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This article is cited in 1 scientific paper (total in 1 paper) A metric description of flexible octahedra S. N. Mikhalev Faculty of Mechanics and Mathematics, Lomonosov Moscow State University, Moscow, Russia
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     § 1. IntroductionFlexible polyhedra are some of the central objects in the metric theory of polyhedra. The first examples of nontrivially flexible polyhedra in $\mathbb{R}^3$ were constructed by Bricard in [1], together with a classification of flexible polyhedra that are combinatorially equivalent to a regular octahedron. Bricard identified three types of flexible octahedra, the first two of which have a simple description, while the third type is rather complicated. The simplicity of the object under investigation (an octahedron has only six vertices!), in combination with the nontrivial nature of the result (both in terms of proof and the description of objects) prompted other researchers to find new ways to ‘reproduce’ Bricard’s results and interpret them; we mention the works by Bennett [2], Lebesgue [3], Stachel [4] and the very recent work [5]. In [6] Bricard’s octahedra arose as a special case of a general result for spaces of arbitrary dimension. Nevertheless, there is a class of problems for which the known descriptions of Bricard octahedra are unsuitable in a certain sense. The development of the metric geometry of polyhedra received a new powerful impetus in the mid-1990s, with Sabitov’s proof of the bellows conjecture on the invariance of the generalised volume of a flexible polyhedron (see [7]–[9]). The method used in that proof is based on the application of the Cayley-Menger relations, and it can be referred to as ‘distance geometry’. It was further developed in [10] and [11] and can potentially be used for the solution of a wide range of problems in the metric geometry of polyhedra. However, in this approach, all metric characteristics of polyhedra must be specified in terms of distances (the lengths of edges and diagonals). Namely, Sabitov’s central result is that for any polyhedron $P$ the quantity $V$ — the square of its generalized volume — is a root of a certain polynomial $Q(V)$ with leading coefficient one whose other coefficients $a_k(l)$ are polynomials in the squares of edge lengths. The coefficients of these polynomials $a_k(l)$ are determined by the combinatorial structure of the polyhedron $P$ alone. The proof of Sabitov’s theorem is constructive: it provides a method for presenting the polynomial $Q(V)$ explicitly. However, this method produces a polynomial of large degree even for polyhedra with a small number of vertices. Therefore, for practical calculations new special methods for constructing polynomials with prescribed properties are required. As shown in [12], in the case of an octahedron (meaning a polyhedron combinatorially equivalent to a regular octahedron, but having arbitrary edge lengths), the minimum possible degree of the polynomial $Q(V)$ is 8, hence However, even in this case, for an arbitrary octahedron (when all edges are denoted by different characters) the polynomial $Q(V)$ contains millions of terms. The work [12] does not present $Q(V)$ itself, but only an algorithm for constructing a matrix whose elements are polynomials in $V$ and edge lengths, with the determinant equal to the polynomial $Q(V)$.On the other hand, if we consider octahedra whose edge lengths satisfy certain relations, then there can be simplifications when we substitute these relations into the expressions for the $a_k(l)$. For example, for Bricard octahedra of the first type, which are characterised by the pairwise equality of opposite edges (six pairs of equalities in total), it turns out that the coefficient $a_7(l)$ contains only 22 terms (in six variables) and $a_k(l)= 0$ for $k<7$. This result was also obtained in [12]. Subsequently, in [13] polynomials $Q(V)$ were found for all cases of metrically symmetric octahedra, including Bricard octahedra of the first two types. Nevertheless, none of the known descriptions enables one to calculate the coefficients $a_k(l)$ of the polynomial $Q(V)$ for Bricard octahedra of the third type. Indeed, for such a calculation it is necessary to have a set of relations characterising a type 3 octahedron, in which edges are expressed in terms of some parameters, in particular in terms of other edges. At the same time, all the known descriptions of type 3 octahedra are largely geometric. There are easily written equations which hold for the edges, but there is no ‘parametrisation’ mentioned above. Without relying on the well-known results of Bricard and his followers, in this work we find a new description of all possible flexible octahedra in a convenient metric notation (Theorems 2–6), obtaining simultaneously conditions for the reducibility of the Cayley-Menger polynomial (for five points), which are also of independent interest (Theorem 1). In § 5 we compare our method with known results. § 2. Definitions and notationConsider a metric simplicial complex $K$ which is combinatorially equivalent to a regular octahedron, with strictly positive numbers (edge lengths) assigned to the edges, as indicated in Figure 1. The edge lengths are assumed to satisfy the strict triangle inequalities on faces. By a nontrivial flexing we mean a continuous family of isometric realisations $P_t\colon K\to \mathbb{R}^3$ such that all the three diagonals of the octahedron vary continuously with $t$ (a diagonal is a line segment connecting two vertices that are not connected by an edge). Our definition of a nontrivial flexing differs from the classical one, in which one assumes that at least one diagonal changes. We thereby exclude from consideration the following known cases of the realisation of the complex $K$: 1) in the form of a doubly covered tetrahedral angle; 2) in the form of a pair of edge-adjacent facets, on each of which three other facets are superimposed. The well-known Gluck theorem [14] implies that in the general case any realisation of the complex $K$ is inflexible, and flexibility is possible only if certain conditions on the metric (edge lengths) of $K$ are met. We find conditions on edge lengths which are necessary for the existence of an isometric realisation of the complex $K$ in the form of a nontrivially flexible octahedron. On the basis of these conditions, we will specify classes of flexible octahedra. In what follows we often use the same notation for metric complexes and their realisations in $\mathbb{R}^3$. In addition, we accept the convention that the same lowercase character denotes both the length of a line segment (a positive number) and the segment itself as a geometric object. Further, two or more consecutive lowercase characters denote the corresponding one-dimensional metric complex (or its isometric image). For example, $pcq$ denotes the triangle $v_1v_2v_3$, and $cfhg$ denotes the edge cycle $v_2v_3v_4v_5$. We denote the squared lengths of the diagonals $v_2v_4$, $v_3v_5$ and $v_1v_6$ by $x$, $y$ and $z$, respectively (we also use the same letters to denote the diagonals themselves as geometric objects). We assume that the edges of the octahedron are known (altogether they determine the metric of the octahedron; their set is denoted by $l$), and the diagonals $x$, $y$ and $z$ are indeterminates. An equator of the octahedron is a four-edge cycle in which no two edges are incident to the same facet. The tetrahedral angle with vertex $v_j$ is the star of $v_j$, viewed as a metric simplicial complex $K_j\subset K$. A diagonal of an arbitrary complex is a line segment connecting two vertices that are not connected by an edge. According to these definitions, the octahedron has three diagonals, three equators (each of which has two diagonals) and six tetrahedral angles (each of which also has two diagonals). Each tetrahedral angle ‘rests’ on an equator, which is its boundary. We denote the edge lengths of an equator in cyclic order by $a$, $b$, $c$ and $d$. We call an equator metrically symmetric if at least one of the following three conditions is satisfied: $a=c$, $b=d$, or $a=b$, $c=d$, or $a=d$, $b=c$. We say that an equator has a zero sum if An equator has a zero sum if and only if at least one expression in brackets in (2.1) is equal to zero. This means that the sum of the lengths of some two edges in the equator is equal to the sum of the lengths of its other two edges. Note that an equator in which the sum of the lengths of three edges is equal to the length of the fourth edge can only be realised in the form of a ‘double-covered segment’; this means that the diagonals of such an equator cannot change in flexing (so the flexing must be trivial). A metrically symmetric equator always has a zero sum; the converse in not true in general. Definition 1. We introduce the classes of equators with edge lengths satisfying certain relations in accordance with Table 1. For example, $sqad$ belongs to the class $S_0$ if and only if $s-q+a-d=0$. Table 1
The classes denoted by the letter $S$ (with an index) consist of equators with a zero sum. We therefore denote by $S$ (without an index) the whole class of equators with a zero sum. Similarly, the classes denoted by the letter $M$ (with an index) consist of metrically symmetric equators, and we denote by $M$ (without an index) the whole class of metrically symmetric equators. Lemma 1. The following equalities hold: $S\cap R_0=M_y\cup M_z$ (for the equator $sqad$), $S\cap R_0=M_x\cup M_z$ (for the equator $bpre$) and $S\cap R_0=M_x\cup M_y $ (for the equator $cfhg$). Proof. Consider, for example, the case of the equator $sqad$. This equator belongs to the class $S$ if and only if $sqad\in S_0 \cup S_y \cup S_z$. It can be checked directly that $S_0\cap R_0=M_y\cup M_z$, $S_y\cap R_0=M_y$ and $S_z\cap R_0=M_z$, which implies the first statement of the lemma. The remaining two cases are proven similarly. Lemma 2. The following equalities hold: $M_y\cap M_z=M_e$ (for the equator $sqad$), $M_x\cap M_z=M_e$ (for the equator $bpre$) and $M_x\cap M_y=M_e$ (for the equator $cfhg$). This is also proved by a simple check. Definition 2. If the equators $sqad$, $bpre$ and $cfhg$ of an octahedron $K$ belong to the classes $\mathcal{M}_1$, $\mathcal{M}_2$ and $\mathcal{M}_3$, respectively, then we say that $K$ belongs to the class $[\mathcal{M}_1, \mathcal{M}_2, \mathcal{M}_3]$. In the case when all equators of the octahedron are metrically symmetric (so that each equator belongs to the class $M_i$, where $i$ is one of the symbols $x$, $y$, $z$, $0$ or $e$), then we use the notation $M_{ijk}$ for the metric class of the octahedron, alongside with $[M_i,M_j,M_k]$. We also introduce the classes $A_1$ and $A_2$ corresponding to flexible Bricard octahedra of types 1 and 2. § 3. Reducibility conditions for the Cayley-Menger polynomialIt is known (see [15]) that the pairwise distances $d_{ij}=d_{ji}$, $1\leqslant i,j\leqslant 5$, between five points in $\mathbb{R}^3$ satisfy the condition
$$
\begin{equation}
\begin{vmatrix} 0& 1& 1& 1& 1& 1\\ 1& 0& d_{12}^2& d_{13}^2& d_{14}^2& d_{15}^2\\ 1& d_{21}^2& 0& d_{23}^2& d_{24}^2& d_{25}^2\\ 1& d_{31}^2& d_{32}^2& 0& d_{34}^2& d_{35}^2\\ 1& d_{41}^2& d_{42}^2& d_{43}^2& 0& d_{45}^2\\ 1& d_{51}^2& d_{52}^2& d_{53}^2& d_{54}^2& 0 \end{vmatrix}=0,
\end{equation}
\tag{3.1}
$$
known as the Cayley-Menger equation. The left-hand side is called the Cayley-Menger determinant or the Cayley-Menger polynomial. Condition (3.1) is necessary for the numbers $d_{ij}$ to be the pairwise distances for some five points. This condition is not sufficient; in order to guarantee the existence of the corresponding five points, it is necessary to impose some additional conditions on the $d_{ij}$. Consider the tetrahedral angle $K_1$ (see Figure 1, (b)). Consider the Cayley-Menger polynomial multiplied by $-1$ for the distances between its vertices as a polynomial in $x$ and $y$:
$$
\begin{equation}
\begin{aligned} \, Q(x,y) &=x^2y^2-2(s^2+q^2)x^2y-2(p^2+r^2)x y^2 \nonumber \\ &\qquad+(s^2-q^2)^2x^2+(p^2-r^2)^2y^2+\dotsb. \end{aligned}
\end{equation}
\tag{3.2}
$$
(Sometimes, as here, we omit some terms and replace them by dots for readability and to save space.) Note that, unlike closed surfaces, which are almost always inflexible, any tetrahedral angle admits a nontrivial flexing, except in some obvious degenerate cases. The Cayley-Menger condition can be interpreted as an equation describing flexing of the tetrahedral angle $K_1$, whose shape is determined by the diagonals $x$ and $y$. This explains the fact that the Cayley-Menger polynomial has degree $2$ in each of the variables $x$ and $y$: for each value of one diagonal there are two possible values of the other (in nondegenerate cases). Namely, the pair of facets $pqc$ and $psg$ can be reflected in the plane $qsy$, and the pair $pqc$ and $qrf$ can be reflected in the plane $prx$. We are interested in the reducibility of the Cayley-Menger polynomial in $\mathbb{R}[x,y]$, that is, the existence of a nontrivial decomposition of this polynomial into rational factors with respect to $x$ and $y$. Such a factorization exists when the edge lengths satisfy certain relations. Our immediate goal is to find these relations and the corresponding factorizations of the Cayley-Menger polynomial. It is worth noting that in the general case the Cayley-Menger determinant is absolutely irreducible (see [15]). We denote the values of the plane angles $pq$, $qr$, $rs$ and $sp$ by $\alpha_1$, $\alpha_2$, $\alpha_3$ and $\alpha_4$, respectively (see Figure 1, (b)). The strict triangle inequalities on the facets imply that $0<\alpha_k<\pi$. We divide the set of all tetrahedral angles into pairwise disjoint classes $\mathrm{I}$, $\mathrm{II}_x$, $\mathrm{II}_y$ and $\mathrm{III}$ as follows. Definition 4. Let The values $\alpha_k$ and $\pi-\alpha_k$ are between 0 and $\pi$, so each angle is uniquely determined by its cosine. For each equality in Definition 4 we can use the cosine theorem to obtain an equivalent formula in terms of edge lengths (signs in formulae correspond to sings in the notation for the corresponding class). Lemma 3. The following statements hold: Additional necessary conditions for the classes introduced above can be obtained from elementary geometric considerations. We denote by $S_1$, $S_2$, $S_3$ and $S_4$ the areas of the triangles $pqc$, $qrf$, $rsh$ and $psg$ (see Figure 1, (b)). We have $S_k>0$ for $k=1,2,3,4$, since the facets are nondegenerate. Lemma 4. The following statements hold: The proof follows by passing from equalities of angles to equalities of the sines of these angles. Theorem 1. Suppose that the tetrahedral angle $K_1$ admits a nontrivial flexing. Then precisely one of the following statements is true: (1) $K_1\in \mathrm{I}$ and the polynomial $Q(x,y)$ is irreducible; (2) $K_1\in \mathrm{II}_x$, the polynomial $Q(x,y)$ is reducible, and for $K_1\in \mathrm{II}_x^{\pm}$ the factorization of $Q(x ,y)$ into irreducible factors has the form
$$
\begin{equation}
\begin{aligned} \, \notag &Q(x,y) =[y-(q\mp s)^2]\cdot\biggl[x^2y-(s\pm q)^2x^2-2(p^2+r^2)xy+(p^2-r^2)^2y \\ \notag &\qquad+\biggl(2(p^2+r^2)(q^2+s^2)\pm 4\biggl(sq(g^2+h^2)-s^3q+ (h^2-r^2)(p^2-g^2)\frac{q}{s}\biggr)\biggr)x \\ \notag &\qquad \pm 2\biggl((p^2-r^2)(p^2-r^2-2g^2+2h^2)qs +2(r^2g^2-p^2h^2)(p^2-r^2-g^2+h^2)\frac{q}{s}\biggr) \\ &\qquad -(p^2-r^2)^2(q^2+s^2)\biggr]; \end{aligned}
\end{equation}
\tag{3.12}
$$
(3) $K_1\in \mathrm{II}_y$, the polynomial $Q(x,y)$ is reducible, and for $K_1\in \mathrm{II}_y^{\pm}$ the factorization of $Q(x,y)$ into irreducible factors has the form
$$
\begin{equation}
\begin{aligned} \, \notag &Q(x,y) =[x-(p\mp r)^2]\cdot\biggl[y^2x-(p\pm r)^2y^2-2(s^2+q^2)xy+(s^2-q^2)^2x \\ \notag &\qquad+\biggl(2(p^2+r^2)(q^2+s^2)\pm 4\biggl( pr(c^2+g^2)-p^3r+ (g^2-s^2)(q^2-c^2)\frac{r}{p}\biggr)\biggr)y \\ \notag &\qquad\pm 2\biggl((q^2-s^2)(q^2-s^2-2c^2+2g^2)pr +2(c^2s^2-q^2g^2)(q^2-s^2-c^2+g^2)\frac{r}{p}\biggr) \\ &\qquad-(q^2-s^2)^2(p^2+r^2)\biggr]; \end{aligned}
\end{equation}
\tag{3.13}
$$
(4) $K_1\in \mathrm{III}$, the polynomial $Q(x,y)$ is reducible, and for $K_1\in \mathrm{III}^{\pm}$
$$
\begin{equation}
\begin{aligned} \, \notag &Q(x,y) =\biggl[xy-(s\pm q)^2x-(p+r)^2y+(p+r)^2(s^2+q^2) \\ \notag &\qquad\mp\frac{1}{prs}\bigl((p+r)qs^2(s^2p+s^2r-2ph^2-2g^2r) +(r^2p+p^2r-g^2r-ph^2)^2q\bigr)\biggr] \\ \notag &\qquad\times\biggl[xy-(s\mp q)^2x-(p-r)^2y+(p-r)^2(s^2+q^2) \\ &\qquad\mp\frac{1}{prs}\bigl((p-r)qs^2(s^2p-s^2r-2ph^2+2g^2r) +(r^2p-p^2r+g^2r-ph^2)^2q\bigr)\biggr]. \end{aligned}
\end{equation}
\tag{3.14}
$$
Remark 1. The four statements in Theorem 1 are mutually exclusive. Remark 2. In (3.14) the irreducibility of the factors is not guaranteed. Proof of Theorem 1. A simple check shows that substituting (3.6) into (3.2) gives (3.12). In a similar way, substituting (3.7) into (3.2) gives (3.13), and substituting (3.8) into (3.2) gives (3.14).
Further, the discriminant $\Delta(x)$ of the quadratic polynomial (3.2) with respect to the variable $y$ has the form where
$$
\begin{equation*}
\begin{aligned} \, P_1(x) &=s^2x^2+(-g^2r^2+s^4-s^2h^2-p^2s^2+p^2r^2-p^2h^2+g^2h^2-g^2s^2-s^2r^2)x \\ &\qquad+p^4h^2-p^2h^2g^2+r^2h^2s^2-p^2h^2s^2+r^4g^2-p^2r^2h^2+p^2g^2s^2+p^2h^4 \\ &\qquad-p^2g^2r^2+r^2g^4-r^2g^2h^2-r^2g^2s^2 \end{aligned}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\begin{aligned} \, P_2(x) &=q^2x^2+(-r^2q^2-q^2f^2-r^2c^2-p^2f^2+c^2f^2+q^4-c^2q^2 -q^2p^2+p^2r^2)x \\ &\qquad+r^4c^2-r^2c^2p^2-f^2r^2c^2-p^2f^2c^2-p^2f^2r^2-c^2q^2r^2-p^2f^2q^2+r^2c^4 \\ &\qquad+q^2f^2r^2+p^4f^2+p^2f^4+q^2c^2p^2. \end{aligned}
\end{equation*}
\notag
$$
The discriminant of $P_1(x)$ is $S_3^2 S_4^2>0$, and the discriminant of $P_2(x)$ is $S_1^2 S_2^2>0$. The leading coefficients of $P_1(x)$ and $P_2(x)$ are positive. Therefore, $\Delta(x)$ is a perfect square if and only if the coefficients of $P_1(x)$ and $P_2(x)$ are proportional, which is equivalent to the identity $q^2P_1( x)-s^2P_2(x)=0$ for all $x$. Equivalently,
$$
\begin{equation}
\begin{cases} (s^2q^2-f^2s^2+s^2r^2)c^2-q^2p^2h^2-p^2r^2s^2-q^2g^2s^2+q^2p^2r^2+f^2q^2s^2 \\ \quad-\,q^2h^2s^2-q^2g^2r^2+p^2f^2s^2+q^2h^2g^2+q^2s^4-s^2q^4=0, \\ r^2s^2c^4+(-p^2f^2s^2-p^2r^2s^2-s^2q^2r^2-f^2s^2r^2+s^2q^2p^2+r^4s^2)c^2 \\ \quad+\,r^2q^2p^2h^2+r^2q^2g^2s^2-q^2p^2g^2s^2+r^2q^2g^2p^2+g^2q^2p^2h^2 \\ \quad+\,s^2q^2p^2h^2-r^2q^2g^4-p^2f^2s^2r^2+q^2h^2g^2r^2-r^4q^2g^2-q^2p^4h^2 \\ \quad+\,p^4f^2s^2-q^2h^2s^2r^2+s^2r^2q^2f^2-s^2p^2f^2q^2-q^2p^2h^4+p^2f^4s^2=0. \end{cases}
\end{equation}
\tag{3.16}
$$
We resolve the system (3.16) with respect to $c^2$ and $f^2$. It is easy to check that this system has precisely four solutions, the pairs $(c^2,f^2)$ from (3.6) and (3.8). A similar argument shows that $\Delta(y)$ is a perfect square if and only if the conditions defining one of the four pairs (3.7) and (3.8) are satisfied. Suppose there is a nontrivial factorization $Q(x,y)=Q_1(x,y)\cdot Q_2(x,y)$. Since the leading coefficient in $Q(x,y)$ is nonzero, one of the following cases occurs: (A) one of the factors is a square trinomial in $x$, and the other is a square trinomial in $y$; (B) both $Q_1$ and $Q_2$ have degree $1$ with respect to $y$; or equivalently, the zeros of $Q(x,y)$ as a quadratic trinomial in $y$ are rational functions of $x$, which in turn is equivalent to the discriminant of $\Delta(x)$ being a perfect square; (C) both $Q_1$ and $Q_2$ have degree $1$ with respect to $x$, or equivalently, the zeros of $Q(x,y)$ as a quadratic trinomial in $x$ are rational functions of $y$, which in turn is equivalent to the discriminant of $\Delta(y)$ being a perfect square. If none of conditions (B) and (C) is satisfied, then (A) holds. However, then nontrivial flexing is impossible for the fixed edge lengths (each diagonal can take no more than two fixed values), which contradicts the assumption of the theorem. Suppose (B) is satisfied, but (C) is not. Then (3.6) holds for one of the signs, while (3.7) and (3.8) do not hold for any signs. Therefore,
$$
\begin{equation}
\begin{cases} \alpha_1=\alpha_4, \\ \alpha_2=\alpha_3 \end{cases} \quad\text{or}\qquad \begin{cases} \alpha_1=\pi-\alpha_4, \\ \alpha_2=\pi-\alpha_3 \end{cases}
\end{equation}
\tag{3.17}
$$
and for none of the systems in (3.4) and (3.5) the equalities in it are satisfied simultaneously. In particular, at least one of the equalities $\alpha_1=\alpha_3$ and $\alpha_2=\alpha_4$ does not hold, and at least one of the equalities $\alpha_1=\pi-\alpha_3$ and $\alpha_2=\pi-\alpha_4$ does not hold. Taking this into account, identities (3.17) imply that
$$
\begin{equation}
\begin{cases} \alpha_1=\alpha_4, \\ \alpha_2=\alpha_3, \\ \alpha_1\ne\alpha_3, \\ \alpha_1\ne\pi-\alpha_3 \end{cases} \quad\text{or}\qquad \begin{cases} \alpha_1=\pi-\alpha_4, \\ \alpha_2=\pi-\alpha_3, \\ \alpha_1\ne\alpha_3, \\ \alpha_1\ne\pi-\alpha_3. \end{cases}
\end{equation}
\tag{3.18}
$$
Hence $K_1\in \mathrm{II}_x$. In addition, the factor of degree $3$ in (3.12) is irreducible; otherwise the roots $x$ of this quadratic trinomial would be rational functions of $y$, which is not the case. Now suppose that (C) is satisfied, but (B) is not. Then (3.7) holds for one of the signs, while (3.6) and (3.8) do not hold for any signs. Arguing in a similar way we find that $K_1\in \mathrm{II}_y$ and the factor of degree $3$ in (3.13) is irreducible. Finally, let both (B) and (C) be satisfied. This can occur in two cases: 1) condition (3.8) is satisfied for one of the signs; then $K_1\in \mathrm{III}$ and (3.14) holds; 2) both (3.6) and (3.7) hold for some sings. Then
$$
\begin{equation}
\begin{cases} \alpha_1=\alpha_4, \\ \alpha_2=\alpha_3 \end{cases} \quad\text{or}\qquad \begin{cases} \alpha_1=\pi-\alpha_4, \\ \alpha_2=\pi-\alpha_3, \end{cases}
\end{equation}
\tag{3.19}
$$
and
$$
\begin{equation*}
\begin{cases} \alpha_1=\alpha_2, \\ \alpha_3=\alpha_4 \end{cases} \quad\text{or}\qquad \begin{cases} \alpha_1=\pi-\alpha_2, \\ \alpha_3=\pi-\alpha_4. \end{cases}
\end{equation*}
\tag{3.20}
$$
This implies again that $K_1\in \mathrm{III}^{+}$ or $K_1\in \mathrm{III}^{-}$. Therefore, (3.8) holds for some sign, which means that (3.14) is satisfied. Thus, we have proved that the reducibility of $Q(x,y)$ is equivalent to the validity of one of statements (2)–(4). We conclude that the class $\mathrm{I}$ consists of those and only those tetrahedral angles for which $Q(x,y)$ is irreducible. This completes the proof. Remark 3. The quantities $c$ and $f$ were chosen as indeterminates with respect to which the system was to be solved. If we choose $g$ and $h$ (which correspond to the edges on the other side of the diagonal $x$) instead, then we arrive at the same linear relations (3.6) between $c^2$ and $g^2$ and between $f^2$ and $h^2$; the only difference is that they will be solved with respect to $g^2$ and $h^2$. Remark 4. Each of the two factors in the decomposition of $Q(x,y)$ corresponds to a continuous family of positions of the tetrahedral angle $K_1$ in the space. For example, it can be a convex tetrahedral angle (the flexing corresponding to one of the factors), or a self-intersecting tetrahedral angle obtained from the original one by reflecting two facets in the ‘diagonal’ plane (the flexing corresponding to the second factor). The linear factor depends on just one variable, $x$ or $y$, according to Theorem 1, and it corresponds to a trivial flexing. That is, one of the factors in (3.12) has the form $y-y_0$. Here the edge lengths are such that the reflected facets are superimposed onto the other two facets, and one vertex of the equator falls on the edge opposite to it (this is possible because the plane angles of adjacent facets are equal). Such a figure with self-impositions is flexed in a trivial way: the diagonal $y$ does not change under flexing. Therefore, we are only interested in the configuration corresponding to the factor of degree $3$. For uniformity of the wording we consider sometimes only the factor of degree $3$ instead of the whole Cayley-Menger polynomial. Remark 5. In all cases the irreducibility of a polynomial in $\mathbb{R}[x,y]$ implies its irreducibility in $\mathbb{C}[x,y]$. This follows from the representation of the discriminant $\Delta(x)$ in the form (3.15) and a similar representation for $\Delta(y)$. Indeed, if $Q(x,y)$ were irreducible over $\mathbb{R}$, but reducible over $\mathbb{C}$, then this would mean that either $\Delta(x)=-P^2(x)$ for $P(x)\in\mathbb{R}[x]$ or $\Delta( y)=-\widetilde{P}^2(y)$. The first case is impossible since the leading coefficient of $x$ in $\Delta(x)$ is $q^2s^2>0$. The second case is also impossible for a similar reason. § 4. Conditions for the flexibility of an octahedronIn what follows we assume that we are given a complex $K$ (see Figure 1, (a)) isometrically realised in $\mathbb{R}^3$ as a flexible octahedron, which we also denote by $K$. Consider an equator of $K$, for example, $cfhg$. It partitions $K$ into two tetrahedral angles, $K_1$ and $K_6$. We refer to these tetrahedral angles, quintuples of points and their Cayley-Menger polynomials as corresponding to the equator in question. Each of the Cayley-Menger conditions corresponding to $cfhg$ defines an algebraic variety on the plane with coordinates $x$ and $y$. The existence of a nontrivial flexing of the octahedron implies that these two varieties intersect in a nontrivial algebraic variety (not reducing to a discrete set of points). For any $f,g\in\mathbb{C}[x,y]$ such that $g$ is not divisible by $f$, the irreducibility of $f$ implies that the set of common zeros of $f$ and $g$ is finite (this is a simple algebraic fact). It follows that either both Cayley-Menger equations are irreducible (and then their coefficients must be pairwise equal), or both are reducible (and then they contain irreducible factors of the same degree and with proportional coefficients). Note also that these irreducible factors must have total degree $2$ or $3$ in $x$ and $y$. Indeed, a linear polynomial in one variable corresponds to a trivial flexing, and a linear polynomial in two variables cannot appear due to the fact that the leading coefficient in (3.2) is equal to $1$; the complete set of possible cases of reducibility is described in Theorem 1. A similar argument works for the other two equators. It follows that for each equator the corresponding tetrahedral angles belong to the same class ($\mathrm{I}$, $\mathrm{II}_x$, $\mathrm{II}_y$ or $\mathrm{III}$). The class of an equator is the class of the tetrahedral angles corresponding to it. Each equator belongs to one of the classes ($\mathrm{I}$, $\mathrm{II}_x$, $\mathrm{II}_y$ or $\mathrm{III}$) independently of the others. We will classify flexible octahedra on the basis of the complete list of possible cases produced by Theorems 2–5 below. Recall that the classes corresponding to Bricard octahedra of types 1 and 2 are defined as $A_1=M_{000}$ and $A_2=M_{0xx}\cup M_{y0y}\cup M_{zz0}$ (see Definition 3). It turns out that the metric symmetry of all three equators implies that the octahedron belongs to (at least) one of the classes $A_1$ or $A_2$: Lemma 5. If all the three equators of a nontrivially flexible octahedron $K$ are metrically symmetric, then $K\in A_1\cup A_2$. Proof. We write out the Cayley-Menger polynomials for the tetrahedral angles $K_1,\dots,K_6$:
$$
\begin{equation*}
\begin{aligned} \, Q_1(x,y) &=x^2y^2-2(s^2+q^2)x^2y-2(p^2+r^2)x y^2 \\ &\qquad+(s^2-q^2)^2x^2+(p^2-r^2)^2y^2+\dotsb , \\ Q_6(x,y) &=x^2y^2-2(a^2+d^2)x^2y-2(b^2+e^2)x y^2 \\ &\qquad+(a^2-d^2)^2x^2+(b^2-e^2)^2y^2+\dotsb , \\ Q_3(x,z) &=x^2z^2-2(a^2+q^2)x^2z-2(c^2+f^2)x z^2 \\ &\qquad+(a^2-q^2)^2x^2+(c^2-f^2)^2z^2+\dotsb , \\ Q_5(x,z) &=x^2z^2-2(s^2+d^2)x^2z-2(g^2+h^2)x z^2 \\ &\qquad+(s^2-d^2)^2x^2+(g^2-h^2)^2z^2+\dotsb , \\ Q_2(y,z) &=y^2z^2-2(b^2+p^2)y^2z-2(c^2+g^2)y z^2 \\ &\qquad+(b^2-p^2)^2y^2+(c^2-g^2)^2z^2+\dotsb , \\ Q_4(y,z) &=y^2z^2-2(e^2+r^2)y^2z-2(f^2+h^2)y z^2 \\ &\qquad+(e^2-r^2)^2y^2+(f^2-h^2)^2z^2+\dotsb. \end{aligned}
\end{equation*}
\notag
$$
By assumption $K\in M_{ijk}$, where $i\in\{0,y,z\}$, $j\in\{0,x,z\}$ and $k\in\{0 ,x,y\}$, so there are $27=3^3$ possible cases. In some of these cases there is nothing to prove (for example, if $i=0$, $j=0$ and $k=0$, then $K\in M_{000}=A_1$). We give the proof in the meaningful cases with $i=0$; the remaining cases are analysed similarly. 1) $K\in M_{00x}$. Subtracting $Q_6(x,y)=0$ from $Q_1(x,y)=0$ and making the substitution
$$
\begin{equation*}
s=a, \quad q=d, \quad b=r, \quad p=e, \quad c=g, \quad f=h,
\end{equation*}
\notag
$$
which corresponds to the class $M_{00x}$ according to Definition 1, we obtain the equation It must hold for an infinite family of pairs $(x,y)$ corresponding to a flexing. This is possible only if $e=r$ (so that $bpre\in M_e$) or $g=h$ (so that $cfhg \in M_e$). Since $M_{0ex}\subset M_{0xx}$ and $M_{00e}\subset M_{000}$, we obtain $K\in A_1\cup A_2$.
2) $K\in M_{00y}$. Subtracting $Q_6(x,y)=0$ from $Q_1(x,y)=0$, we obtain Hence $a=d$ (so that $sqad\in M_e$) or $g=f$ (so that $cfhg\in M_e$). Since $M_{e0y}\subset M_{y0y}$ and $M_{00e}\subset M_{000}$, we obtain $K\in A_1\cup A_2$.3) $K\in M_{0x0}$. Subtracting $Q_5(x,y)=0$ from $Q_3(x,y)=0$, we obtain Hence $p=r$ (so that $bpre\in M_e$) or $g=h$ (so that $cfhg\in M_e$). Since $M_{0e0}\subset M_{000}$ and $M_{0xe}\subset M_{0xx}$, we obtain $K\in A_1\cup A_2$.4) $K\in M_{0z0}$. Subtracting $Q_5(x,y)=0$ from $Q_3(x,y)=0$ we obtain Hence $e=r$ (so that $bpre\in M_e$) or $d=a$ (so that $sqad\in M_e$). Since $M_{0e0}\subset M_{000}$ and $M_{ez0}\subset M_{zz0}$, we obtain $K\in A_1\cup A_2$.5) $K\in M_{0xy}$. Subtracting $Q_6(x,y)=0$ from $Q_1(x,y)=0$ we obtain Hence $g=f$ (so that $K\in M_{0xe}\subset M_{0xx}$) or $a=d$ (so that $K\in M_{exy}$, but this does not mean that $K\in A_1\cup A_2$ yet). We make the substitution corresponding to the class $M_{exy}$, namely,
$$
\begin{equation*}
s=d, \quad q=d, \quad a=d, \quad b=p, \quad e=r, \quad c=f\quad\text{and} \quad h=g
\end{equation*}
\notag
$$
in the equation $Q_2(y,z)=Q_4(y,z)$. Then we obtain Hence $p=r$ (and then $K\in M_{eey}\subset M_{y0y}$).
6) $K\in M_{0zx}$. Subtracting $Q_5(x,y)=0$ from $Q_3(x,y)=0$ we obtain Hence $e=r$ (so that $K\in M_{0ex}\subset M_{0xx}$) or $a=d$ (so that $K\in M_{ezx}$). Making the substitution corresponding to the class $M_{ezx}$ in the equation $Q_2(y,z)=Q_4(y,z)$, we obtain Hence $g=h$ (so that $K\in M_{eze}\subset M_{zz0}$).7) $K\in M_{0zy}$. Subtracting $Q_5(x,y)=0$ from $Q_3(x,y)=0$ we obtain Hence $g=f$ and $e=r$ (so that $K\in M_{0ee}\subset M_{000}$) or $g=f$ and $a=d$ (so that $K\in M_{eze}\subset M_{zz0}$).Theorem 2. If at least two equators belong to the class $\mathrm{I}$, then the octahedron belongs to the class $A_1$ or $A_2$. Proof. We can assume without loss of generality that $cfhg, bpre\in \mathrm{I}$.
The equator $cfhg$ corresponds to the polynomials $Q_1(x,y)$ and $Q_6(x,y)$. They are irreducible by assumption, so their coefficients are proportional. We obtain four conditions, which are conveniently arranged into two systems of equations:
$$
\begin{equation*}
\begin{cases} s^2+q^2=a^2+d^2, \\ (s^2-q^2)^2=(a^2-d^2)^2 \end{cases}\quad\text{and}\qquad \begin{cases} p^2+r^2=b^2+e^2, \\ (p^2-r^2)^2=(b^2-e^2)^2. \end{cases}
\end{equation*}
\notag
$$
From these systems we obtain
$$
\begin{equation*}
sqad\in M_0\cup M_y\quad\text{and} \quad bpre\in M_0\cup M_x.
\end{equation*}
\notag
$$
Repeating this argument for the equator $bpre$ corresponding to the irreducible polynomials $Q_3(x,z)$ and $Q_5(x,z)$, we obtain
$$
\begin{equation*}
sqad\in M_0\cup M_z \quad\text{and} \quad cfhg\in M_0\cup M_x.
\end{equation*}
\notag
$$
Using Lemma 5 we finally deduce that $K\in A_1\cup A_2$. Theorem 3. If at least two equators belong to the class $\mathrm{II}$, then the octahedron belongs to the class $A_1$ or $A_2$. Proof. We can assume without loss of generality that $bpre, cfhg\in \mathrm{II}$. There are four possible cases.
Case (A): $bpre, cfhg\in \mathrm{II}_x$. We write out the Cayley-Menger determinants for the tetrahedral angles $K_1$, $K_6$, $K_3$ and $K_5$:
$$
\begin{equation*}
\begin{gathered} \, Q_1(x,y)=x^2y-(s\pm q)^2x^2-2(p^2+r^2)xy+(p^2-r^2)^2y+\dotsb, \\ Q_6(x,y)=x^2y-(a\pm d)^2x^2-2(b^2+e^2)xy+(b^2-e^2)^2y+\dotsb, \\ Q_3(x,z)=x^2z-(a\pm q)^2x^2-2(c^2+f^2)xz+(c^2-f^2)^2z+\dotsb, \\ Q_5(x,z)=x^2z-(s\pm d)^2x^2-2(h^2+g^2)xz+(h^2-g^2)^2z+\dotsb. \end{gathered}
\end{equation*}
\notag
$$
(The signs are chosen independently, so there are 16 possible cases.)
Equating the coefficients at $x^2$, $xy$ and $y$ in the first pair of equations and at $x^2$, $xz$ and $z$ in the second pair we obtain six conditions, which are conveniently arranged into three systems of equations:
$$
\begin{equation}
\begin{gathered} \, \begin{cases} (s\pm q)^2=(a\pm d)^2, \\ (a\pm q)^2=(s\pm d)^2, \end{cases} \qquad \begin{cases} p^2+r^2=b^2+e^2, \\ (p^2-r^2)^2=(b^2-e^2)^2, \end{cases} \\ \begin{cases} c^2+f^2=h^2+g^2, \\ (c^2-f^2)^2=(h^2-g^2)^2. \end{cases} \end{gathered}
\end{equation}
\tag{4.1}
$$
The systems for $bpre$ and $cfhg$ in (4.1) imply that
$$
\begin{equation*}
bpre\in M_0\cup M_x\quad\text{and} \quad cfhg\in M_0\cup M_x.
\end{equation*}
\notag
$$
It remains to check that the equator $sqad$ is also metrically symmetric. Without additional conditions systems (4.1) imply only that the equator $sqad$ has a zero sum, but does not mean its metric symmetry. Some sign patterns in (4.1) lead to degeneracy (for example, the identity $s+q+a+d=0$ implies that the edges are zero, and $s-q-a-d=0$ corresponds to a trivial flexing). These degeneracies do not occur only in four cases, namely, when
$$
\begin{equation}
\begin{cases} (s+\varepsilon_1 q)^2=(a+\varepsilon_1 d)^2, \\ (a+\varepsilon_2 q)^2=(s+\varepsilon_2 d)^2, \end{cases}
\end{equation}
\tag{4.2}
$$
where $\varepsilon_1=\pm1$ and $\varepsilon_2=\pm1$.
Recall that the reducibility of Cayley-Menger polynomials is ensured by conditions (3.6), which have the following form in the cases under consideration:
$$
\begin{equation*}
\begin{gathered} \, c^2=q^2+p^2+\varepsilon_1 \frac{q}{s}(g^2-p^2-s^2), \qquad f^2=q^2+r^2+\varepsilon_1 \frac{q}{s}(h^2-r^2-s^2), \\ c^2=a^2+b^2+\varepsilon_1 \frac{a}{d}(g^2-b^2-d^2), \qquad f^2=a^2+e^2+\varepsilon_1 \frac{a}{d}(h^2-e^2-d^2), \\ b^2=a^2+c^2+\varepsilon_2 \frac{a}{q}(p^2-q^2-c^2), \qquad e^2=a^2+f^2+\varepsilon_2 \frac{a}{q}(r^2-q^2-f^2), \\ b^2=d^2+g^2+\varepsilon_2 \frac{d}{s}(p^2-s^2-g^2), \qquad e^2=d^2+h^2+\varepsilon_2 \frac{d}{s}(r^2-s^2-h^2). \end{gathered}
\end{equation*}
\notag
$$
We add the equations in the left-hand column (the formulae for $c^2$ and $b^2$) after multiplying them by $-\varepsilon_1 asd$, $\varepsilon_1\varepsilon_2 qsd$, $\varepsilon_1\varepsilon_2 qsd$ and $- \varepsilon_2 asq$, respectively. Carrying out calculations and taking into account that $\varepsilon_1^2=1$ and $\varepsilon_2^2=1$ we obtain
$$
\begin{equation}
2sqad(s-\varepsilon_1 q+\varepsilon_1 \varepsilon_2a-\varepsilon_2d)=0 \quad \Longrightarrow\quad s-\varepsilon_1 q+\varepsilon_1 \varepsilon_2a-\varepsilon_2d=0.
\end{equation}
\tag{4.3}
$$
Combining (4.2) and (4.3) and considering the resulting system of equations
$$
\begin{equation}
\begin{cases} (s+\varepsilon_1 q-a-\varepsilon_1 d)(s+\varepsilon_1 q+a+\varepsilon_1 d)=0, \\ (s-\varepsilon_2 q-a+\varepsilon_2 d)(s+\varepsilon_2 q+a+\varepsilon_2 d)=0, \\ s-\varepsilon_1 q+\varepsilon_1 \varepsilon_2a-\varepsilon_2d=0, \end{cases}
\end{equation}
\tag{4.4}
$$
we obtain $sqad\in M_e\subset M_0$ for $\varepsilon_1=\varepsilon_2=1$ and $sqad\in M_0$ for $\varepsilon_1=-\varepsilon_2=\pm1$, while the equator is degenerate when $\varepsilon_1=\varepsilon_2=-1$.
Thus, in case (A) we obtain
$$
\begin{equation*}
sqad\in M_0, \qquad bpre\in M_0\cup M_x\quad\text{and} \quad cfhg\in M_0\cup M_x.
\end{equation*}
\notag
$$
Applying Lemma 5, we conclude that $K\in A_1\cup A_2$. Case (B): $bpre\in \mathrm{II}_z$ and $cfhg\in \mathrm{II}_y$. We write out the Cayley-Menger polynomials for the tetrahedral angles $K_1$, $K_6$, $K_3$ and $K_5$:
$$
\begin{equation*}
\begin{gathered} \, Q_1(x,y)=y^2x-(p\pm r)^2y^2-2(s^2+q^2)xy+(s^2-q^2)^2x+\dotsb, \\ Q_6(x,y)=y^2x-(b\pm e)^2y^2-2(a^2+d^2)xy+(a^2-d^2)^2x+\dotsb, \\ Q_3(x,z)=z^2x-(c\pm f)^2z^2-2(a^2+q^2)zx+(a^2-q^2)^2x+\dotsb \end{gathered}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
Q_5(x,z)=z^2x-(h\pm g)^2z^2-2(s^2+d^2)zx+(s^2-d^2)^2x+\dotsb.
\end{equation*}
\notag
$$
Equating the coefficients at $y^2$ and $xy$ in the first pair of equations and at $z^2$ and $zx$ in the second pair of equations, we obtain four conditions that are conveniently arranged into two systems of equations:
$$
\begin{equation}
\begin{cases} (p\pm r)^2=(b\pm e)^2, \\ (h\pm g)^2=(c\pm f)^2 \end{cases} \quad\text{and}\quad \begin{cases} s^2+q^2=a^2+d^2, \\ s^2+d^2=a^2+q^2. \end{cases}
\end{equation}
\tag{4.5}
$$
The system for $sqad$ in (4.5) shows that $sqad\in M_0$. We use the previously introduced notation $S_1$, $S_2$, $S_3$ and $S_4$ for the areas of the triangles $pqc$, $qrf$, $rsh$ and $psg$, and we also denote by $S_5$, $S_6$, $ S_7$ and $S_8$ the areas of the triangles $bac$, $aef$, $edh$ and $dbg$, respectively; see Figure 1, (a). By Lemma 4,
$$
\begin{equation*}
\begin{gathered} \, K_1\in \mathrm{II}_y \quad \Longrightarrow \quad pS_2=rS_1, \quad rS_4=pS_3, \\ K_6\in \mathrm{II}_y \quad \Longrightarrow \quad eS_5=bS_6, \quad eS_8=bS_7, \\ K_3\in \mathrm{II}_z \quad \Longrightarrow \quad fS_1=cS_2, \quad cS_6=fS_5 \end{gathered}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
K_5\in \mathrm{II}_z \quad \Longrightarrow \quad gS_3=hS_4, \qquad gS_7=hS_8.
\end{equation*}
\notag
$$
Multiplying the equalities we obtain
$$
\begin{equation*}
pS_2\cdot eS_5\cdot fS_1\cdot cS_6=rS_1\cdot bS_6 \cdot cS_2 \cdot fS_5 \quad \Longrightarrow \quad br=pe \quad \Longrightarrow \quad bpre\in R_0
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
pS_2\cdot fS_1\cdot gS_3\cdot rS_4=rS_1\cdot cS_2 \cdot hS_4 \cdot pS_3 \quad \Longrightarrow \quad ch=fg \quad \Longrightarrow \quad cfhg\in R_0.
\end{equation*}
\notag
$$
The equations in the first system in (4.5) imply that the equators $bpre$ and $cfhg$ have zero sums. Therefore, from Lemma 1 we obtain $bpre \in M_x \cup M_z$ and $cfhg \in M_x\cup M_y$. Thus, in case (B) we have
$$
\begin{equation*}
sqad\in M_0, \qquad bpre\in M_x\cup M_z\quad\text{and} \quad cfhg\in M_x\cup M_y.
\end{equation*}
\notag
$$
Using Lemma 5 we conclude that $K\in A_1\cup A_2$. Case (C): $bpre\in \mathrm{II}_z$ and $cfhg\in \mathrm{II}_x$. In this case Theorem 1) implies that
It follows that $y$ is rationally expressed in terms of $z$, but not vice versa. Therefore, $sqad\in \mathrm{II}_z$ (we use Theorem 1 here). Since $bpre\in \mathrm{II}_z$in this case, we arrive at case (A) considered above, with respect to the equators $sqad$ and $bpre$. Case (D): $bpre\in \mathrm{II}_x$ and $cfhg\in \mathrm{II}_y$. Then we obtain that
It follows that $z$ is rationally expressed in terms of $y$, but not vice versa. Therefore, $sqad\in \mathrm{II}_y$. Since we also have $cfhg\in \mathrm{II}_y$, we arrive at case (A) with respect to the equators $sqad$ and $cfhg$. Theorem 4. There is no octahedron with an equator in each of the classes $\mathrm{I}$, $\mathrm{II}$ and $\mathrm{III}$. Proof. Suppose the converse. Without loss of generality we can assume that ${sqad\in \mathrm{I}}$, $bpre\in \mathrm{II}$ and $cfhg\in \mathrm{III}$.
Let $bpre\in \mathrm{II}_x$. Then $z$ is a rational function of $x$. Furthermore, $x$ is a rational function of $y$ (this follows from the fact that $cfhg\in \mathrm{III}$). Therefore, $z$ is rationally expressed in terms of $y$, which contradicts the condition $sqad\in \mathrm{I}$. Similarly, let $bpre\in \mathrm{II}_z$. Then $x$ is a rational function of $z$. Furthermore, $y$ is a rational function in $x$. Therefore, $y$ is rationally expressed in terms of $z$, which contradicts the condition $sqad\in \mathrm{I}$. We consider the relations
$$
\begin{equation}
qge=pfd, \qquad bhq=cdr, \qquad rga=sfb\quad\text{and} \quad ahp=ces,
\end{equation}
\tag{4.6}
$$
which we refer to as the conditions of metric proportionality of the octahedron. Note that each of the four equations in (4.6) is a consequence of the other three. Lemma 6. Suppose that the edges of the octahedron satisfy the conditions of metric proportionality, one equator belongs to the class $M_0$, and the remaining two equators have zero sums. Then these two equators are metrically symmetric. Proof. We can assume without loss of generality that $squad\in M_0$. Let where $\varepsilon_k=\pm1$, $k=1,\dots ,6$.
If $\varepsilon_2=-1$, then the third equation in (4.7) implies that $\varepsilon_1=\varepsilon_3=1$. If $\varepsilon_2=1$, then $\varepsilon_3=- \varepsilon_1$. In both cases $\varepsilon_3= -\varepsilon_1\cdot \varepsilon_2$. We find similarly that $\varepsilon_6=-\varepsilon_4\cdot \varepsilon_5$. Therefore, (4.7) can be rewritten as
$$
\begin{equation}
\begin{cases} s=a, \\ q=d, \\ b-\varepsilon_1 p-\varepsilon_2 r+\varepsilon_1\cdot\varepsilon_2 e=0, \\ c-\varepsilon_4 f-\varepsilon_5 h+\varepsilon_4\cdot\varepsilon_5 g=0. \end{cases}
\end{equation}
\tag{4.8}
$$
Substituting $s=a$ and $q=d$ into (4.6) we obtain
$$
\begin{equation}
\begin{cases} ge=pf, \\ bh=cr, \\ rg=fb \end{cases} \quad \Longrightarrow\qquad \begin{cases} be=pr, \\ cf=hg. \end{cases}
\end{equation}
\tag{4.9}
$$
Now (4.8) and (4.9) imply that
$$
\begin{equation}
\begin{cases} b-\varepsilon_1 p-\varepsilon_2 r+\varepsilon_1\cdot\varepsilon_2 e=0, \\ be=pr, \\ c-\varepsilon_4 f-\varepsilon_5 h+\varepsilon_4\cdot\varepsilon_5 g=0, \\ cf=hg \end{cases} \quad \Longrightarrow\quad \begin{cases} (p-\varepsilon_2 e)(r-\varepsilon_1 e)=0, \\ (f-\varepsilon_5 g)(\varepsilon_5 h+\varepsilon_4 f)=0. \end{cases}
\end{equation}
\tag{4.10}
$$
It is easy to verify that the vanishing of any pair of expressions in brackets in (4.10) implies the metric symmetry of the equators $bpre$ and $cfhg$. Lemma 7. Assume that the edges of the octahedron satisfy the conditions of metric proportionality, not all of its equators are metrically symmetric, but all of them have a zero sum, namely, Remark 6. Formulae (4.11) are written by taking account of the argument at the beginning of the proof of Lemma 6. Proof of Lemma 7. If it is known that the denominators in (4.12) are distinct from zero, then these formulas can obtained by solving the following systems of equations with respect to the indeterminates ($d$, $q$), ($p$, $e$), ($f$, $g$):
$$
\begin{equation}
\begin{gathered} \, \begin{cases} \varepsilon_1 \cdot (q-\varepsilon_2 d)=s-\varepsilon_2 a, \\ bh\cdot q-cr\cdot d=0, \end{cases} \qquad \begin{cases} \varepsilon_3 \cdot (p-\varepsilon_4 e)=b-\varepsilon_4 r, \\ ah\cdot p-cs\cdot e=0, \end{cases} \\ \begin{cases} \varepsilon_5 \cdot (f-\varepsilon_6 g)=c-\varepsilon_6 h, \\ sb\cdot f-ra\cdot g=0. \end{cases} \end{gathered}
\end{equation}
\tag{4.13}
$$
Therefore, it remains to check that the denominators in (4.12) are distinct from zero in any admissible case (not all combinations of signs $\varepsilon_k=\pm1$, $k=1,\dots ,6$, in (4.11) are admissible).
Let $s=a$. Then (4.14) implies that $q=d$, and the equator $sqad$ belongs to the class $M_0$. By Lemma 6 this implies the metric symmetry of the equators $bpre$ and $cfhg$. Therefore, the case $s=a$ is impossible, since not all equators are metrically symmetric by assumption. Hence we can assume that $s\ne a$. Then (4.14) implies that $q\ne d$, and therefore (4.6) implies that $bh\ne cr$. So the denominators in the formulae for $d$ and $q$ in (4.12) do not vanish. Thus, we have proved that in all admissible cases the denominator in the formulae for $d$ and $q$ in (4.12) is not zero. For the remaining denominators in (4.12) the argument is quite similar. Let
$$
\begin{equation}
\begin{gathered} \, P_{0}(a,b,c,s,r,h)=B_{1,1,1}, \qquad P_{1}(a,b,c,s,r,h)=B_{1,-1,-1}, \\ P_{2}(a,b,c,s,r,h)=B_{-1,1,-1}\quad\text{and} \quad P_{3}(a,b,c,s,r,h)=B_{-1,-1,1}, \end{gathered}
\end{equation}
\tag{4.15}
$$
where
$$
\begin{equation*}
\begin{aligned} \, B_{\varepsilon_1, \varepsilon_2, \varepsilon_3} &=abc(s^2+r^2+h^2)-srh(a^2+b^2+c^2) \\ &\qquad +\varepsilon_1\cdot(sbc(h^2+r^2-s^2)-arh(b^2+c^2-a^2)) \\ &\qquad+\varepsilon_2\cdot(arc(s^2+h^2-r^2)-sbh(a^2+c^2-b^2)) \\ &\qquad+\varepsilon_3\cdot(abh(s^2+r^2-h^2)-src(a^2+b^2-c^2)). \end{aligned}
\end{equation*}
\notag
$$
Theorem 5. If at least two equators belong to class $\mathrm{III}$, then the following statements hold: (1) the third equator also belongs to class $\mathrm{III}$; (2) the conditions of metric proportionality (4.6) are met; (3) all equators have a zero sum; (4) if not all equators are metrically symmetric, then $P_{k}(a,b,c,s,r,h)=0$ for some $k=0,1,2,3$. Proof. Without loss of generality we assume that $cfhg, bpre\in \mathrm{III}$. Then the Cayley-Menger equations for $K_1$ and $K_3$ have the form and
Each condition is a product of two factors which are identical in structure. The flexing of a tetrahedral angle in a particular position is described by one of the two (irreducible) factors. Therefore, $y$ is a linear fractional function of $x$, and $x$ is a linear fractional function of $z$. Then $y$ is a linear fractional function of $z$, which implies that $sqad\in \mathrm{III}$ by Theorem 1. Statement (1) is proved. By Lemma 4,
$$
\begin{equation*}
\begin{cases} afS_1=cqS_6, \\ hrS_6=feS_3, \\ pqS_3=srS_1. \end{cases}
\end{equation*}
\notag
$$
Multiplying these equations we obtain
$$
\begin{equation*}
qrf(ahp-ces)S_1S_3S_6=0 \quad \Longrightarrow\quad ahp=ces.
\end{equation*}
\notag
$$
The remaining equalities in statement (2) are obtained similarly.
We write down the conditions for the tetrahedral angles $K_1$ and $K_6$ to belong to class $\mathrm{III}$ (see relations (3.8) in Lemma 3):
$$
\begin{equation}
\begin{gathered} \, (c^2-p^2-q^2)rs=\lambda pq(h^2-r^2-s^2), \qquad (f^2-q^2-r^2)sp=\lambda qr(g^2-p^2-s^2), \\ (c^2-b^2-a^2)ed=\widetilde{\lambda} ba(h^2-e^2-d^2)\quad\text{and} \quad (f^2-a^2-e^2)db=\widetilde{\lambda} ae(g^2-b^2-d^2). \end{gathered}
\end{equation}
\tag{4.16}
$$
Here $\lambda=\pm1$ and $\widetilde{\lambda}=\pm1$, where the value $+1$ corresponds to pairwise equal plane angles at the vertex of a tetrahedral angle, and the value $-1$ corresponds to the angles summing up pairwise to $\pi$.
Now we write out the Cayley-Menger polynomials for $K_1$ and $K_6$ (their reducibility is ensured by relations (4.16)):
$$
\begin{equation*}
Q_1(x,y)\,{=}\,\bigl[xy\,{-}\,(s\,{+}\,\lambda q)^2x\,{-}\,(p\,{+}\,r)^2y+\dotsb \bigr]\,{\cdot}\,\bigl[xy\,{-}\,(s\,{-}\,\lambda q)^2x\,{-}\,(p\,{-}\,r)^2y+\dotsb \bigr]
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
Q_6(x,y)\,{=}\,\bigl[xy\,{-}\,(d\,{+}\,\widetilde{\lambda} a)^2x\,{-}\,(b\,{+}\,e)^2y+\dotsb \bigr]\,{\cdot}\,\bigl[xy\,{-}\,(d\,{-}\,\widetilde{\lambda} a)^2x\,{-}\,(b\,{-}\,e)^2y+\dotsb \bigr].
\end{equation*}
\notag
$$
For the octahedron to be nontrivially flexible, one of the factors in $Q_1(x,y)$ must coincide with one of the factors in $Q_6(x,y)$ identically for all $x$ and $y$. This, in particular, implies the pairwise equality of the coefficients at $x$ and coefficients at $y$ in the factors selected (this is how the necessary conditions for flexibility are obtained). It can be seen that out of the four a priori possible combinations, only two can be realised: the first factor in $Q_1$ coincides with the first factor in $Q_6$, or the second factor in $Q_1$ coincides with the second factor in $Q_6$. In the remaining two cases the flexing is degenerate. For example, if we pair the first factor $Q_1$ with the second factor in $Q_6$, then we obtain This implies that the equator $bpre$ can only be realised as a doubly covered segment, and so its diagonals cannot change under flexing. We set $k=1$ if the first factors in $Q_1$ and $Q_6$ coincide, and $k=-1$ if the second factors coincide.Further, of the four combinations of the parameters $\lambda=\pm1$ and $\widetilde{\lambda}=\pm1$ only ones with $\lambda=\widetilde{\lambda}$ are admissible: it is easy to see that for $\lambda=-\widetilde{\lambda}$ the equator $sqad$ can only be realised as a doubly covered segment. Therefore, the conditions (of equality or complementarity) on the plane angles at vertices of the tetrahedral angles $K_1$ and $K_6$ (connected by the diagonal $z$) must be the same. Let $k=1$. Equating the coefficients of $x$ of the first factors in $Q_1$ and $Q_6$ we obtain which implies that where $m=\pm1$. The parameter $m$ therefore reflects the choice of one of the two possible necessary conditions on the equator $sqad$ in this case. In a similar way we introduce the parameter $n=\pm1$, which is responsible for the choice of one of the possible conditions based on the equality of the coefficients of $y$. In the case $k= -1$ the parameters $m$ and $n$ are defined similarly.Therefore, for the polynomials $Q_1$ and $Q_6$ corresponding to the vertices of the diagonal $z$, all possible cases are described by the parameters $\lambda$, $k$, $m$ and $n$. Each parameter takes values $\pm1$ independently of the others, so there are $2^4=16$ cases in total. We treat the polynomials corresponding to the vertices of the diagonals $x$ and $y$ similarly. It is convenient to introduce 12 indexed parameters as follows: $\lambda_1$, $k_1$, $m_1$, $n_1$: the parameters for $K_2$ and $K_4$ (the diagonal $x$); $\lambda_2$, $k_2$, $m_2$, $n_2$: the parameters for $K_3$ and $K_5$ (the diagonal $y$); $\lambda_3$, $k_3$, $m_3$, $n_3$: the parameters for $K_1$ and $K_6$ (the diagonal $z$). We refer to a set of values of these parameters, considering the values $\pm1$ separately, as a configuration. The total number $\varkappa$ of configurations is $2^{12}=4096$. Some configurations may lead to degenerate foldings or trivial flexings. For each configuration the reducibility of the Cayley-Menger polynomials is ensured by conditions on plane angles:
$$
\begin{equation}
\begin{alignedat}{2} &(a^2-b^2-c^2)pg=\lambda_1bc(s^2-p^2-g^2), &\ \quad &(q^2-c^2-p^2)gb= \lambda_1cp(d^2-b^2-g^2), \\ &(a^2-e^2-f^2)rh=\lambda_1ef(s^2-r^2-h^2), &\ \quad &(q^2-f^2-r^2)he= \lambda_1fr(d^2-e^2-h^2), \\ &(p^2-q^2-c^2)af=\lambda_2qc(e^2-a^2-f^2), &\ \quad &(b^2-c^2-a^2)fq= \lambda_2ca(r^2-q^2-f^2), \\ &(p^2-s^2-g^2)dh=\lambda_2sg(e^2-d^2-h^2), &\ \quad &(b^2-g^2-d^2)hs= \lambda_2gd(r^2-s^2-h^2), \\ &(c^2-p^2-q^2)rs=\lambda_3pq(h^2-r^2-s^2), &\ \quad &(f^2-q^2-r^2)sp= \lambda_3qr(g^2-p^2-s^2), \\ &(c^2-b^2-a^2)ed=\lambda_3ba(h^2-e^2-d^2), &\ \quad &(f^2-a^2-e^2)db= \lambda_3ae(g^2-b^2-d^2). \end{alignedat}
\end{equation}
\tag{4.17}
$$
The Cayley-Menger polynomials themselves have the form
$$
\begin{equation*}
\begin{aligned} \, Q_2(z,y) &=\bigl[zy-(g+\lambda_1 c)^2z-(b+p)^2y+\dotsb \bigr] \\ &\qquad\times\bigl[zy-(g-\lambda_1 c)^2z-(b-p)^2y+\dotsb \bigr], \\ Q_4(z,y) &=\bigl[zy-(h+\lambda_1 f)^2z-(e+r)^2y+\dotsb \bigr] \\ &\qquad\times\bigl[zy-(h-\lambda_1 f)^2z-(e-r)^2y+\dotsb \bigr], \\ Q_3(z,x) &=\bigl[zx-(f+\lambda_2 c)^2z-(q+a)^2x+\dotsb \bigr] \\ &\qquad\times\bigl[zx-(f-\lambda_2 c)^2z-(q-a)^2x+\dotsb \bigr], \\ Q_5(z,x) &=\bigl[zx-(h+\lambda_2 g)^2z-(s+d)^2x+\dotsb \bigr] \\ &\qquad\times\bigl[zx-(h-\lambda_2 g)^2z-(s-d)^2x+\dotsb \bigr], \\ Q_1(x,y) &=\bigl[xy-(s+\lambda_3 q)^2x-(p+r)^2y+\dotsb \bigr] \\ &\qquad\times\bigl[xy-(s-\lambda_3 q)^2x-(p-r)^2y+\dotsb \bigr] \end{aligned}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\begin{aligned} \, Q_6(x,y) &=\bigl[xy-(d+\lambda_3 a)^2x-(b+e)^2y+\dotsb \bigr] \\ &\qquad\times\bigl[xy-(d-\lambda_3 a)^2x-(b-e)^2y+\dotsb \bigr], \end{aligned}
\end{equation*}
\notag
$$
or, in short,
$$
\begin{equation}
Q_j=Q_j^{+}\cdot Q_j^{-}\quad\text{for }j=1,\dots ,6.
\end{equation}
\tag{4.18}
$$
The parameters $k_1$, $k_2$ and $k_3$ determine the choice of one of the two factors in each of the six expressions $Q_j$. Equating these factors to zero we obtain equations describing a flexing of the octahedron. Equating the coefficients at the first powers of $x$, $y$ and $z$ in these equations we obtain the following system of six necessary conditions for each configuration $\varkappa$:
$$
\begin{equation}
\begin{cases} g+\lambda_1 k_1 c+m_1 (h+\lambda_1 k_1 f)=0, \\ f+\lambda_2 k_2 c+m_2 (h+\lambda_2 k_2 g)=0, \\ s+\lambda_3 k_3 q+m_3 (d+\lambda_3 k_3 a)=0, \\ b+k_1 p+n_1 (e+k_1 r)=0, \\ q+k_2 a+n_2 (s+k_2 d)=0, \\ p+k_3 r+n_3 (b+k_3 e)=0. \end{cases}
\end{equation}
\tag{4.19}
$$
The left-hand side of each equation in (4.19) is the algebraic sum (with plus or minus signs) of the edge lengths of an equator. We say that the configuration is admissible if each equation in the system involves exactly two edges with plus sign and two edges with minus sign. For all other arrangements of signs we obtain either a degenerate folding or a degenerate flexing. Therefore, we consider only admissible configurations; then each equation in the system describes a zero-sum condition for the corresponding equator. Thus, statement (3) is proved. Now let $\varkappa$ be an admissible configuration. If the first two equations in (4.19) are independent (not proportional, that is, represent different zero-sum conditions), then the equator $cfhg$ is metrically symmetric: $cfhg\in M_0\,{\cup} \, M_x\,{\cup}\, M_y\,{=}\,M$. If the first two equations in (4.19) are proportional, then we only know that the equator $cfhg$ has a zero sum: $cfhg\,{\in}\, S_0\,{\cup}\, S_x\,{\cup}\, S_y\,{=}\,S$ (the class of zero-sum equators is larger than the class of metrically symmetric equators). A similar observation can be made for the remaining two pairs of equations in (4.19). Next we find conditions on the parameters of a configuration which are necessary and sufficient for the equations in (4.19) to be pairwise proportional. Lemma 8. The equations in (4.19) are pairwise proportional if and only if the parameters of the configuration satisfy the relations To prove this lemma we rewrite (4.19) as
$$
\begin{equation}
\begin{cases} s+\lambda_3 k_3 q+m_3\lambda_3 k_3 a+m_3 d=0, \\ s+n_2 q+n_2 k_2 a+k_2 d=0, \\ b+k_1 p+n_1 k_1 r+n_1 e=0, \\ b+n_3 p+n_3 k_3 r+k_3 e=0, \\ c+m_1 f+\lambda_1 k_1 m_1 h+\lambda_1 k_1 g=0, \\ c+\lambda_2 k_2 f+\lambda_2 k_2 m_2 h+m_2 g=0. \end{cases}
\end{equation}
\tag{4.21}
$$
(We take into account that each parameter is equal to $\pm1$, so its square is $1$.) Remark 7. Lemma 8 implies that among all configurations with fixed $[\lambda, k]$, one can specify just one configuration for which the equations in (4.19) are pairwise proportional (the parameters $[m,n]$ are uniquely expressed in terms of $[\lambda, k]$). Here and below we use square brackets to denote sets of same-name variables with indices from 1 to 3. For such a configuration we can discard duplicate equations in (4.21) and rewrite the system (4.19) in the form Now we move on to the proof of statement (4) of Theorem 5. Proposition. For any admissible configuration $\varkappa=[\lambda,k,m,n]$ one of the following statements is true: (1) the equations in (4.19) are pairwise proportional; (2) there is an admissible configuration $\varkappa'=[\lambda,k,m',n']$ (with the same $\lambda$ and $k$ as in the original configuration $\varkappa$) such that the equations in (4.19) are pairwise proportional; (3) the system (4.19) yields the condition that at least one of the equators belongs to class $M_0$. This is proved by enumerating all possible configurations. This proposition implies that we need only consider configurations with pairwise proportional equations with a zero sum (case (1)). Indeed, in case (2) we can change the parameters $[m,n]$ to obtain a more general case, that is, when the system specifies more general necessary conditions, but with the same $[\lambda,k]$, and therefore with the same conditions (4.17) and flexing equations (4.18). In case (3), all equators are metrically symmetric by Lemma 6, which contradicts the assumption of statement (4) of Theorem 5. Consider the set of polynomials $T_k$, $k=1,\dots ,8$, and note that their linear combinations are reducible:
$$
\begin{equation}
\begin{cases} T_1=arh-sbh-src+srh, \\ T_2=abh-abc-sbc+arc, \\ T_3=arh-sbh+src-srh, \\ T_4=abh-abc+sbc-arc, \\ T_5=arh+sbh-src-srh, \\ T_6=abh+abc-sbc-arc, \\ T_7=arh+sbh+src+srh, \\ T_8=abh+abc+sbc+arc \end{cases} \Longrightarrow \begin{cases} T_1-T_2=(s+q)(b-r)(c-h), \\ T_3+T_4=(s-q)(b+r)(c-h), \\ T_4-T_6=(s-q)(b-r)(c+h). \end{cases}
\end{equation}
\tag{4.23}
$$
Now consider all admissible configurations with pairwise proportional zero-sum equations (4.19). The list of these configurations is formed during the enumeration process in the proof of the above proposition. According to Remark 7, each configuration in this list is determined by its parameters $[\lambda,k]$, the parameters $[m,n]$ are calculated using formulae (4.21), and system (4.19) can be rewritten in the form (4.22). By assumption not all equators are metrically symmetric, but (4.22) has the form (4.11), so Lemma 7 is applicable. Formulae (4.12) take the form
$$
\begin{equation}
\begin{cases} d=-\dfrac{bh(s+\lambda_3 k_3 k_2 a)}{\lambda_3 k_3 cr+k_2 bh}, \quad p=-\dfrac{cs(b+k_1 k_3 r)}{k_1 cs+k_3 ah}, \quad f=-\dfrac{ra(c+\lambda_1 k_1 \lambda_2 k_2 h)}{\lambda_2 k_2 ra+\lambda_1 k_1 sb}, \\ q=- \dfrac{cr(s+\lambda_3 k_3 k_2 a)}{\lambda_3 k_3 cr+k_2 bh}, \quad e=- \dfrac{ah(b+k_1 k_3 r)}{k_1 cs+k_3 ah}, \quad g=- \dfrac{sb(c+\lambda_1 k_1 \lambda_2 k_2 h)}{\lambda_2 k_2 ra+\lambda_1 k_1 sb}. \end{cases}
\end{equation}
\tag{4.24}
$$
For each configuration under consideration we substitute (4.24) into (4.17). After multiplication by a common denominator and factorisation, each equation in (4.17) takes the form $T_l\cdot P_k=0$ for some $l=1,\dots ,8$ and $ k=0,\dots ,3$ (see (4.15) and (4.23)). We summarise the results in Table 2 by grouping together the configurations with identical sets $[\lambda_1, \lambda_2, \lambda_3]$. Table 2
For each configuration we obtain 12 equations of the form $T_l\cdot P_k=0$ (in accordance with the number of equations in (4.17)), but we do not need all of them. In each case one or two equations are selected. Typically, the first and second equations of the first column in (4.17) are used. If only one equation is taken, then Table 2 shows its ordinal number in the first column of (4.17). For the configuration in the first line, the assumption that $P_0\ne 0$ implies that $T_1=0$ and $T_2=0$. Then (4.23) implies that $b=r$ or $c=h$. Taking (4.22) into account we obtain $bpre\in M_0$ or $cfhg\in M_0$. Therefore, by Lemma 6 all equators are metrically symmetric, which contradicts the assumption. It follows that $P_0=0$ for this configuration. For other configurations with two equations of the form $T_l\cdot P_k=0$ the same argument shows that $P_k=0$ for the corresponding $k$. For configurations with one equation of the form $T_l\cdot P_k=0$ we have $T_7\ne 0$, $T_8\ne 0$, which implies that $P_k=0$ (for the corresponding $k$). The last column of Table 2 gives the number $k$ for which the polynomial $P_k$ vanishes for each configuration. Statement (4) is proved, which finishes the proof of Theorem 5. Remark 8. In Theorem 5 statement (4) does not follow from (2) and (3). Indeed, the conditions in (2) and (3) define a variety given by (4.24) with free parameters $a$, $ b$, $c$, $h$, $s$ and $r$, whereas the condition $P_k=0$ in statement (4) specifies a relation between these parameters. Thus, for each configuration we have a manifold of dimension 5 in the space of edge lengths, which coincides with the dimension ‘predicted’ by Theorem 8.1 in [6]. Remark 9. The configurations corresponding to the first four combinations $[\lambda_1, \lambda_2, \lambda_3]$ from Table 2 have some additional properties. Firstly, the factor $P_k$ appears after substitution (4.24) not only in those equations in (4.17) which are listed in Table 2, but also in all other equations in this system. Therefore, the reducibility conditions for all six Cayley-Menger polynomials are satisfied. Secondly, the flexibility conditions (the vanishing factors in (4.18), chosen in accordance with the parameters $k_1$, $k_2$ and $k_3$ of the configuration in question) turn out to coincide pairwise. (Note that the substitution (4.24) ensures the equality of the coefficients at the first powers of $x$, $y$ and $z$, but the constant terms become equal only if we additionally assume that $P_k=0$.) In the case of pairwise equality of all coefficients we obtain three equations relating the three diagonals (instead of six equations in the general case). Such a system of equations has a continuous family of solutions (with the exception of some degenerate cases), which gives reason to expect that foldings admitting flexible realisations form a full-dimensional subset of the space of all foldings corresponding to any of the chosen configurations. Remark 10. The configurations corresponding to the last four combinations $[\lambda_1, \lambda_2, \lambda_3]$ in Table 2 are in fact impossible: in these cases the remaining equations in (4.17) give additional conditions on edge lengths which turn out to be inconsistent. We have a proof of this fact, but it is rather lengthy and we do not present it here. The ‘emptiness’ of these cases does not affect the validity of Theorem 5. Remark 11. One can take another polynomial for $B_{\varepsilon_1, \varepsilon_2, \varepsilon_3}$, which is similar in structure, but depends on different variables (the edge lengths of any other pair of opposite facets). Then the statement of Theorem 5 remains the same, but in (4.13) we have to regroup the equations into three pairs in a different way and solve the equations for other variables (not involved in the expression $B_ {\varepsilon_1, \varepsilon_2, \varepsilon_3}$). The classification in Table 2 remains the same, but formulae (4.24) and the expressions for $P_k$ are different. Recall that we introduced before the classes $A_1$ and $A_2$ corresponding to Bricard’s classes $\mathrm{I}$ and $\mathrm{II}$ (see Definition 3). Now we can give the definition of the class $A_3$ corresponding to the class $\mathrm{III}$ of Bricard octahedra (with the exception of the octahedra belonging simultaneously to the class $\mathrm{III}$ and at least one of the classes $\mathrm{I}$ or $\mathrm{II}$). Definition 5. We say that an octahedron belongs to the class $A_3$ if the conditions of metric proportionality (4.6) are satisfied, all equators have a zero sum, not all of them are metrically symmetric, and $P_k=0$ for some $k=0,1,2,3$ (the classes of equators and the value of $k$ are chosen according to one of the rows of Table 2). We finally obtain the following classification of flexible octahedra. Proof. Theorems 2–5 imply that at least one of the following conditions is satisfied: 1) $K\in A_1 \cup A_2$; 2) $K$ has all three equators in the class $\mathrm{III}$. In the latter case it can occur that all equators are metrically symmetric, but then Lemma 5 yields $K\in A_1 \cup A_2$. If not all equators are metrically symmetric, then using Theorem 5 we obtain $K\in A_3$. Remark 12. To each octahedron in the class $A_3$ there corresponds a row of Table 2, and therefore also Therefore, there are characterising relations for octahedra of all three types. In these relations some edges are expressed in terms of other edges, which are viewed as parameters. In the case of octahedra of types $A_1$ and $A_2$ these parameters are free (the relations are pairwise equalities of edges), and in the case of octahedra of type $A_3$ the relations are expressed by the fractional rational functions (4.24) and the edges serving as parameters satisfy the relation $P_k=0$ for some $k=0,1,2,3$. § 5. Comparison with known results, and unsolved problemsHere we compare the obtained result with results in classical works. In [1] Bricard considered a tetrahedral angle with the metric characterised by the four plane angles at the vertex and spatial realisation characterised by the values of two dihedral angles adjacent to their common face. Denoting by $t$ and $u$ the tangents to the halves of these dihedral angles, Bricard arrived at the equation in which the coefficients depend on the metric of the tetrahedral angle (the values of the plane angles of its facets). The essence of Bricard’s method [1] was to use this equation. By writing the equations for three of the six tetrahedral angles of the octahedron, Bricard found conditions on the metric in terms of plane angles for which the system of equations has a continuous family of solutions. This is the phenomenon of flexibility.In the case of the tetrahedral angle $K_1$ (see Figure 1, (b)) we can use the well-known formula expressing the cosine of the dihedral angle between two facets of a tetrahedron in terms of the areas of these facets and edge lengths. This allows us to obtain formulae expressing the squares of the quantities $t$ and $u$ in terms of the squares of the lengths of the diagonals $x$ and $y$, and also to express the coefficients of equation (5.1) in terms of edge lengths. After some transformations (5.1) becomes the Cayley-Menger equation (see (3.1) and (3.2)). This allows us to ‘translate’ Bricard’s calculations from the language of the ‘geometry of angles’ into the language of the ‘geometry of distances’. In this sense, the approach of our work is not fundamentally new. However, firstly, in the most complex case of a type 3 octahedron an attempt to translate the corresponding equations directly into the language of distance geometry leads to a combination of equations that are difficult to use together. In particular, it is not at all obvious how the equations obtained in this way can be used to find the coefficients $a_k(l)$ of the polynomial $Q(V)$ for the volume of the octahedron. Secondly, Bricard’s approach has so far demonstrated its effectiveness only for octahedra. For more complex polyhedra an approach using Cayley-Menger polynomials can be more efficient. Therefore, our work aims at expanding the research methodology of the section of geometry referred to as ‘solution of polyhedra’ by Sabitov. Finally, in his study of octahedra of the third type Bricard considered only one of the cases that arose and wrote that the remaining cases are treated similarly. Had Bricard considered (or at least written down) all possible cases (as done in our work: see Table 2), his article would have become significantly lengthier. We now turn to another classical work related to flexible polyhedra, the one by Bennett [2]. There, in the case of an octahedron of the third type, new necessary conditions appeared for edge lengths, which are not present in Bricard’s work. These are conditions of the form $a\pm b\pm c\pm d=0$ for the edges of each of the three equators and four conditions of the form $a\cdot b\cdot c=p\cdot q\cdot r$. All these conditions appear in our Theorem 5; we called them ‘zero-sum conditions’ and ‘metric proportionality’, respectively. However, Bennett did not observe that only three of the four conditions of metric proportionality are independent (each of the four conditions is a consequence of the other three). Therefore, the dimension of the variety specified by Bennett’s seven conditions in the 12-dimensional space of edges is 6, rather than 5 as Bennett claimed when he wrote that the dimension of the family of flexible octahedra of the third type that he constructed coincides with the dimension of a manifold in the space of edges. However, the assertion about full dimension still turns out to be true, since the edge lengths of an octahedron of the third type satisfy yet another equation, which is found in our work. This is the condition $P_k=0$ from statement (4) of Theorem 5), and it does not follow from the zero sum equations for the edge lengths of the equators and from the conditions of metric proportionality (see Remark 8). Dwelling on the idea that the two approaches via the ‘geometry of angles’ and ‘geometry of distances’ are essentially equivalent, we can observe that the condition $P_k=0$ is not entirely new (although it is not present in Bennett’s work). Namely, it turns out that if we write condition (13) on p. 19 of [1] in terms of edge lengths (using necessarily the conditions of metric proportionality, which Bricard apparently did not notice), then the expression $P_k$ is included in the resulting expression as a factor, but there is also other factors with an even larger number of monomials. This is a manifestation of the nontriviality of the metric description of foldings that admit flexible realisations, and of a certain ‘lack of order’ in the previously accumulated knowledge on the structure of these objects. Generally speaking, the relationship between the conditions which are necessary or sufficient for the flexibility of type 3 octahedra still remains open. For example, there is a geometric description by Lebesgue [3] of type 3 octahedra via the condition that their equators are quadrangles circumscribed about three concentric circles, but it is not known whether this condition is also necessary. Without dwelling on other papers, we can say that since 1897 no work has appeared with a description of flexible octahedra that would make it possible to use the methods of the ‘geometry of distances’. This paper is intended to fill this gap. Finally, we mention two problems that can possibly be solved using the new metric description of flexible octahedra. 1. The conditions found in this work are necessary for flexibility, just like the Bricard conditions. It would be interesting to find necessary and sufficient conditions for the existence of a flexible realisation of a fixed complex. We illustrate the nontriviality of this problem with two observations: (a) even for a complex corresponding to a Bricard octahedron of type 1, in many cases there is both a flexible and an inflexible realisation of it (in the form of a convex polyhedron); (b) on the other hand there are flexible Bricard octahedra of the first type that do not have convex realisations. Some sufficient conditions for realisability were obtained in [17]. However, this question is still far from being fully explored. 2. It would be interesting to check Sabitov’s conjecture that for any Bricard octahedron of the third type all coefficients $a_k(l)$ of the polynomial $Q(V)$ vanish, and therefore the equation for the volume should take the form $V^8=0$. Apart from being of independent interest, this would immediately imply that the corresponding folding cannot be realised in the form of a convex octahedron. Our metric description of the class $A_3$ by means of formulae (4.24) allows one to find the coefficients $a_k(l)$ for octahedra of the third type by a mere substitution of (4.24) into $a_k(l)$. This work was carried out using the equipment of the Center for Collective Use of Ultra-High-Performance Computing Resources at Lomonosov Moscow State University. AcknowledgementsThe author expresses his sincere gratitude to I. Kh. Sabitov and D. I. Sabitov for their valuable comments and advice, as well as to the referee, whose comments helped us to significantly improve this work (in particular, to eliminate some gaps in proofs). 
Citation in format AMSBIB
\Bibitem{Mik23} |
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