that polyhedron;

$\bullet$ if $V$ is the result of gluing a vertex $W$ of $\mathscr{T}$ with a
vertex of $\mathscr{B}$ and/or $\overline{\mathscr{B}}$, then we assign the
vertex $V\in\mathscr{S}$ the designation that $W$ had on $\mathscr{T}$;

$\bullet$ if $V$ is the result of gluing a vertex $W$ of $\mathscr{B}$ with a
vertex of $\overline{\mathscr{B}}$, then we assign the vertex
$V\in\mathscr{S}$ the designation that $W$ had on $\mathscr{B}$.

For example, in the process of construction of $\mathscr{S}$, we first glued
the vertices $T_{3}\in\mathscr{T}$ and $A_{1}\in\mathscr{B}$, and then glued
the resulting point to the vertex $\overline{A}_{1}\in\overline{\mathscr{B}}$.
In accordance with what has been said, we denote the resulting vertex of
$\mathscr{S}$ by $T_{3}$. Another example: at the last stage of construction
of $\mathscr{S}$, we glued together the vertices $C_{2}\in\mathscr{B}$ and
$\overline{C}_{2}\in\overline{\mathscr{B}}$. Hence, we denote the resulting
vertex of $\mathscr{S}$ by $C_{2}$.

##  4\. Construction of a modified Steffen polyhedron $\mathscr{M}$

By construction, the edge $T_{1}T_{4}$ of the Steffen polyhedron $\mathscr{S}$
has length 17. Direct calculations show that the internal dihedral angle at
this edge is equal to $\arccos(45/287)\approx 80^{\circ}59^{\prime}$. Let us
change the length of $T_{1}T_{4}$ so that it becomes equal to
$\sqrt{334}\approx 18.28$. We do not change neither the lengths of the
remaining edges of $\mathscr{S}$, nor the designations of its vertices. The
resulting polyhedron is called a modified Steffen polyhedron and is denoted by
$\mathscr{M}$. Direct calculation shows that the dihedral angle of
$\mathscr{M}$ at the edge $T_{1}T_{4}$ is equal to $90^{\circ}$. It is clear
directly from the construction of $\mathscr{M}$ that it is combinatorially
equivalent to a sphere, has only triangular faces, and is flexible. The
development of $\mathscr{M}$ is shown in Fig. 4. The reader can scan it, print
it on a larger scale on thick paper and glue together a model of
$\mathscr{M}$. This will simplify the understanding of our further
constructions. We especially emphasize that by construction the vertex
$C_{2}\in\mathscr{M}$ coincides in $\mathbb{R}^{3}$ with the point $Z$
constructed in § 3.

Figure 4. The development of a modified Steffen polyhedron $\mathscr{M}$. It
must be bent so that the internal dihedral angle of $\mathscr{M}$ is less than
$180^{\circ}$ at the edges drawn with solid lines and is greater than
$180^{\circ}$ at the edges drawn with dotted lines. The sides of the
development with identical vertices need to be glued in pairs; more precisely,
one must glue in pairs the sides connected to each other by circular arcs, two
sides $T_{2}C_{2}$, and two sides $T_{3}C_{2}$.

We associate with $\mathscr{M}$ the following Cartesian coordinate system in
$\mathbb{R}^{3}$. The origin is at the point $X$, which was defined in § 3 as
the midpoint of the segment $T_{1}T_{4}$. We choose the $x$-axis so that it
passes through the point $T_{3}$ and $T_{3}$ has a positive $x$-coordinate. We
choose the $y$-axis so that it passes through the point $T_{2}$ and $T_{2}$
has a positive $y$-coordinate. And we choose the $z$-axis so that it passes
through the point $T_{1}$ and $T_{1}$ has a positive $z$-coordinate. We use
this and only this coordinate system in §§ 4–7.

Vertex $A_{2}$ is connected by edges of $\mathscr{M}$ with its three vertices
$T_{1}$, $T_{2}$ and $C_{2}$, whose coordinates are known to us directly from
the constructions of $\mathscr{M}$ and of the coordinate system. They are
shown in Table 1. Using Mathematica we solve symbolically the system

$\begin{cases}|A_{2}T_{1}|^{2}&=10^{2},\\\ |A_{2}T_{2}|^{2}&=5^{2},\\\
|A_{2}C_{2}|^{2}&=12^{2}\end{cases}$ (1)

of three algebraic equations of the second degree with respect to the
coordinates of $A_{2}$, and we obtain two exact (i.e., expressed in radicals)
solutions. It is obvious that the two points in $\mathbb{R}^{3}$ corresponding
to these solutions are symmetrical to each other with respect to the plane
passing through the vertices $T_{1}$, $T_{2}$ and $C_{2}$. Considering the
model of $\mathscr{M}$, we conclude that vertex $A_{2}$ corresponds to the
solution whose $z$-coordinate is the largest (since only in this case the
inner dihedral angle of $\mathscr{M}$ at edge $A_{2}T_{2}$ is less than
$180^{\circ}$; the latter agrees with the development shown in Fig. 4). We put
the exact values of the coordinates of vertex $A_{2}$ found in this way on
Table 1. The approximate values of the coordinates of vertices shown in Table
1 are not used in our reasoning and are presented solely to make it easier for
the reader to imagine the spatial form of $\mathscr{M}$. For example, the fact
that the approximate value of the $x$-coordinate of point $A_{2}$ is negative
suggests that $A_{2}$ (and therefore $\mathscr{M}$) is not contained in the
quarter of the space consisting of the points with positive $x$\- and
$y$-coordinates.

|  $x$-coordinate |  $y$-coordinate |  $z$-coordinate
---|---|---|---
$T_{1}$ |  $-$$0$ |  $-$$0$ |  $-$$\frac{\sqrt{167}}{\sqrt{2}}\approx 9.13$
$T_{2}$ |  $-$$0$ |  $-$$\frac{11}{\sqrt{2}}\approx 7.77$ |  $-$$0$
$T_{3}$ |  $-$$\frac{11}{\sqrt{2}}\approx 7.77$ |  $-$$0$ |  $-$$0$
$T_{4}$ |  $-$$0$ |  $-$$0$ | $-\frac{\sqrt{167}}{\sqrt{2}}\approx-9.13$
$C_{2}$ |  $-$$\frac{11+3\sqrt{31}}{2\sqrt{2}}\approx 9.79$ |  $-$$\frac{11+3\sqrt{31}}{2\sqrt{2}}\approx 9.79$ |  $-$$0$
$A_{2}$ |  $-$$\frac{\omega_{1}\sqrt{2}-2(200-33\sqrt{31})\sqrt{\rho}}{1230304047998}\approx-1.19$ |  $-$$\frac{\omega_{2}\sqrt{2}+2\sqrt{\rho}}{15573468962}\approx 8.89$ |  $-$$\frac{167\omega_{3}\sqrt{2}+22\sqrt{\rho}}{15573468962\sqrt{167}}\approx 4.72$
$B_{1}$ |  $-$$\frac{\omega_{2}\sqrt{2}-2\sqrt{\rho}}{15573468962}\approx 2.79$ |  $-$$\frac{\omega_{1}\sqrt{2}+2(200-33\sqrt{31})\sqrt{\rho}}{1230304047998}\approx 0.05$ |  $-$$\frac{167\omega_{3}\sqrt{2}-22\sqrt{\rho}}{15573468962\sqrt{167}}\approx-0.46$
$\overline{A}_{2}$ |  $-$$\frac{\omega_{1}\sqrt{2}+2(200-33\sqrt{31})\sqrt{\rho}}{1230304047998}\approx 0.05$ |  $-$$\frac{\omega_{2}\sqrt{2}-2\sqrt{\rho}}{15573468962}\approx 2.79$ | $-\frac{167\omega_{3}\sqrt{2}-22\sqrt{\rho}}{15573468962\sqrt{167}}\approx 0.46$
$\overline{B}_{1}$ |  $-$$\frac{\omega_{2}\sqrt{2}+2\sqrt{\rho}}{15573468962}\approx 8.89$ |  $-$$\frac{\omega_{1}\sqrt{2}-2(200-33\sqrt{31})\sqrt{\rho}}{1230304047998}\approx-1.19$ | $-\frac{167\omega_{3}\sqrt{2}+22\sqrt{\rho}}{15573468962\sqrt{167}}\approx-4.72$
Table 1. Exact and approximate values of the coordinates of the vertices of
$\mathscr{M}$. All decimal places in approximate values are correct, i.e., are
written without taking into account rounding rules. To shorten expressions for
exact values, the following notation is used:
$\rho=167(1712315512948039256+297671463726717927\sqrt{31})$,
$\omega_{1}=237(670333576-497644539\sqrt{31})$,
$\omega_{2}=3(26431711823-892912093\sqrt{31})$,
$\omega_{3}=2798420941-176443707\sqrt{31}$.

Similarly, we find the exact coordinates of the vertex $B_{1}$. To do this, we
use the fact that $B_{1}$ is connected by edges of $\mathscr{M}$ with its
three vertices $T_{1}$, $T_{3}$ and $C_{2}$. But this time, from two solutions
to the corresponding system of three algebraic equations of the second degree,
similar to system (1), we choose the one with the smallest $z$-coordinate. We
put the exact values of the coordinates of $B_{1}$ found in this way on Table
1.

We check the correctness of the above described method for recognizing the
coordinates of $A_{2}$ among solutions of system (1) and the coordinates of
$B_{1}$ among solutions of a similar system by calculating the length of
$A_{2}B_{1}$ through the coordinates of its ends given in Table 1. Symbolic
calculations in Mathematica show that this length is indeed equal to 11.

The coordinates of the vertices $\overline{A}_{2}$ and $\overline{B}_{1}$ can
be found in the same way as we found the coordinates of the vertices $A_{2}$
and $B_{1}$. However, these calculations can be avoided using the following
lemma.

######  Lemma 2.

The modified Steffen polyhedron $\mathscr{M}$ transforms into itself under the
action of rotation of the entire space $\mathbb{R}^{3}$ by $180^{\circ}$
around the line $XY$ passing through the points $X$ and $Y$, constructed in §
3, i.e. under the rotation of $\mathbb{R}^{3}$ defined by the matrix

$L=\begin{pmatrix}0&1&0\\\ 1&0&0\\\ 0&0&-1\end{pmatrix}.$

###### Proof.

Using Table 1, we directly verify that $L(T_{1})=T_{4}$, $L(T_{2})=T_{3}$,
$L(T_{3})=T_{2}$, $L(T_{4})=T_{1}$, and $L(C_{2})=C_{2}$. Therefore, the
distances from the point $L(A_{2})$ to the vertices $T_{3}$, $T_{4}$, and
$C_{2}$ are the same as the distances from the point $\overline{B}_{1}$ to the
specified vertices. Indeed,
$|L(A_{2})T_{3}|=|L(A_{2})L(T_{2})|=|A_{2}T_{2}|=|A_{2}B_{2}|=5$ and
$|\overline{B}_{1}T_{3}|=|\overline{B}_{1}\overline{A}_{1}|=5$;
$|L(A_{2})T_{4}|=|L(A_{2})L(T_{1})|=|A_{2}T_{1}|=|A_{2}C_{1}|=10$ and
$|\overline{B}_{1}T_{4}|=|\overline{B}_{1}\overline{C}_{1}|=10$;
$|L(A_{2})C_{2}|=|L(A_{2})L(C_{2})|=|A_{2}C_{2}|=12$ and
$|\overline{B}_{1}C_{2}|=12$. In addition, both points $L(A_{2})$ and
$\overline{B}_{1}$ lie on the same side of the plane passing through the
vertices $T_{3}$, $T_{4}$, and $C_{2}$. Indeed, according to Fig. 4, the inner
dihedral angle of $\mathscr{M}$ at both edges
$\overline{B}_{1}T_{3}=\overline{B}_{1}\overline{A}_{1}$ and
$A_{2}T_{2}=A_{2}B_{2}$ (and hence the edge
$L(A_{2})L(T_{2})=L(A_{2})L(B_{2})$) is less than $180^{\circ}$. Hence,
$L(A_{2})=\overline{B}_{1}$.

Reasoning in a similar way, we can prove that the distances from $L(B_{1})$ to
the vertices $T_{2}=L(T_{3})$, $T_{4}=L(T_{1})$, $C_{2}=L(C_{2})$ are the same
as distances from $\overline{A}_{2}$ to the specified vertices. In addition,
by analogy with what was said above, it can be proven that both points
$L(B_{1})$ and $\overline{A}_{2}$ lie on the same side of the plane passing
through $T_{2}$, $T_{4}$, and $C_{2}$. Hence, $L(B_{1})=\overline{A}_{2}$.

Thus, we are convinced that the set of vertices of $\mathscr{M}$ (as well as
the sets of its edges and faces) is mapped onto itself by $L$. This means that
$\mathscr{M}$ is mapped onto itself by $L$. ∎

The proof of Lemma 2 yields $\overline{B}_{1}=L(A_{2})$ and
$\overline{A}_{2}=L(B_{1})$. Using these relations, we can find coordinates of
$\overline{A}_{2}$ and $\overline{B}_{1}$ without additional calculations and
thereby complete filling out Table 1.

So, $\mathscr{M}$ has 9 vertices, 21 edges and 14 faces, and has the symmetry
described in Lemma 2. The study of flexible and nonrigid frameworks (and
therefore polyhedra) with symmetry is in itself an interesting and nontrivial
problem, see, e.g., [17] and literature cited there. But we will achieve our
goals in the most direct and elementary way, without using general results on
the effect of symmetry on the rigidity and flexibility of frameworks.

##  5\. Do the polyhedra $\mathscr{M}$ and $\mathscr{S}$ have self-
intersections?

To solve Problem 1 we must be able to answer to decide whether a given
polyhedron has self-intersections. To simplify and unify our reasoning, in § 5
we describe an algorithm that makes it possible to guarantee the absence of
self-intersections for the polyhedra $\mathscr{S}$, $\mathscr{M}$ and
$\mathscr{P}$ (the latter will be built in § 6) and, of course, not only for
them.

The problem of finding intersections and self-intersections of polyhedral
surfaces arises in a variety of problems in mathematics and applied
mathematics, computer graphics and video games. There are many variations of
this problem and dozens of algorithms designed to solve them, see, for
example, [18], [19], [20], [21], [22] and the literature mentioned there.
However, existing algorithms are not suitable for us for the following
reasons:

$\bullet$ we want to solve the problem of finding self-intersections without
using of floating point arithmetic, i.e., we allow ourselves to use only
symbolic calculations (which we de facto execute in Mathematica);

$\bullet$ we want the algorithm to be logically as transparent as possible,
and therefore we do not analyze “exceptional cases”; instead, the algorithm
must notify us of the occurrence of each such case (i.e., we are not
interested in either complete automation of calculations or optimization of
their complexity and speed) .

Therefore, we are forced to develop our own algorithm, the description of
which we proceed to.

Recall that, according to §2, a polyhedron is the image $f(M)$ of an abstract
two-dimensional manifold $M$, glued from a finite number of Euclidean
triangles $\Delta_{k}$, $k=1,\dots,n$, under the action of a continuous map
$f:M\to\mathbb{R}^{3}$ whose restriction to each triangle $\Delta_{k}$ is a
linear isometric embedding. The point ${\bf x}\in f(M)$ is a self-intersection
point of $f(M)$ if its complete inverse image $f^{-1}({\bf x})\subset M$
consists from more than one point.

Our algorithm is based on the following considerations.

Let ${\bf x}\in f(M)$ be a self-intersection point of $f(M)$ and let $u,v\in
M$ be two distinct points in $f^{-1}({\ bfx})$. Denote by $\Delta_{k_{i}}$,
$i=1,2$, two different Euclidean triangles from which $M$ is glued, such that
$u\in\Delta_{k_{1}}$ and $v\in\Delta_{k_{2}}$. Recall that in § 2 we agreed to
call the triangles $f(\Delta_{k_{1}})$ and $f(\Delta_{k_{2}})$ faces of
$f(M)$. Fig. 5 illustraits why, for any mutual arrangement of
$f(\Delta_{k_{1}})$ and $f(\Delta_{k_{2}})$, at least on one of these
triangles has a point $\widetilde{\bf x}$ of intersection of a face and an
edge of $f(M)$. Thus, the problem of whether a polyhedron $f(M)$ has self-
intersections is reduced to the problem of whether a closed triangle and a
closed segment in $\mathbb{R}^{3}$ intersect.

Figure 5. Self-intersection point ${\bf x}$ of a polyhedron $f(M)$ belonging
to faces $f(\Delta_{k_{i}})\subset f(M)$, $i=1,2$. The point $\widetilde{\bf
x}$ is a point of intersection of a face and an edge of $f(M)$.

To solve the last problem, we use the oriented volume of a tetrahedron. As is
known, if ${\bf x}_{k}=(x_{k,1},x_{k,2},x_{k,3})$, $k=0,1,2,3$, are position
vectors of the vertices of a tetrahedron in $\mathbb{R}^{3}$, then the
oriented volume $\operatorname{Vol}({\bf x}_{0},{\bf x}_{1},{\bf x}_{2},{\bf
x}_{3})$ of this tetrahedron can be calculated using one of the following
formulas:

$\operatorname{Vol}({\bf x}_{0},{\bf x}_{1},{\bf x}_{2},{\bf
x}_{3})={\frac{1}{6}}{\begin{vmatrix}1&x_{0,1}&x_{0,2}&x_{0,3}\\\
1&x_{1,1}&x_{1,2}&x_{1,3}\\\ 1&x_{2,1}&x_{2,2}&x_{2,3}\\\
1&x_{3,1}&x_{3,2}&x_{3,3}\end{vmatrix}}={\frac{1}{6}}{\begin{vmatrix}x_{1,1}-x_{0,1}&x_{1,2}-x_{0,2}&x_{1,3}-x_{0,3}\\\
x_{2,1}-x_{0,1}&x_{2,2}-x_{0,2}&x_{2,3}-x_{0,3}\\\
x_{3,1}-x_{0,1}&x_{3,2}-x_{0,2}&x_{3,3}-x_{0,3}\end{vmatrix}}.$ (2)

Recall that the oriented volume of a tetrahedron is equal to its “usual”
volume if the triple of vectors ${\bf x}_{1}-{\bf x}_{0}$, ${\bf x}_{2}-{\bf
x}_{0}$, ${\bf x}_{3}-{\bf x}_{0}$ is right-oriented; is equal to the number
opposite to the “usual” volume, if the triple of vectors ${\bf x}_{1}-{\bf
x}_{0}$, ${\bf x}_{2}-{\bf x}_{0}$, ${\bf x}_{3}-{\bf x}_{0}$ is left-
oriented, and is equal to zero if the vectors ${\bf x}_{1}-{\bf x}_{0}$, ${\bf
x}_{2}-{\bf x}_{0}$ and ${\bf x}_{3}-{\bf x}_{0}$ lie in the same plane.

So, let a closed triangle $\Delta\subset\mathbb{R}^{3}$ be defined by the
coordinates of its vertices ${\bf y}_{k}=(y_{k,1},y_{k,2},y_{k,3})$, $k=1,2,3$
and a closed segment $I\subset\mathbb{R}^{3}$ is given by the coordinates of
its vertices ${\bf z}_{j}=(z_{j,1},z_{j,2},z_{j,3})$, $j=1,2$. If both
determinants

$6\operatorname{Vol}({\bf y}_{1},{\bf y}_{2},{\bf y}_{3},{\bf z}_{1}),\qquad
6\operatorname{Vol}({\bf y}_{1},{\bf y}_{2},{\bf y}_{3},{\bf z}_{2})$ (3)

are nonzero and have the same sign, then ${\bf z}_{1}$ and ${\bf z}_{2}$ lie
srictly to one side of the plane containing $\Delta$ (see the left side of
Fig. 6).

Figure 6. Various cases of mutual arrangement of a triangle $\Delta$ with
vertices ${\bf y}_{1},{\bf y}_{2},{\bf y}_{3}$ and a segment $I$ with
endpoints ${\bf z}_{1}$, ${\bf z}_{2}$.

Consequently, in this case $I$ does not intersect the plane containing
$\Delta$, and hence $I\cap\Delta=\varnothing$. If both determinants (3) are
nonzero and have different signs, then ${\bf z}_{1}$ and ${\bf z}_{2}$ lie to
opposite sides of the plane containing $\Delta$ (see the central and right
parts of Fig. 6). In this case, to figure out whether $I$ and $\Delta$
intersect, we need to calculate three more determinants:

$6\operatorname{Vol}({\bf z}_{1},{\bf z}_{2},{\bf y}_{1},{\bf y}_{2}),\qquad
6\operatorname{Vol}({\bf z}_{1},{\bf z}_{2},{\bf y}_{2},{\bf y}_{3}),\qquad
6\operatorname{Vol}({\bf z}_{1},{\bf z}_{2},{\bf y}_{3},{\bf y}_{1}).$ (4)

If all determinants (4) are non-zero and have the same sign, then
$I\cap\Delta\neq\varnothing$. This case is schematically shown in the central
part of Fig. 6. To understand why $I\cap\Delta\neq\varnothing$, for each point
${\bf x}$ lying on the boundary of the triangle $\Delta$, we denote by
$\pi({\bf x})$ the half-plane bounded by the line containing $I$ and passing
through ${\bf x}$. The fact that all three determinants in (4) have the same
sign, implies that, when ${\bf x}$ goes around the boundary of $\Delta$ once,
moving all the time in the same direction, the half-plane $\pi({\bf x})$ also
always rotates in the same direction and makes a complete turn around the
straight line containing $I$. This means that the boundary of $\Delta$ is
linked to the staight line containing $I$. Taking into account that the ends
of $I$ lie on opposite sides of the plane containing $\Delta$, we conclude
that $I\cap\Delta\neq\varnothing$.

Similarly, if all three determinants in (4) are nonzero, but not all of them
have the same sign, then after ${\bf x}$ circles the boundary of $\Delta$
once, moving all the time in the same direction, the half-plane $\pi({\bf x})$
will return to its original position, but will not make a complete turn around
the straight line containing $I$. This means that the boundary of $\Delta$ and
the line containing $I$ are not linked curves. Therefore, in this case
$I\cap\Delta=\varnothing$.

This algorithm does not work in “exceptional” cases, when at least one of the
determinants in (3) and (4) is zero. In principle, it can be improved by
introducing new functions, somewhat similar to the determinants in (3) and
(4), the use of which will allow us to fully understand each of the
“exceptional” cases. But we expect that for the polyhedra $\mathscr{M}$,
$\mathscr{S}$ and $\mathscr{P}$, i.e., for the polyhedra of interest to us,
there will be no “exceptional” cases at all or they will be very few so that
we can study each of them individually. Therefore, in order to avoid
complicating the algorithm, when encountering an “exceptional” case, our
algorithm only states that the question of the existence of intersections of
$\Delta$ and $I$ requires additional study.

More specifically, our algorithm can be formulated as follows:

Step 1. [Preliminary work for compiling lists of vertices, edges and faces.]

Enumerate the vertices of the abstract manifold $M$ in an arbitrary way.
Denote the $r$th vertex of $M$ by $v_{r}$. The set $S$ of all unordered pairs
$s=(v_{p},v_{q})$ of vertices of $M$ such that $v_{p}$ and $v_{q}$ are
connected by an edge of $M$ is called the list of edges of $M$. The set $T$ of
all unordered triples $t=(v_{i},v_{j},v_{k})$ of vertices of $M$ such that
$v_{i}$, $v_{j}$ and $v_{k}$ are vertices of some face of $M$ is called the
list of faces of $M$. Fix a linear order $<$ on the Cartesian product $T\times
S$ of $T$ and $S$. Generate a list of all vertices ${\bf
x}_{r}=f(v_{r})\in\mathbb{R}^{3}$ of the polyhedron $f(M)$, in which every
${\bf x}_{r}$ is given by its coordinates in $\mathbb{R}^{3}$. Finally, create
an empty auxiliary file.

Step 2. [This step begins exhaustion of all pairs $(t,s)\in T\times S$ and
verification whether the intersection $f(t)\cap f(s)$ is empty. The first time
we perform Step 2, we set $(t,s)$ equal to the least element of the ordered
set $(T\times S,<)$. When Step 2 is repeated, the selection of $(t,s)$ occurs
in Step 5. In order to bring the notation closer to those previously used in §
5, we assume that the face $t\subset M$ has vertices $u_{1}$, $u_{2}$,
$u_{3}$, and the edge $s\subset M$ has vertices $w_{1}$, $w_{2}$. We denote
the face $f(t)\subset f(M)$ by $\Delta$, denote its vertices by ${\bf
y}_{k}=f(u_{k})\in\mathbb{R}^{3}$, $k=1,2,3$, and write these vertices in
coordinates in the form ${\bf
y}_{k}=(y_{k,1},y_{k,2},y_{k,3})\in\mathbb{R}^{3}$. Similarly, we denote the
edge $f(s)\subset f(M)$ by $I$, denote its ends by ${\bf
z}_{j}=f(w_{j})\in\mathbb{R}^{3}$, $j=1,2$, and write these ends in
coordinates in the form ${\bf
z}_{j}=(z_{j,1},z_{j,2},z_{j,3})\in\mathbb{R}^{3}$.]

Using the notation just introduced, we can describe Step 2 as follows:

$\bullet$ if $t$ and $s$ are not incident to each other, i.e., if $u_{k}\neq
w_{j}$ for all $k=1,2,3$ and all $j=1,2$, then go to Step 3;

$\bullet$ if $t$ and $s$ have exactly one common point, i.e., if the equality
$u_{k}=w_{j}$ holds for only one pair of indices $k$, $j$, then go to Step 4;

$\bullet$ if $t$ and $s$ have more than one common point (i.e., if for any
$j=1,2$ there is $k=1,2,3$ for which the equality $u_{k}=w_{j}$ holds or,
which is the same, if $s$ is a side of $t$), then directly from the
definition, it is clear that the pair $(s,t)$ does not generate self-
intersections of $f(M)$, and we go to Step 5.

Step 3. [The case when $t$ and $s$ are not incident to each other.]

Calculate two determinants $6\operatorname{Vol}({\bf y}_{1},{\bf y}_{2},{\bf
y}_{3},{\bf z}_{1})$ and $6\operatorname{Vol}({\bf y}_{1},{\bf y}_{2},{\bf
y}_{3},{\bf z}_{2})$ (see formulas (2), (3)). Then,

$\bullet$ if at least one of these two determinants is equal to zero, then
write the message “The question whether $f(t)$ and $f(s)$ do intersect
requires additional study” to the auxiliary file, and go to Step 5;

$\bullet$ if both determinants are non-zero and have the same sign, then we
conclude that $f(t)\cap f(s)=\varnothing$, write nothing to the auxiliary
file, and go to Step 5;

$\bullet$ if both determinants are non-zero and have different signs, then
calculate three more determinants $6\operatorname{Vol}({\bf z}_{1},{\bf
z}_{2},{\bf y}_{1},{\bf y}_{2})$, $6\operatorname{Vol}({\bf z}_{1},{\bf
z}_{2},{\bf y}_{2},{\bf y}_{3})$, $6\operatorname{Vol}({\bf z}_{1},{\bf
z}_{2},{\bf y}_{3},{\bf y}_{1})$ (see formulas (2), (4)), and proceed as
follows:

$\diamond$ if at least one of these three determinants is equal to zero, then
write the message “The question whether $f(t)$ and $f(s)$ do intersect
requires additional study” to the auxiliary file, and go to Step 5;

$\diamond$ if all three determinants are non-zero and have the same sign, then
we conclude that $f(t)\cap f(s)\neq\varnothing$, write the message “$f(t)$ and
$f(s)$ intersect” to the auxiliary file, and go to Step 5;

$\diamond$ if all three determinants are non-zero but not all have the same
sign, then we conclude that $f(t)\cap f(s)=\varnothing$, write nothing to the
auxiliary file, and go to Step 5.

Step 4. [The case when $t$ and $s$ have exactly one common point.]

Calculate two determinants $6\operatorname{Vol}({\bf y}_{1},{\bf y}_{2},{\bf
y}_{3},{\bf z}_{1})$ and $6\operatorname{Vol}({\bf y}_{1},{\bf y}_{2},{\bf
y}_{3},{\bf z}_{2})$ (see formulas (2), (3)). Taking into account that at
least one of them is equal to zero, proceed as follows:

$\bullet$ if one of these determinants is not equal to zero, then we conclude
that $f(s)$ does not lie in the plane containing $f(t)$ and the pair $(s,t)$
does not contribute to self-intersections of $f(M)$; in this case write
nothing to the auxiliary file, and go to Step 5.

$\bullet$ if both of these determinants are equal to zero, then we conclude
that $f(s)$ lies in the plane containing $f(t)$; in this case write the
message “The question whether $f(t)$ and $f(s)$ do intersect requires
additional study” to the auxiliary file, and go to Step 5.

Step 5. [Move to the next pair $(t,s)$ or terminate the algorithm.]

Do this:

$\bullet$ if $(t,s)$ is not the maximal element of the linearly ordered set
$(T\times S,<)$, then replace $(t,s)$ with the next element in the order and
go to Step 2;

$\bullet$ if $(t,s)$ is the maximal element of $(T\times S,<)$, then output
the auxiliary file and finish the algorithm.

The main result of § 5 is the following lemma.

######  Lemma 3.

Modified Steffen polyhedron $\mathscr{M}$ is combinatorially equivalent to
some partition of the sphere, has only triangular faces, has no self-
intersections, and is flexible.

###### Proof.

It follows directly from the construction of $\mathscr{M}$ that it is
combinatorially equivalent to the triangulation of the sphere shown in Fig. 4,
and is flexible.

We checked the absence of self-intersections using the algorithm described
above, implemented in Mathematica using exclusively symbolic calculations.
Table 1 provides us with the list of vertices of $\mathscr{M}$ and expressions
in radicals for all their coordinates. Lists of edges and faces of
$\mathscr{M}$ can be easily compiled using Fig. 4. The calculation time was
about 0.01 of a second. Our algorithm found no self-intersections of
$\mathscr{M}$ or “exceptional” cases requiring additional study. Based on
this, we consider Lemma 3 proven. ∎

Although this is not necessary for solution of Problem 1, we have verified
that Steffen polyhedron $\mathscr{S}$ has no self-intersections. To do this,
we applied to $\mathscr{S}$ reasoning and calculations similar to those given
in the proof of Lemma 3. Our algorithm found no self-intersections of
$\mathscr{S}$ or “exceptional” cases requiring additional study.

##  6\. Construction of polyhedron $\mathscr{P}$

Let $\mathscr{M}$ be the modified Steffen polyhedron constructed in § 4. Let
$K:\mathbb{R}^{3}\to\mathbb{R}^{3}$ be the rotation around the $z$-axis by
$90^{\circ}$ such that $K(T_{3})=T_{2}$. In the coordinate system constructed
in § 4 and associated with $\mathscr{M}$, the rotation $K$ corresponds to the
matrix

$K=\begin{pmatrix}0&-1&0\\\ 1&0&0\\\ 0&0&1\end{pmatrix}.$ (5)

Let us consider the images of $\mathscr{M}$ under the action of the maps $K$,
$K^{2}=K\circ K$ and $K^{3}=K\circ K^{2}$, and glue them along the coincided
faces. We denote the resulting polyhedron by $\mathscr{P}$. It has 26
vertices, 72 edges and 48 faces.

Informally speaking, the main idea of constructing $\mathscr{P}$ is to
“surround” the edge $T_{1}T_{4}$ of $\mathscr{M}$ with isometric copies of
$\mathscr{M}$ so that $T_{1}T_{4}$ is no longer an edge of $\mathscr{P}$ and,
thus, the question “whether the dihedral angle attached to $T_{1}T_{4}$ is
constant” makes no sense. However, for what follows, such a general idea of
$\mathscr{P}$ will not be sufficient. Therefore, we are forced to go into
detail.

We denote the vertices of polyhedron $K(\mathscr{M})$ with the same letters as
the corresponding vertices of $\mathscr{M}$, but we provide them with a prime.
Moreover, we use the same convention as in § 3, namely: if a vertex of
$K(\mathscr{M})$ coincides with a vertex of $\mathscr{M}$ so that they must be
glued together and must be considered as a single vertex of $\mathscr{P}$,
then for this vertex we always use the notation that it had in $\mathscr{M}$.
For example, vertex $T^{\prime}_{1}=K(T_{1})$ coincides with $T_{1}$ and in
$\mathscr{P}$ is denoted by $T_{1}$; vertex $T^{\prime}_{4}=K(T_{4})$
coincides with $T_{4}$ and is denoted by $T_{4}$; finally, vertex
$T^{\prime}_{3}=K(T_{3})$ coincides with $T_{2}$ and is denoted by $T_{2}$.
Using these notations we can say that $\mathscr{M}$ and $K(\mathscr{M})$ are
glued along the coinciding faces $T_{1}T_{2}T_{4}$ and
$T^{\prime}_{1}T^{\prime}_{3}T^{\prime}_{4}$, which after gluing “disappear”
so that they are not faces of $\mathscr{P}$.

Similarly, we denote the vertices of polyhedra $K^{2}(\mathscr{M})$ and
$K^{3}(\mathscr{M})$ with the same letters as the corresponding vertices of
$\mathscr{M}$, but we provide their with two and three primes, respectively.
Moreover, if some vertex of $\mathscr{P}$ appears in our constructions several
times, then we assign to it the name that contains the minimum number of
primes. For example, in this notation $K^{3}(\mathscr{M})$ and $\mathscr{M}$
are glued along the coincided faces
$T^{\prime\prime\prime}_{1}T^{\prime\prime\prime}_{2}T^{\prime\prime\prime}_{4}$
and $T_{1}T_{3}T_{4}$; and vertex $T^{\prime\prime\prime}_{2}$ of
$K^{3}(\mathscr{M})$ receives the designation $T_{3}$ in $\mathscr{P}$.

Using these notations and Table 1 we can easily find the coordinates of any
vertex of $\mathscr{P}$. For example,

$\displaystyle A^{\prime\prime\prime}_{2}$ $\displaystyle=$ $\displaystyle
K^{3}(A_{2})=\begin{pmatrix}0&1&0\\\ -1&0&0\\\ 0&0&1\end{pmatrix}A_{2}=$
$\displaystyle=$
$\displaystyle\biggl{(}\frac{\omega_{2}\sqrt{2}+2\sqrt{\rho}}{15573468962},-\frac{\omega_{1}\sqrt{2}-2(200-33\sqrt{31})\sqrt{\rho}}{1230304047998},\frac{167\omega_{3}\sqrt{2}+22\sqrt{\rho}}{15573468962\sqrt{167}}\biggr{)}\approx(8.89,1.19,4.72),$

where the expressions $\omega_{1}$, $\omega_{2}$, $\omega_{3}$ and $\rho$ are
defined in the caption to Table 1.

The list of all vertices of $\mathscr{P}$ along with their coordinates is
given in Table 2.

|  $x$-coordinate |  $y$-coordinate |  $z$-coordinate
---|---|---|---
$T_{1}$ |  $-$$0$ |  $-$$0$ |  $-$$\frac{\sqrt{167}}{\sqrt{2}}\approx 9.13$
$T_{2}$ |  $-$$0$ |  $-$$\frac{11}{\sqrt{2}}\approx 7.77$ |  $-$$0$
$T_{3}$ |  $-$$\frac{11}{\sqrt{2}}\approx 7.77$ |  $-$$0$ |  $-$$0$
$T_{4}$ |  $-$$0$ |  $-$$0$ | $-\frac{\sqrt{167}}{\sqrt{2}}\approx-9.13$
$C_{2}$ |  $-$$\frac{11+3\sqrt{31}}{2\sqrt{2}}\approx 9.79$ |  $-$$\frac{11+3\sqrt{31}}{2\sqrt{2}}\approx 9.79$ |  $-$$0$
$A_{2}$ |  $-$$\frac{\omega_{1}\sqrt{2}-2(200-33\sqrt{31})\sqrt{\rho}}{1230304047998}\approx-1.19$ |  $-$$\frac{\omega_{2}\sqrt{2}+2\sqrt{\rho}}{15573468962}\approx 8.89$ |  $-$$\frac{167\omega_{3}\sqrt{2}+22\sqrt{\rho}}{15573468962\sqrt{167}}\approx 4.72$
$B_{1}$ |  $-$$\frac{\omega_{2}\sqrt{2}-2\sqrt{\rho}}{15573468962}\approx 2.79$ |  $-$$\frac{\omega_{1}\sqrt{2}+2(200-33\sqrt{31})\sqrt{\rho}}{1230304047998}\approx 0.05$ |  $-$$\frac{167\omega_{3}\sqrt{2}-22\sqrt{\rho}}{15573468962\sqrt{167}}\approx-0.46$
$\overline{A}_{2}$ |  $-$$\frac{\omega_{1}\sqrt{2}+2(200-33\sqrt{31})\sqrt{\rho}}{1230304047998}\approx 0.05$ |  $-$$\frac{\omega_{2}\sqrt{2}-2\sqrt{\rho}}{15573468962}\approx 2.79$ | $-\frac{167\omega_{3}\sqrt{2}-22\sqrt{\rho}}{15573468962\sqrt{167}}\approx 0.46$
$\overline{B}_{1}$ |  $-$$\frac{\omega_{2}\sqrt{2}+2\sqrt{\rho}}{15573468962}\approx 8.89$ |  $-$$\frac{\omega_{1}\sqrt{2}-2(200-33\sqrt{31})\sqrt{\rho}}{1230304047998}\approx-1.19$ | $-\frac{167\omega_{3}\sqrt{2}+22\sqrt{\rho}}{15573468962\sqrt{167}}\approx-4.72$
$T^{\prime}_{2}$ | $-\frac{11}{\sqrt{2}}\approx-7.77$ |  $-$$0$ |  $-$$0$
$C^{\prime}_{2}$ | $-\frac{11+3\sqrt{31}}{2\sqrt{2}}\approx-9.79$ |  $-$$\frac{11+3\sqrt{31}}{2\sqrt{2}}\approx 9.79$ |  $-$$0$
$A^{\prime}_{2}$ | $-\frac{\omega_{2}\sqrt{2}+2\sqrt{\rho}}{15573468962}\approx-8.89$ |  $-$$\frac{\omega_{1}\sqrt{2}-2(200-33\sqrt{31})\sqrt{\rho}}{1230304047998}\approx-1.19$ |  $-$$\frac{167\omega_{3}\sqrt{2}+22\sqrt{\rho}}{15573468962\sqrt{167}}\approx 4.72$
$B^{\prime}_{1}$ | $-\frac{\omega_{1}\sqrt{2}+2(200-33\sqrt{31})\sqrt{\rho}}{1230304047998}\approx-0.05$ |  $-$$\frac{\omega_{2}\sqrt{2}-2\sqrt{\rho}}{15573468962}\approx 2.79$ |  $-$$\frac{167\omega_{3}\sqrt{2}-22\sqrt{\rho}}{15573468962\sqrt{167}}\approx-0.46$
$\overline{A}^{\prime}_{2}$ | $-\frac{\omega_{2}\sqrt{2}-2\sqrt{\rho}}{15573468962}\approx-2.79$ |  $-$$\frac{\omega_{1}\sqrt{2}+2(200-33\sqrt{31})\sqrt{\rho}}{1230304047998}\approx 0.05$ | $-\frac{167\omega_{3}\sqrt{2}-22\sqrt{\rho}}{15573468962\sqrt{167}}\approx 0.46$
$\overline{B}^{\prime}_{1}$ | $-\frac{\omega_{1}\sqrt{2}-2(200-33\sqrt{31})\sqrt{\rho}}{1230304047998}\approx 1.19$ |  $-$$\frac{\omega_{2}\sqrt{2}+2\sqrt{\rho}}{15573468962}\approx 8.89$ | $-\frac{167\omega_{3}\sqrt{2}+22\sqrt{\rho}}{15573468962\sqrt{167}}\approx-4.72$
$T^{\prime\prime}_{2}$ | $0$ | $-\frac{11}{\sqrt{2}}\approx-7.77$ |  $-$$0$
$C^{\prime\prime}_{2}$ | $-\frac{11+3\sqrt{31}}{2\sqrt{2}}\approx-9.79$ | $-\frac{11+3\sqrt{31}}{2\sqrt{2}}\approx-9.79$ |  $-$$0$
$A^{\prime\prime}_{2}$ | $-\frac{\omega_{1}\sqrt{2}-2(200-33\sqrt{31})\sqrt{\rho}}{1230304047998}\approx 1.19$ | $-\frac{\omega_{2}\sqrt{2}+2\sqrt{\rho}}{15573468962}\approx-8.89$ |  $-$$\frac{167\omega_{3}\sqrt{2}+22\sqrt{\rho}}{15573468962\sqrt{167}}\approx 4.72$
$B^{\prime\prime}_{1}$ | $-\frac{\omega_{2}\sqrt{2}-2\sqrt{\rho}}{15573468962}\approx-2.79$ | $-\frac{\omega_{1}\sqrt{2}+2(200-33\sqrt{31})\sqrt{\rho}}{1230304047998}\approx-0.05$ |  $-$$\frac{167\omega_{3}\sqrt{2}-22\sqrt{\rho}}{15573468962\sqrt{167}}\approx-0.46$
$\overline{A}^{\prime\prime}_{2}$ | $-\frac{\omega_{1}\sqrt{2}+2(200-33\sqrt{31})\sqrt{\rho}}{1230304047998}\approx-0.05$ | $-\frac{\omega_{2}\sqrt{2}-2\sqrt{\rho}}{15573468962}\approx-2.79$ | $-\frac{167\omega_{3}\sqrt{2}-22\sqrt{\rho}}{15573468962\sqrt{167}}\approx 0.46$
$\overline{B}^{\prime\prime}_{1}$ | $-\frac{\omega_{2}\sqrt{2}+2\sqrt{\rho}}{15573468962}\approx-8.89$ | $-\frac{\omega_{1}\sqrt{2}-2(200-33\sqrt{31})\sqrt{\rho}}{1230304047998}\approx 1.19$ | $-\frac{167\omega_{3}\sqrt{2}+22\sqrt{\rho}}{15573468962\sqrt{167}}\approx-4.72$
$C^{\prime\prime\prime}_{2}$ |  $-$$\frac{11+3\sqrt{31}}{2\sqrt{2}}\approx 9.79$ | $-\frac{11+3\sqrt{31}}{2\sqrt{2}}\approx-9.79$ |  $-$$0$
$A^{\prime\prime\prime}_{2}$ |  $-$$\frac{\omega_{2}\sqrt{2}+2\sqrt{\rho}}{15573468962}\approx 8.89$ | $-\frac{\omega_{1}\sqrt{2}-2(200-33\sqrt{31})\sqrt{\rho}}{1230304047998}\approx 1.19$ |  $-$$\frac{167\omega_{3}\sqrt{2}+22\sqrt{\rho}}{15573468962\sqrt{167}}\approx 4.72$
$B^{\prime\prime\prime}_{1}$ |  $-$$\frac{\omega_{1}\sqrt{2}+2(200-33\sqrt{31})\sqrt{\rho}}{1230304047998}\approx 0.05$ | $-\frac{\omega_{2}\sqrt{2}-2\sqrt{\rho}}{15573468962}\approx-2.79$ |  $-$$\frac{167\omega_{3}\sqrt{2}-22\sqrt{\rho}}{15573468962\sqrt{167}}\approx-0.46$
$\overline{A}^{\prime\prime\prime}_{2}$ |  $-$$\frac{\omega_{2}\sqrt{2}-2\sqrt{\rho}}{15573468962}\approx 2.79$ | $-\frac{\omega_{1}\sqrt{2}+2(200-33\sqrt{31})\sqrt{\rho}}{1230304047998}\approx-0.05$ | $-\frac{167\omega_{3}\sqrt{2}-22\sqrt{\rho}}{15573468962\sqrt{167}}\approx 0.46$
$\overline{B}^{\prime\prime\prime}_{1}$ |  $-$$\frac{\omega_{1}\sqrt{2}-2(200-33\sqrt{31})\sqrt{\rho}}{1230304047998}\approx-1.19$ | $-\frac{\omega_{2}\sqrt{2}+2\sqrt{\rho}}{15573468962}\approx-8.89$ | $-\frac{167\omega_{3}\sqrt{2}+22\sqrt{\rho}}{15573468962\sqrt{167}}\approx-4.72$

|  |  |
Table 2. Exact and approximate values of the coordinates of the vertices of
$\mathscr{P}$. The expressions $\omega_{1}$, $\omega_{2}$, $\omega_{3}$ and
$\rho$ are defined in the caption to Table 1. All decimal places in
approximate values are correct.

We formulate the properties of $\mathscr{P}$ related to our research in the
following lemma.

######  Lemma 4.

Polyhedron $\mathscr{P}$ is combinatorially equivalent to some triangulation
of the sphere, has no self-intersections, and is flexible.

###### Proof.

It follows directly from the construction of $\mathscr{P}$ that it is
combinatorially equivalent to a triangulation of the sphere and is flexible.

We checked the absence of self-intersections using the algorithm that was
described in § 5 and was implemented in Mathematica using exclusively symbolic
calculations. Table 2 provides us with the list of vertices of $\mathscr{P}$
and expressions in radicals for all their coordinates. Lists of edges and
faces of $\mathscr{P}$ are given in Tables 3 and 4, respectively. The
calculation time was less than 0.1 of a second. Our algorithm found no self-
intersections of $\mathscr{P}$ or “exceptional” cases requiring additional
study. ∎

$\mathscr{P}\cap\mathscr{M}$AAAAAAA |  $\mathscr{P}\cap K(\mathscr{M})$AAAAAAA |  $\mathscr{P}\cap K^{2}(\mathscr{M})$AAAAAAA |  $\mathscr{P}\cap K^{3}(\mathscr{M})$AAAAAAA |
---|---|---|---|---
$T_{1}T_{2}$ | $T_{1}T^{\prime}_{2}$ | $T_{1}T^{\prime\prime}_{2}$ | $\varnothing$ | 7
$T_{1}T_{3}$ | $\varnothing$ | $\varnothing$ | $\varnothing$ | 7
$T_{1}A_{2}$ | $T_{1}A^{\prime}_{2}$ | $T_{1}A^{\prime\prime}_{2}$ | $T_{1}A^{\prime\prime\prime}_{2}$ | 5
$T_{1}B_{1}$ | $T_{1}B^{\prime}_{1}$ | $T_{1}B^{\prime\prime}_{1}$ | $T_{1}B^{\prime\prime\prime}_{1}$ | 5
$T_{2}T_{4}$ | $T^{\prime}_{2}T_{4}$ | $T^{\prime\prime}_{2}T_{4}$ | $\varnothing$ | 7
$T_{2}C_{2}$ | $T^{\prime}_{2}C^{\prime}_{2}$ | $T^{\prime\prime}_{2}C^{\prime\prime}_{2}$ | $T_{3}C^{\prime\prime\prime}_{2}$ | 6
$T_{2}A_{2}$ | $T^{\prime}_{2}A^{\prime}_{2}$ | $T^{\prime\prime}_{2}A^{\prime\prime}_{2}$ | $T_{3}A^{\prime\prime\prime}_{2}$ | 5
$T_{2}\overline{A}_{2}$ | $T^{\prime}_{2}\overline{A}^{\prime}_{2}$ | $T^{\prime\prime}_{2}\overline{A}^{\prime\prime}_{2}$ | $T_{3}\overline{A}^{\prime\prime\prime}_{2}$ | 5
$T_{3}T_{4}$ | $\varnothing$ | $\varnothing$ | $\varnothing$ | 7
$T_{3}C_{2}$ | $T_{2}C^{\prime}_{2}$ | $T^{\prime}_{2}C^{\prime\prime}_{2}$ | $T^{\prime\prime}_{2}C^{\prime\prime\prime}_{2}$ | 6
$T_{3}B_{1}$ | $T_{2}B^{\prime}_{1}$ | $T^{\prime}_{2}B^{\prime\prime}_{1}$ | $T^{\prime\prime}_{2}B^{\prime\prime\prime}_{1}$ | 5
$T_{3}\overline{B}_{1}$ | $T_{2}\overline{B}^{\prime}_{1}$ | $T^{\prime}_{2}\overline{B}^{\prime\prime}_{1}$ | $T^{\prime\prime}_{2}\overline{B}^{\prime\prime\prime}_{1}$ | 5
$T_{4}\overline{A}_{2}$ | $T_{4}\overline{A}^{\prime}_{2}$ | $T_{4}\overline{A}^{\prime\prime}_{2}$ | $T_{4}\overline{A}^{\prime\prime\prime}_{2}$ | 5
$T_{4}\overline{B}_{1}$ | $T_{4}\overline{B}^{\prime}_{1}$ | $T_{4}\overline{B}^{\prime\prime}_{1}$ | $T_{4}\overline{B}^{\prime\prime\prime}_{1}$ | 5
$C_{2}A_{2}$ | $C^{\prime}_{2}A^{\prime}_{2}$ | $C^{\prime\prime}_{2}A^{\prime\prime}_{2}$ | $C^{\prime\prime\prime}_{2}A^{\prime\prime\prime}_{2}$ | 5
$C_{2}B_{1}$ | $C^{\prime}_{2}B^{\prime}_{1}$ | $C^{\prime\prime}_{2}B^{\prime\prime}_{1}$ | $C^{\prime\prime\prime}_{2}B^{\prime\prime\prime}_{1}$ | 5
$C_{2}\overline{A}_{2}$ | $C^{\prime}_{2}\overline{A}^{\prime}_{2}$ | $C^{\prime\prime}_{2}\overline{A}^{\prime\prime}_{2}$ | $C^{\prime\prime\prime}_{2}\overline{A}^{\prime\prime\prime}_{2}$ | 5
$C_{2}\overline{B}_{1}$ | $C^{\prime}_{2}\overline{B}^{\prime}_{1}$ | $C^{\prime\prime}_{2}\overline{B}^{\prime\prime}_{1}$ | $C^{\prime\prime\prime}_{2}\overline{B}^{\prime\prime\prime}_{1}$ | 5
$A_{2}B_{1}$ | $A^{\prime}_{2}B^{\prime}_{1}$ | $A^{\prime\prime}_{2}B^{\prime\prime}_{1}$ | $A^{\prime\prime\prime}_{2}B^{\prime\prime\prime}_{1}$ | 5
$\overline{A}_{2}\overline{B}_{1}$ | $\overline{A}^{\prime}_{2}\overline{B}^{\prime}_{1}$ | $\overline{A}^{\prime\prime}_{2}\overline{B}^{\prime\prime}_{1}$ | $\overline{A}^{\prime\prime\prime}_{2}\overline{B}^{\prime\prime\prime}_{1}$ | 5

|  |  |  |
Table 3. List of all 72 edges of polyhedron $\mathscr{P}$. The sign $\varnothing$ marks cells that are intentionally left empty because the edge corresponding to such a cell has already appeared (in another row) in one of the columns located to the left. The right column contains the numbers of lemmas, from the proof of which it follows that the dihedral angle at each edge in this row is not constant. $\mathscr{P}\cap\mathscr{M}$AAAAAAA |  $\mathscr{P}\cap K(\mathscr{M})$AAAAAAA |  $\mathscr{P}\cap K^{2}(\mathscr{M})$AAAAAAA |  $\mathscr{P}\cap K^{3}(\mathscr{M})$AAAAAAA
---|---|---|---
$T_{1}T_{2}A_{2}$ | $T_{1}T^{\prime}_{2}A^{\prime}_{2}$ | $T_{1}T^{\prime\prime}_{2}A^{\prime\prime}_{2}$ | $T_{1}T_{3}A^{\prime\prime\prime}_{2}$
$T_{1}T_{3}B_{1}$ | $T_{1}T_{2}B^{\prime}_{1}$ | $T_{1}T^{\prime}_{2}B^{\prime\prime}_{1}$ | $T_{1}T^{\prime\prime}_{2}B^{\prime\prime\prime}_{1}$
$T_{1}A_{2}B_{1}$ | $T_{1}A^{\prime}_{2}B^{\prime}_{1}$ | $T_{1}A^{\prime\prime}_{2}B^{\prime\prime}_{1}$ | $T_{1}A^{\prime\prime\prime}_{2}B^{\prime\prime\prime}_{1}$
$T_{2}T_{4}\overline{A}_{2}$ | $T^{\prime}_{2}T_{4}\overline{A}^{\prime}_{2}$ | $T^{\prime\prime}_{2}T_{4}\overline{A}^{\prime\prime}_{2}$ | $T_{3}T_{4}\overline{A}^{\prime\prime\prime}_{2}$
$T_{2}C_{2}A_{2}$ | $T^{\prime}_{2}C^{\prime}_{2}A^{\prime}_{2}$ | $T^{\prime\prime}_{2}C^{\prime\prime}_{2}A^{\prime\prime}_{2}$ | $T_{3}C^{\prime\prime\prime}_{2}A^{\prime\prime\prime}_{2}$
$T_{2}C_{2}\overline{A}_{2}$ | $T^{\prime}_{2}C^{\prime}_{2}\overline{A}^{\prime}_{2}$ | $T^{\prime\prime}_{2}C^{\prime\prime}_{2}\overline{A}^{\prime\prime}_{2}$ | $T_{3}C^{\prime\prime\prime}_{2}\overline{A}^{\prime\prime\prime}_{2}$
$T_{3}T_{4}\overline{B}_{1}$ | $T_{2}T_{4}\overline{B}^{\prime}_{1}$ | $T^{\prime}_{2}T_{4}\overline{B}^{\prime\prime}_{1}$ | $T^{\prime\prime}_{2}T_{4}\overline{B}^{\prime\prime\prime}_{1}$
$T_{3}C_{2}B_{1}$ | $T_{2}C^{\prime}_{2}B^{\prime}_{1}$ | $T^{\prime}_{2}C^{\prime\prime}_{2}B^{\prime\prime}_{1}$ | $T^{\prime\prime}_{2}C^{\prime\prime\prime}_{2}B^{\prime\prime\prime}_{1}$
$T_{3}C_{2}\overline{B}_{1}$ | $T_{2}C^{\prime}_{2}\overline{B}^{\prime}_{1}$ | $T^{\prime}_{2}C^{\prime\prime}_{2}\overline{B}^{\prime\prime}_{1}$ | $T^{\prime\prime}_{2}C^{\prime\prime\prime}_{2}\overline{B}^{\prime\prime\prime}_{1}$
$T_{4}\overline{A}_{2}\overline{B}_{1}$ | $T_{4}\overline{A}^{\prime}_{2}\overline{B}^{\prime}_{1}$ | $T_{4}\overline{A}^{\prime\prime}_{2}\overline{B}^{\prime\prime}_{1}$ | $T_{4}\overline{A}^{\prime\prime\prime}_{2}\overline{B}^{\prime\prime\prime}_{1}$
$C_{2}A_{2}B_{1}$ | $C^{\prime}_{2}A^{\prime}_{2}B^{\prime}_{1}$ | $C^{\prime\prime}_{2}A^{\prime\prime}_{2}B^{\prime\prime}_{1}$ | $C^{\prime\prime\prime}_{2}A^{\prime\prime\prime}_{2}B^{\prime\prime\prime}_{1}$
$C_{2}\overline{A}_{2}\overline{B}_{1}$ | $C^{\prime}_{2}\overline{A}^{\prime}_{2}\overline{B}^{\prime}_{1}$ | $C^{\prime\prime}_{2}\overline{A}^{\prime\prime}_{2}\overline{B}^{\prime\prime}_{1}$ | $C^{\prime\prime\prime}_{2}\overline{A}^{\prime\prime\prime}_{2}\overline{B}^{\prime\prime\prime}_{1}$

|  |  |
Table 4. List of all 48 faces of polyhedron $\mathscr{P}$.

##  7\. Special flex $\mathscr{P}(t)$ of polyhedron $\mathscr{P}$

Let us start by discussing the flex of the modified Steffen polyhedron
$\mathscr{M}$.

In order to define such a flex, we need to specify the positions of all
vertices of $\mathscr{M}$ as continuous functions of some parameter $t$ so
that the distance between any two vertices connected by an edge of
$\mathscr{M}$ does not depend on $t$, and the distance between at least two
vertices not connected by an edge is not constant. We denote each of these
functions by the same symbol as the corresponding vertex of $\mathscr{M}$. For
example, the vertex $C_{2}\in\mathscr{M}$ corresponds to the function
$C_{2}(t)$. The values of these functions for a fixed $t$ are declared to be
the vertices of the polyhedron $\mathscr{M}(t)$ and, by definition, we put
that the correspondence “the vertex of $\mathscr{M}$ $\leftrightarrow$ the
value of the function of the same name” defines the combinatorial equivalence
of polyhedra $\mathscr{M}$ and $\mathscr{M}(t)$. Finally, since the choice of
parameter $t$ is arbitrary, we assume without loss of generality that
$\mathscr{M}(0)=\mathscr{M}$.

The functions corresponding to the vertices of $\mathscr{M}$ are defined as
follows.

We assume that the vertices $T_{j}$, $j=1,\dots,4$, do not change their
positions in space, i.e., by definition, we assume $T_{j}(t)=T_{j}$ for all
$t$, where $T_{j}$ has the coordinates given in Table 1.

Since the vertex $C_{2}(t)$ is connected in $\mathscr{M}(t)$ by edges to the
vertices $T_{2}(t)\equiv T_{2}$ and $T_{3}(t)\equiv T_{3}$, then it lies on
the circle $\gamma$, determined by the following conditions: $\gamma$ lies in
a plane perpendicular to the segment $T_{2}T_{3}$; the center of $\gamma$
coincides with the middle point of $T_{2}T_{3}$; the radius of $\gamma$ is
equal to $\sqrt{|A_{1}C_{2}|^{2}-|A_{1}B_{2}|^{2}/4}=3\sqrt{31}/2$. It is the
movement of $C_{2}(t)$ along $\gamma$ that sets the flex of $\mathscr{M}$
(here we do not care whether $\mathscr{M}(t)$ has self-intersections). Indeed,
the positions of the five vertices $C_{2}(t)$, $T_{j}(t)$ ($j=1,\dots,4$) of
$\mathscr{M}(t)$ uniquely determine the position of every of its other four
vertices $A_{2}(t)$, $B_{1}(t)$, $\overline{A}_{2}(t)$ and
$\overline{B}_{1}(t)$, because each of the latter vertices is connected in
$\mathscr{M}(t)$ by three edges to some of the five vertices $C_{2}(t)$,
$T_{j}(t)$ ($j=1,\dots,4$). Taking into account that the position of each
vertex of $\mathscr{M}(t)$ changes continuously and is known for $t=0$ from
Table 1, we are convinced that the positions of $A_{2}(t)$, $B_{1}(t)$,
$\overline{A}_{2}(t)$ and $\overline{B}_{1}(t)$ are uniquely determined by the
positions of $C_{2}(t)$, $T_{j}(t)$ ( $j=1,\dots,4$) for all $t$ sufficiently
close to zero (namely, until the three edges mentioned above are not in the
same plane). That is why we say that the flex $\mathscr{M}(t)$ of
$\mathscr{M}$ is defined by the position of $C_{2}(t)$.

Let us discuss the flex of polyhedron $\mathscr{P}$.

By construction, $\mathscr{P}$ is obtained by gluing four copies of
$\mathscr{M}$, namely, by gluing together polyhedra $\mathscr{M}$,
$K(\mathscr{M})$, $K^{2}(\mathscr{M})$ and $K^{3}(\mathscr{M})$. Here, as
before, $K:\mathbb{R}^{3}\to\mathbb{R}^{3}$ is the rotation around $z$-axis by
$90^{\circ}$; in particular, the matrix of $K$ in the coordinate system
constructed in § 4 which is associated with $\mathscr{P}$ is given by (5).
From what was said at the beginning of § 7 it is clear that, similarly to the
movement of vertex $C_{2}\in\mathscr{M}$ along the circle $\gamma$, each of
the vertices $C^{\prime}_{2}\in K(\mathscr{M})$, $C^{\prime\prime}_{2}\in
K^{2}(\mathscr{M})$ and $C^{\prime\prime\prime}_{2}\in K^{3}(\mathscr{M})$ can
be moved along the corresponding circle $K(\gamma)$, $K^{2}(\gamma)$,
$K^{3}(\gamma)$ independently of each other and of the motion of $C_{2}$.
After the positions of the vertices $C_{2}(t)$, $C^{\prime}_{2}(t)$,
$C^{\prime\prime}_{2}(t)$ and $C^{\prime\prime\prime}_{2}(t)$ are given (and
taking into account that the positions of the vertices $T_{1}$, $T_{2}$,
$T_{3}$, $T_{4}$, $T^{\prime}_{2}$ and $T^{\prime\prime}_{2}$ do not change
during the flex and are known to us from Table 2), we can uniquely find the
positions of all other vertices of $\mathscr{P}(t)$. To reflect the
possibility of independent changing the positions of $C_{2}$,
$C^{\prime}_{2}$, $C^{\prime\prime}_{2}$ and $C^{\prime\prime\prime}_{2}$, we
say that $\mathscr{P}$ admits a 4-parameter flex.

However, in accordance with the definition of a flexible polyhedron given in §
1, when proving Theorem 1 we need some special flex $\mathscr{P}(t)$ of
$\mathscr{P}$, depending on one real parameter $t$. We distinguish it from the
4-parameter flex of $\mathscr{P}$ just described by the following three
conditions:

$\bullet$ $\mathscr{P}(0)=\mathscr{P}$;

$\bullet$ for $t=0$, the speed of $C_{2}(t)$ is equal to $(0,0,1)$;

$\bullet$ for all $t$, $\mathscr{P}(t)$ is invariant under the rotation
$K:\mathbb{R}^{3}\to\mathbb{R}^{3}$ around $z$-axis by $90^{\circ}$.

We denote the vertices of $\mathscr{P}(t)$ by analogy with the corresponding
vertices of $\mathscr{P}$. For example, the vertices $C_{2}(t)$ and
$A^{\prime\prime\prime}_{2}(t)$ of $\mathscr{P}(t)$ correspond to the vertices
$C_{2}=C_{2}(0)$ and
$A^{\prime\prime\prime}_{2}=A^{\prime\prime\prime}_{2}(0)$ of $\mathscr{P}$.

Recall that Lemma 3 from § 5 and Lemma 4 from § 6 state that polyhedra
$\mathscr{M}$ and $\mathscr{P}$ have no self-intersections. Now we draw the
reader’s attention to the fact that without any additional calculations we can
state that for all $t$ sufficiently close to zero polyhedra $\mathscr{M}(t)$
and $\mathscr{P}(t)$ also do not have self-intersections. To prove this,
observe that non of the determinants (3) and (4), calculated in the proofs of
Lemmas 3 and 4, were equal to zero. Hence, they are nonzero for all $t$
sufficiently close to zero. Having fixed any of these $t$ and repeating the
proofs of Lemmas 3 and 4 as applied to $\mathscr{M}(t)$ and $\mathscr{P}(t)$,
we come to the conclusion that both $\mathscr{M}(t)$ and $\mathscr{P}(t)$ have
no self-intersections.

Let us begin to study whether $\mathscr{P}(t)$ has a dihedral angle, the value
of which remains constant for all $t$ sufficiently close to zero. For
different dihedral angles we will need different arguments.

######  Lemma 5.

The dihedral angle at the edge $A_{2}(t)T_{1}(t)\equiv A_{2}(t)T_{1}$ of
polyhedron $\mathscr{P}(t)$ is not constant as a function of $t$.

###### Proof.

In $\mathscr{P}$, the edge $A_{2}T_{1}$ is incident to faces $A_{2}B_{1}T_{1}$
and $A_{2}B_{2}T_{1}$. Both of these faces belong to the Bricard octahedron
$\mathscr{B}$, which participated in the construction of the modified Steffen
polyhedron $\mathscr{M}$, which, in turn, was used to construct polyhedron
$\mathscr{P}$ (see §§ 2, 3, 4, 6). The flex $\mathscr{P}(t)$ of $\mathscr{P}$
obviously gives rise to the flex $\mathscr{B}(t)$ of the Bricard octahedron
$\mathscr{B}$ (indeed, the length of each edge of $\mathscr{B}(t)$ does not
depend on $t$, while the length of the diagonal $C_{2}(t)T_{1}$ is obviously
nonconstant as function of $t$). According to Lemma 1, during the flex of the
Bricard octahedron $\mathscr{B}$ the value of each of its dihedral angles does
not remain constant. Hence, the value of the dihedral angle at edge
$A_{2}(t)T_{1}$, considered either as a dihedral angle of the Bricard
octahedron $\mathscr{B}(t)$, or as a dihedral angle of $\mathscr{P}(t)$, is
also nonconstant. ∎

Note that in Lemma 5 the edge $A_{2}(t)T_{1}$ can be replaced by any edge of
$\mathscr{P}(t)$, which is incident to two faces of any Bricard octahedron
participating in the construction of $\mathscr{P}$ (i.e., the Bricard
octahedra $\mathscr{B}$ and $\overline{\mathscr{B}}$ in notation of § 3, as
well as their images under rotations $K$, $K^{2}$ and $K^{3}$). There are a
total of 7 such edges on each part of $\mathscr{P}$ which corresponds to a
single Bricard octahedron. Hence, there are 56 such edges on $\mathscr{P}$.
Each row of Table 3, consisting entirely of such edges, ends with the number
5. We use this notation to fix the fact that the dihedral angles at the edges
in such rows are nonconstant follows from the proof of Lemma 5. Similarly,
with numbers 6 and 7 in the last column we mark those rows of Table 3 for
which the inconstancy of dihedral angles at the edges in these rows follows
from the proofs of Lemmas 6 and 7, respectively.

So, we already know that the dihedral angles at 56 edges do not remain
constant during the special flex of $\mathscr{P}$. Lemmas 6 and 7 are devoted
to the study of the remaining 16 dihedral angles. But before we formulate and
prove them, we need to do preparatory work.

Let ${\bf p}_{j}(t)$, $j=1,2,3,4$, be points which may change their position
in space in such a way that the distance between any two of them, except
perhaps ${\bf p}_{3}(t)$ and ${\bf p}_{4}(t)$, do not depend on $t$. Then, as
is known (see, for example, [23, Section 3.2]), their velocities ${\bf
v}_{j}(t)$ satisfy the relations

$({\bf p}_{j}(t)-{\bf p}_{k}(t))\cdot({\bf v}_{j}(t)-{\bf v}_{k}(t))=0,\qquad
j,k=1,2,3,4;\ \\{j,k\\}\neq\\{3,4\\},$ (6)

where $\cdot$ denotes the standard scalar product in $\mathbb{R}^{3}$. It is
easy to understand that if the inequality

$({\bf p}_{3}(0)-{\bf p}_{4}(0))\cdot({\bf v}_{3}(0)-{\bf v}_{4}(0))\neq 0$
(7)

holds true, then the derivative, calculated at $t=0$, of the dihedral angle
between the triangles ${\bf p}_{1}(t),{\bf p}_{2}(t),{\bf p}_{3}(t)$ and ${\bf
p}_{1}(t),{\bf p}_{2}(t),{\bf p}_{4}(t)$ is not equal to zero, which means the
value of this dihedral angle is not constant as a function of $t$ on some
sufficiently small interval in $\mathbb{R}$ containing $0$.

######  Lemma 6.

Dihedral angles at edges $C_{2}(t)T_{2}(t)\equiv C_{2}(t)T_{2}$ and
$C_{2}(t)T_{3}(t)\equiv C_{2}(t)T_{3}$ of polyhedron $\mathscr{P}(t)$ are not
constant as functions of $t$ on some sufficiently small interval in
$\mathbb{R}$ containing 0.

###### Proof.

By definition, put ${\bf p}_{1}(t)=C_{2}(t)$, ${\bf p}_{2}(t)\equiv T_{2}$,
${\bf p}_{3}(t)=A_{2}(t)$, ${\bf p}_{4}(t)=\overline{A}_{2}(t)$, ${\bf
p}_{5}(t)\equiv T_{1}$, ${\bf p}_{6}(t)\equiv T_{4}$, and denote by ${\bf
v}_{j}(t)$, $j=1,\dots,6$, the velocity of the point ${\bf p}_{j}(t)$. To
prove the statement of Lemma 6 it is sufficient to prove inequality (7) for
$t=0$. We do this using symbolic computation in Mathematica.

The coordinates of the points ${\bf p}_{1}(0)=C_{2}$, ${\bf p}_{2}(0)=T_{2}$,
${\bf p}_{3}(0)=A_{2}$, ${\bf p}_{4}(0)=\overline{A}_{2}$, ${\bf
p}_{5}(0)=T_{1}$ and ${\bf p}_{6}(0)=T_{4}$ in radicals we take from Table 1.
The velocity vectors ${\bf v}_{1}(0)=(0,0,1)$ and ${\bf v}_{2}(0)={\bf
v}_{5}(0)={\bf v}_{6}(0)=(0,0,0)$ are known to us by construction.

The components of the velocity vector ${\bf v}_{3}(0)$ must satisfy the
following system of linear algebraic equations, each of which is similar to
equation (6):

$\begin{cases}({\bf p}_{1}(0)-{\bf p}_{3}(0))\cdot({\bf v}_{1}(0)-{\bf
v}_{3}(0))=0,\\\ ({\bf p}_{2}(0)-{\bf p}_{3}(0))\cdot({\bf v}_{2}(0)-{\bf
v}_{3}(0))=0,\\\ ({\bf p}_{5}(0)-{\bf p}_{3}(0))\cdot({\bf v}_{5}(0)-{\bf
v}_{3}(0))=0.\\\ \end{cases}$ (8)

Using Mathematica, we solve (8) and find the values of the components of ${\bf
v}_{3}(0)$ in radicals. However, they are too long to be written here. So, we
present their approximate numerical values only: ${\bf
v}_{3}(0)\approx(-0.4602,-0.1074,-0.0914)$, where all decimal places are
correct (i.e., we do not apply rounding rules).

Similarly, the components of the velocity vector ${\bf v}_{4}(0)$ must satisfy
the system of equations

$\begin{cases}({\bf p}_{1}(0)-{\bf p}_{4}(0))\cdot({\bf v}_{1}(0)-{\bf
v}_{4}(0))=0,\\\ ({\bf p}_{2}(0)-{\bf p}_{4}(0))\cdot({\bf v}_{2}(0)-{\bf
v}_{4}(0))=0,\\\ ({\bf p}_{6}(0)-{\bf p}_{4}(0))\cdot({\bf v}_{6}(0)-{\bf
v}_{4}(0))=0.\\\ \end{cases}$ (9)

Using Mathematica, we solve (9) and find the values of the components of ${\bf
v}_{4}(0)$ in radicals. As before, they are too long to be written here and we
present their approximate numerical values only: ${\bf
v}_{4}(0)\approx(-0.0470,-0.0004,0.0004)$.

Finally, using Mathematica we evaluate the expression

$({\bf p}_{3}(0)-{\bf p}_{4}(0))\cdot({\bf v}_{3}(0)-{\bf v}_{4}(0)).$ (10)

The result, written in radicals, is too long to be given here; but, again with
the help of Mathematica, we are convinced that it is non-zero in the
corresponding extension of the field of rational numbers. The numerical value
of expression (10) is equal to $-0.5253$; this gives us additional confidence
in the adequacy of our symbolic calculations. Thus, the statement of Lemma 6
about the inconstancy of the dihedral angle at the edge $C_{2}(t)T_{2}$ of
$\mathscr{P}(t)$ is proven.

The statement of Lemma 6 about the inconstancy of the dihedral angle at the
edge $C_{2}(t)T_{3}$ of $\mathscr{P}(t)$ can be proven in the same way as we
have just proved a similar statement for the angle at the edge
$C_{2}(t)T_{2}$. But we prefer to avoid these additional calculations by
noting that the rotation $L$ defined in the statement of Lemma 2 maps the
polyhedron $\mathscr{P}$ onto itself. To present this reasoning in more
detail, we introduce the notation ${\bf q}_{1}(t)=C_{2}(t)$, ${\bf
q}_{2}(t)\equiv T_{3}$, ${\bf q}_{3}(t)=\overline{B}_{1}(t)$, ${\bf
q}_{4}(t)={B}_{1}(t)$, ${\bf q}_{5}(t)\equiv T_{4}$, ${\bf q}_{6}(t)\equiv
T_{1}$. Denote by ${\bf w}_{j}(t)$, $j=1,\dots,6$, the velocity of the point
${\bf q}_{j}(t)$. Lemma 2 implies that $L({\bf q}_{j}(0))={\bf p}_{j}(0)$ for
all $j=1,\dots,6$. We have already reflected this fact in the Table 1. Taking
into account that ${\bf w}_{1}(0)=(0,0,1)$ and ${\bf w}_{2}(0)={\bf
w}_{5}(0)={\bf w}_{6}(0)=(0,0,0)$, and writing equations for finding ${\bf
w}_{3}(0)$ and ${\bf w}_{4}(0)$ by analogy with (8) and (9), we obtain $L({\bf
w}_{j}(0))=-{\bf v}_{j}(0)$, $j=1,\dots,6$. Hence, $({\bf q}_{3}(0)-{\bf
q}_{4}(0))\cdot({\bf w}_{3}(0)-{\bf w}_{4}(0))=-({\bf p}_{3}(0)-{\bf
p}_{4}(0))\cdot({\bf v}_{3}(0)-{\bf v}_{4}(0))\neq 0$. As we already know,
this inequality implies that the dihedral angle at the edge $C_{2}(t)T_{3}$ is
nonconstant. ∎

Note that in Lemma 6 the edges $C_{2}T_{2}$ and $C_{2}T_{3}$ can be replaced
by their images under the action of rotations $K$, $K^{2}$ and $K^{3}$. This
means that Lemma 6 guarantees us that the eight dihedral angles of
$\mathscr{P}(t)$ are not constant as functions of $t$. This fact is reflected
in the right column of Table 3.

######  Lemma 7.

Dihedral angles at edges $T_{1}T_{3}$ and $T_{2}T_{4}$ of polyhedron
$\mathscr{P}(t)$ are not constant as functions of $t$ on some sufficiently
small interval in $\mathbb{R}$ containing 0.

###### Proof.

Lemma 7 can be proven similarly to Lemma 6. The proof is left to the reader. ∎

Note that in Lemma 7 the edges $T_{1}T_{3}$ and $T_{2}T_{4}$ can be replaced
by their images under the action of rotations $K$, $K^{2}$ and $K^{3}$. This
means that Lemma 7 guarantees us that the eight dihedral angles of
$\mathscr{P}(t)$ are not constant as functions of $t$. This fact is reflected
in the right column of Table 3.

Thus, Lemmas 5, 6 and 7 imply that all 72 dihedral angles of $\mathscr{P}(t)$
are not constant as functions of $t$ on some sufficiently small interval in
$\mathbb{R}$ containing 0.

##  8\. Proof of Theorem 1 and concluding remarks

###### Proof.

All components of the proof of Theorem 1 have already appeared in the previous
Sections. We just need to put them together.

Let $\mathscr{P}$ be the polyhedron constructed in § 6. According to Lemma 3,
$\mathscr{P}$ has no self-intersections, is homeomorphic to a sphere, has only
triangular faces, and is flexible. Thus, statement (i) of Theorem 1 is
fulfilled for $\mathscr{P}$.

In § 7 a special flex $\mathscr{P}(t)$ of $\mathscr{P}$ is constructed.
According to Lemmas 5, 6 and 7, there is a sufficiently small neighborhood of
0 in $\mathbb{R}$ in which none of the dihedral angles of $\mathscr{P}(t)$ is
constant as a function of $t$. Reducing the range of $t$ to the segment
$[0,\beta)$ completely contained in this neighborhood, we obtain the family of
polyhedra $\\{\mathscr{P}(t)\\}_{t\in[0,\beta)}$, whose existence proves
statement (ii) of Theorem 1. ∎

Let us formulate two open Problems 2 and 3, closely related to Problem 1.

######  Problem 2.

Is there a one-parametric flexible polyhedron in $\mathbb{R}^{3}$, without
boundary and without self-intersections, for which all dihedral angles change
during the flex?

The concept of a one-parametric flexible polyhedron is intuitively obvious.
This is exactly how it was used in [13]. The definition of a $p$-parametric
flexible polyhedron requires clarification. It is strictly formulated in [24].
For our purposes, it is sufficient to have an intuitive understanding that
flexible Steffen polyhedron $\mathscr{S}$ is one-parametric, and flexible
polyhedron $\mathscr{P}$ constructed in § 6 is 4-parametric.

Problem 2 seems interesting to us due to the fact that in mathematics there
are many situations where the behavior of an object changes greatly depending
on whether it depends on one parameter or on several.

It is obvious that the statement “small diagonal $AB$ of a polyhedron
$\mathscr{R}$ does not change during the flex” is equivalent to the statement
“the dihedral angle at the common edge of those two faces of $\mathscr{R}$
which contain $A$ and $B$ does not change during the flex”. Therefore, the
following Problem 3 is a generalization of Problem 1.

######  Problem 3.

Is there a flexible polyhedron in $\mathbb{R}^{3}$, without boundary and
without self-intersections, for which all diagonals change during the flex?

Examples show that Problem 3 is non-trivial. Indeed, Bricard octahedra have
self-intersections and have no diagonals whose lengths do not change during
the flex (see Lemma 1). On the other hand, Steffen flexible polyhedron
$\mathscr{S}$, modified Steffen flexible polyhedron $\mathscr{M}$, and
flexible polyhedron $\mathscr{P}$ constructed in § 6 have no self-
intersections, but have diagonals whose lengths do not change during the flex
(namely, $T_{2}T_{3}$ for $\mathscr{S}$ and $\mathscr{M}$; and, for example,
$T_{1}T_{4}$ and $T_{2}T_{3}$ for $\mathscr{P}$).

An additional interest to Problem 3 is added by the fact that, unlike Problems
1 and 2, it makes sense not only for orientable polyhedra. Note also that
Problem 3 can be considered as a special case of the problem of which lengths
of diagonals of a flexible polyhedron must be fixed so that it ceases to be
flexible. Various aspects of the latter problem have been studied, for
example, in [24], [25], [26].

Note that analogues of Problems 1– 3 can be posed not only in
$\mathbb{R}^{3}$, but also in any space of constant curvature of dimension
$\geq 3$, as well as in spaces with an indefinite metric.

In conclusion, let us clarify that all symbolic calculations performed in the
preparation of this article were performed using the computer software system
Mathematica 12.1 [15], license 3322–8225.

### Acknowledgments

The authors thank N.P. Dolbilin, A.A. Gaifullin and I.Kh. Sabitov for their
interest in this study and useful discussions.

### Funding

The work was carried out within the State Task to the Sobolev Institute of
Mathematics (namely, V.A. Alexandrov was supported by Project FWNF–2022–0006
and E.P. Volokitin was supported by Project FWNF–2022–0005).

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