    # Mono-monostatic polyhedra with uniform point masses have at least 8 vertices

The monostatic property of convex polyhedra (i.e. the property of having just one stable or unstable static equilibrium point) has been in the focus of research ever since Conway and Guy published the proof of the existence of the first such object, followed by the constructions of Bezdek and Reshetov. These examples establish F≤ 14, V≤ 18 as the respective upper bounds for the minimal number of faces and vertices for a homogeneous mono-stable polyhedron. By proving that no mono-stable homogeneous tetrahedron existed, Conway and Guy established for the same problem the lower bounds for the number of faces and vertices as F, V ≥ 5 and the same lower bounds were also established for the mono-unstable case. It is also clear that the F,V ≥ 5 bounds also apply for convex, homogeneous point sets with unit masses at each point (also called polyhedral 0-skeletons) and they are also valid for mono-monostatic polyhedra with exactly on stable and one unstable equilibrium point (both homogeneous and 0-skeletons). Here we present an algorithm by which we improve the lower bound to V≥ 8 vertices (implying f ≥ 6 faces) on mono-unstable and mono-monostable 0-skeletons. Our algorithm appears to be less well suited to compute the lower bounds for mono-stability. We point out these difficulties in connection with the work of Dawson and Finbow who explored the monostatic property of simplices in higher dimensions.

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## 1. Introduction

### 1.1. Mono-stability and polyhedra

The minimal number of positions where a rigid body could be at rest, also called static equilibria, have been investigated at least since Archimedes developed his famous design for ships . Having one stable position (i.e. being mono-stable) is of obvious advantage for ships, however, for rigid bodies under gravity, supported on a rigid (frictionless) surface this property, facilitating self-righting, could also be of advantage.

Beyond being useful, mono-stable bodies also appear to be of particular mathematical interest, because it is still unclear what are the minimal numbers of faces and vertices for a homogeneous, convex, mono-stable polyhedron. The first such object with faces and vertices has been described by Conway and Guy in 1967 . This construction was improved by Bezdek  to and later by Reshetov  to . These values define the best known upper bounds, so we have . Even less is known about the lower bounds: the only result is due to Conway  who proved that a homogeneous tetrahedron has at least two stable equilibria, so we have . The gap between the upper and lower bounds (illustrated in Figure 1) is substantial. Figure 1. Illustration of some monostatic polyhedra and their location on the overlay of the [F,V] (face and vertex number) grid and [S,U] (number of stable and unstable equilibria) grid: Conway’s polyhedron  PC in classes (F,V)=(19,34),(S,U)=(1,4), Bezdek’s polyhedron , PB in classes (F,V)=(18,18),(S,U)=(1,3), Reshetov’s polyhedron,  PR in classes (F,V)=(14,24),(S,U)=(1,2). Mono-unstable polyhedra P2,P3 in classes (F,V)=(18,18), (S,U)=(1,2),(1,3). White squares correspond on the [F,V] grid to combinatorial classes where, according to Steinitz [17, 18] we find polyhedra, on the [S,U] grid the same squares correspond to equilibrium classes where we find polyhedra with S=F,U=V .

### 1.2. Generalizations

The above problem has many generalizations of which we list a few below and we also briefly summarize the related results:

1. Instead of looking at homogeneous polyhedra, one may look at -skeletons

which are polyhedra with mass uniformly distributed on their

-dimensional faces (see Figure 2). If then we have the original problem (homogeneous polyhedron). A 0-skeleton is a homogeneous point set, with unit mass at each point (i.e. a polyhedron with unit masses at the vertices). We will denote the minimal facet and vertex numbers for mono-stable -skeletons by , respectively. In case of simplices, the center of mass of a homogeneous body and a 0-skeleton coincide  , so Conway’s result for the tetrahedron implies that for 0-skeletons we also have and upper bounds are not known. For simplicity and consistency of notation, in the homogeneous case we will drop the lower index and use , as before. Figure 2. Illustration of homogeneous material distributions for polyhedra: (a) 0-skeleton, (b) 1-skeleton, (c) 2-skeleton. The case h=3, i.e., uniform mass distribution over the entire volume is not shown for simple graphical reasons.
2. Instead of looking at mono-stable polyhedra one may look at mono-unstable polyhedra, i.e. for polyhedra which have only one vertex on which they can be balanced in an (unstable) equilibrium position. For the homogeneous case, such polyhedra were first exhibited in  with an example at In the same paper, Conway’s result for the nonexistence of a mono-stable tetrahedron was extended to the nonexistence of a mono-unstable polyhedron, so for the minimal numbers for the faces and vertices of a mono-unstable polyhedron we have and for 0-skeletons we have .

3. Instead of looking at polyhedra one may look at polytopes in higher dimensions. Dawson and Finbow [4, 5, 6] published a series of papers where they proved that in dimensions mono-stable simplices exist, thus they proved for that and they also proved that for we have

4. Instead of looking at mono-stable or mono-unstable objects (collectively called monostatic) one may look at mono-monostatic objects which have just one stable and one unstable equilibrium point. Among piecewise smooth convex bodies the first such object, called Gömböc, was identified in 2007 . Little is known about the minimal number of faces and and vertices for a polyhedron in this class. These lower bounds we will denote by . Obviously, we have

 (1) F∗h≥FSh,FUh,V∗h≥VSh,VUh,

however, despite a related prize , there are no other bounds known for either or .

### 1.3. Main result, strategy of the proof and outline of the paper

Our goal is to prove the statement claimed in the title of the paper. In fact, we will prove a stronger claim which, using the current notation can be stated as

###### Theorem 1.

.

Using (1), it is immediately clear that from Theorem 1 we have

###### Lemma 1.

which is the statement claimed in the title of the paper. We will prove Theorem 1 in four steps:

1. In section 2 we prove Theorem 2 which connects the vertex geometry (i.e. the collection of the vectors pointing from the center of mass to the vertices) of a polytope, to the number of its unstable equilibria. (We will also introduce the dual statement for faces and stable equilibria but we will not use the latter in the current paper for computations.)

2. In section 3 we show that if then Theorem 2 can be converted into a system of quadratic inequalities and we construct the general system for -dimensional 0-skeletons. The Theorem also admits to develop such systems for other cases, such as homogeneous polyhedra, or polytopes in other dimensions, however, in the current paper we only investigate 3 dimensional 0-skeletons.

3. In section 4 we develop and solve this system for the tetrahedron, serving as an illustrative example of our method. Here we prove that in a manner which is independent of the original proof of Conway.

4. In section 5 we develop an optimization algorithm which can prove the unsolvability of the system of quadratic inequalities constructed in Section 3 by doing exact (rational) arithmetic and by using this code we prove Theorem 1.

The same algorithm could, in principle, also solve such a system (instead of proving its unsolvability). If the solution for was convex and all systems with were proven to be unsolvable then, in this case the algorithm would provide the exact value as . Since we did not find such a solution, Theorem 1 provides a lower bound for . Analogous computations for other mass distributions (e.g. the homogeneous case) and other dimensions are, in principle, possible but have not been performed in this paper. Section 6 discusses why the dual case (i.e. the algorithm to compute ) is not pursued in the current paper and we establish the link between our formulae and those of Dawson . We also draw some conclusions.

## 2. The geometric construction

### 2.1. Problem statement

Let be a -dimensional convex polytope with faces , and vertices , .

We identify the faces by the face vectors orthogonal to face with being equal to the orthogonal distance of the face from the center of mass .

The vertices are identified by vertex vectors , () with origin at the center of mass and we assume to be generic in the sense that if then .

###### Definition 1.

We say that is an (unstable) static equilibrium point of if the support plane of at with respect to does not intersect . More precisely: we speak about a nondegenerate equilibrium point if , and the equilibrium is degenerate if also contains some edges of . The term ‘unstable equilibrium’ refers to a nondegenerate one henceforth unless stated otherwise. We denote the number of unstable static equilibrium points of by and we have .

Our main result connects the vertex geometry (i.e. the set of position vectors ) of with the number of its unstable static equilibria via the following formula:

###### Theorem 2.

Let be a convex polytope. Then

 (2) U(P)=V∑i=1⎢⎢ ⎢ ⎢⎣12+∑Vj=1sign((ri−rj)ri)2(V−1)⎥⎥ ⎥ ⎥⎦.

We also phrase an analogous, though more restricted result for stable equilibria.

###### Definition 2.

We say that face carries a (nondegenerate) stable static equilibrium point of if the orthogonal projection of the center of mass onto the plane of is contained in the interior of . The equilibrium is degenerate if but the term ‘stable equilibrium’ refers to a nondegenerate one henceforth unless stated otherwise. We denote the number of stable static equilibrium points of by and we have .

###### Theorem 3.

Let be a convex polytope . Then

 (3) S(P)=F∑i=1⎢⎢ ⎢ ⎢⎣12+∑Fj=1sign((qj−qi)qj)2(F−1)⎥⎥ ⎥ ⎥⎦.

Next we prove Theorems 2 and 3. In Subsection 6.1 we will show that in case of homogeneous simplices the statement of Theorem 3 follows from a result of Dawson .

### 2.2. Proof of Theorem 2

Let and denote two vertices and let denote the center of mass of the polytope . It appears to be intuitively clear that, for any fixed value of , if the angle between the rays and is sufficiently small, only one of the vertices may carry a static equilibrium. We express this observation by the following

###### Definition 3.

Let be two vertices of , with respective position vectors We say that is in the shadow of (or, alternatively, is shadowing ) if the existence of prohibits to be a static equilibrium point.

This intuition can be clarified by the following simple

###### Lemma 2.

Vertex is in the shadow of vertex if and only if

 (4) (ri−rj)ri<0.
###### Proof.

Condition (4) implies that the support plane of at will intersect , so can not be an equilibrium point. ∎

###### Corollary 1.

If vertex shadows vertex , then

 (5) |ri|<∣∣rj∣∣.

By Lemma 2, but

###### Lemma 3.

If the vertex is not in the shadow of any other vertex then has an unstable equilibrium point at .

###### Proof.

The support plane at defines two half-spaces. If the vertex is not in the shadow of any other vertex then the center of mass and all other vertices will be contained in the same half-space. This implies that the support plane of at does not intersect , so is an unstable equilibrium point of . ∎

###### Remark 1.

There are three possible shadowing relationships between two vertices and :

1. Vertex is shadowing vertex .

2. Vertex is shadowing vertex .

3. Vertex and vertex are not shadowing each other.

###### Definition 4.

The elements of the vertex shadowing matrix are defined as follows: If then if is shadowing and if is not shadowing . If then . Based on Lemma 2 we have

 (6) si,j=sign((ri−rj)ri).

Now we conclude the proof of Theorem 2.

###### Proof.

Based on Definition 4, for the absolute value of the row-sums of the shadowing matrix we can write

 (7) ∣∣ ∣∣V∑j=1si,j∣∣ ∣∣≤V−1

with equality if and only if is an unstable static equilibrium point of . According to the formula (2) of Theorem 2 in this case we add one to the value of . ∎

### 2.3. Proof of Theorem 3

###### Definition 5.

Let be two faces of , with respective face vectors and let denote the projection of the center of mass onto the plane of the face . We say that is in the shadow of (or, alternatively, is shadowing ) if the existence of prohibits to be a static equilibrium point.

###### Lemma 4.

Face is in the shadow of face if and only if

 (8) (qj−qi)qj<0.
###### Proof.

Condition (8) implies that the support plane of at will intersect , so can not be an equilibrium point. ∎

###### Corollary 2.

If face shadows face , then

 (9) ∣∣qj∣∣<|qi|.

By Lemma 4, but

###### Lemma 5.

If the face is not in the shadow of any other face then has a stable equilibrium point on the face .

###### Proof.

If is not in the shadow of any other face then will not tip to any other face if positioned on , so carries a stable equilibrium point. ∎

###### Definition 6.

The elements of the face shadowing matrix are defined as follows: If then if is shadowing and if is not shadowing . If then . Based on Lemma 4 we have

 (10) ¯si,j=sign((qj−qi)qj).

Now we conclude the proof of Theorem 3.

###### Proof.

Based on Definition 6, for the absolute value of the row-sums of the shadowing matrix we can write

 (11) ∣∣ ∣∣F∑j=1si,j∣∣ ∣∣≤F−1

with equality if and only if is a stable static equilibrium point of . According to the formula (3) of Theorem 3 in this case we add one to the value of .

## 3. The Algorithm

### 3.1. Problem statement

Formula (2) of Theorem 2 is, by default, capable to compute the shadowing matrix of a given polytope , based on position vectors , .

However, one can also use this formula to find monostatic polytopes with unstable equilibrium point.

### 3.2. Notation and definition of equations

###### Definition 7.

Let us denote the vertex at largest distance from the center of mass by .

###### Lemma 6.

Vertex always carries an unstable equilibrium.

###### Proof.

All vertices with are on the same side of the support plane at , so none of the vertices , may shadow . Via Lemma 2 this provides the proof. ∎

###### Lemma 7.

Let be a mono-unstable polytope with . Then, for any there exists at least one such that .

###### Proof.

We prove Lemma 7 by contradiction. If the statement of the Lemma is false then we have for some a row of the shadowing matrix with positive entries (except for the main diagonal). This means that vertex is not shadowed by any other vertex, i.e. it is an equilibrium (in addition to the equilibrium located at ). However, having equilibria contradicts the statement of the lemma. ∎

### 3.3. Material distribution

Since polytopes have multi-level geometric structure, material homogeneity may be interpreted in various ways. In subsection 1.2 we introduced -skeletons where material is uniform on the -faces of the polyhderon. Figure 2(a)-(c) illustrate -skeletons for 3-dimensional polyhedra. While Theorems 2 and 3 are valid irrespective of material distribution, if we want to utilize these results to find monostatic polytopes (as mentioned in subsection 3.1), the difficulty of the resulting algebraic system will strongly depend on the material distribution. In this paper we concentrate on 0-skeletons which yield the simplest balance equations. We also mention that computing for the homogeneous distribution in dimensions is identical to the problem of -skeletons if we only investigate simplices; however, the equilibria of that class of polyhedra are well understood. The dual case, corresponding to Theorem 3 is more challenging: none of the () homogeneous skeletons yields convenient algebraic equations. Nevertheless, as we will point out in Subsection 6.1, one may define a somewhat artificial distribution for which computations appear to be straightforward.

### 3.4. Control parameters

The problem is defined by 3 integer parameters: is the dimension of the polytope, defines the material distribution and provides the number of vertices. For brevity, we will characterize the problem with the triplet . If is not specified, then the triplet will refer to the problem of finding the smallest value of for which a monostatic polyhedron exists.

Now we regard the following set of equations:

 (12) si,j=−1,i=2,3,…V,j∈{1,2,…i−1}

(note the condition that will be justified below) and proceed by

###### Definition 8.

The -expansion of (12) is constructed in the following manner:

1. We construct systems, each consisting of inequalities with index running from to , defining systems by using admissible permutations of .

2. Each system is supplemented by moment equations guaranteeing that the center of mass of the -homogeneous polytope is at the origin.

So the -expansion of (12) in -dimension will consist of systems, each having inequalities and equations to which, for brevity, we will henceforth refer as a system of inequalities or system.

### 3.5. Possible scenarios

Solving the systems of inequalities and equations for the scalar coordinates , (, ) of the position vectors may yield two types of result:

1. If none of the systems has a solution, that implies the existence of a row for in the shadowing matrix with only positive entries. This is the necessary and sufficient condition for not to be monostatic, so it proves the nonexistence of -dimensional monostatic polytopes with vertices.

2. If any of the systems has a solution, that implies that for , in each row of the shadowing matrix there is a negative element. According to Lemma 7, this is the necessary and sufficient condition for to be monostatic. However, we remark that our algorithm does not include the condition that the solution should be convex, so this has to be checked additionally.

###### Remark 2.

The algorithm is only testing the lower triangle of the shadowing matrix, characterized by . It is easy to see that this is sufficient: If the problem has a solution, this means that there exists a monostatic, -dimensional, -homogeneous polytope with vertices. Based on Lemma 7, the shadowing matrix of a monostatic polytope has the property that for , in each row there will be at least one negative element . If all vertices are labelled such that , no negative entries in the upper triangle of can occur by Corollary 1. In other words, always carries an unstable equilibrium, is shadowed by , is shadowed either by or , etc., which finally yields possible shadowing systems.

### 3.6. Initial conditions

In dimensions a rigid body has

translational degrees of freedom and

rotational degrees of freedom. By fixing the origin at the translations are eliminated. In addition, we fix coordinates to eliminate the rotations and also to specify one element of a family of similar polytopes.

In dimensions we have , so, in addition to fixing the origin, we fix the coordinates of one point. In dimensions we have , so, in addition to fixing the origin, we fix all 3 coordinates of the first point and one additional coordinate of the second point. In dimensions we have , so, in addition to fixing the origin, we fix all 4 coordinates of the first point and 3 coordinates of the second point.

### 3.7. General system of inequalities for 0-skeletons in 3 dimensions, i.e. the (d,h,V)=(3,0,V) problem.

Equation (12) is equivalent to a system of inequalities which can be written as

 (13) d∑k=1r2i,k≤d∑k=1ri,krj,k,(i=2,…,V;j=1,…,i−1)

expressing that the -th point is shadowed by the -th point.

If we consider the problem then the balance equations for the center of mass can be written as:

 (14) V∑i=1ri,k=0,k=1,2,…d.

We remark that in case of a simplex, the balance equations for the and cases coincide. Using the considerations of subsection 3.6, we fix =4 coordinates:

 (15) r1,1=1r1,2=0r1,3=0r2,3=0,

which, by Definition 7 implies

 (16) |ri,k|<1,i>1,k=1,2,…d.

Henceforth, it is our goal to prove the unsolvability the system (13-15) for various values of .

## 4. Tetrahedron: the (d,h,V)=(3,3,4) and (d,h,V)=(3,0,4) problems

###### Lemma 8.

There exists no mono-unstable tetrahedron.

###### Remark 3.

The statement of the Lemma is equivalent to the claim that the -expansion of (13) has no solution. We will prove the latter.

Next, using Definition 8 we construct the -expansion of (13) in two steps.

In the first step, after substituting the initial conditions (15) we obtain the following groups of inequalities:

 (17) r22,1+r22,2≤r2,1
 (18)
 (19) ⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩r24,1+r24,2+r24,3≤r4,1r24,1+r24,2+r24,3≤r2,1r4,1+r2,2r4,2r24,1+r24,2+r24,3≤r3,1r4,1+r3,2r4,2+r3,3r4,3⎫⎪ ⎪ ⎪⎬⎪ ⎪ ⎪⎭

In the second step, we have possibilities to pick one inequality each from the three groups (17-19) and complementing these with the substitution of (14) yields 6 systems, each consisting of 3 equations and 3 inequalities. Here the moment balance equations for the and problem coincide since the tetrahedron is a simplex, so the above system describes both problems. Here follows the expanded system:

 (20) ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩(a)r22,1+r22,2≤r2,1(b)r23,1+r23,2+r23,3≤r3,1(c)r24,1+r24,2+r24,3≤r4,1(d)1+r2,1+r3,1+r4,1=0(e)r2,2+r3,2+r4,2=0(f)r3,3+r4,3=0⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭
 (21) ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩(a)r22,1+r22,2≤r2,1(b)r23,1+r23,2+r23,3≤r3,1(c)r24,1+r24,2+r24,3≤r2,1r4,1+r2,2r4,2(d)1+r2,1+r3,1+r4,1=0(e)r2,2+r3,2+r4,2=0(f)r3,3+r4,3=0⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭
 (22) ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩(a)r22,1+r22,2≤r2,1(b)r23,1+r23,2+r23,3≤r3,1(c)r24,1+r24,2+r24,3≤r3,1r4,1+r3,2r4,2+r3,3r4,3(d)1+r2,1+r3,1+r4,1=0(e)r2,2+r3,2+r4,2=0(f)r3,3+r4,3=0⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭
 (23) ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩(a)r22,1+r22,2≤r2,1(b)r23,1+r23,2+r23,3≤r2,1r3,1+r2,2r3,2(c)r24,1+r24,2+r24,3≤r4,1(d)1+r2,1+r3,1+r4,1=0(e)r2,2+r3,2+r4,2=0(f)r3,3+r4,3=0⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭
 (24) ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩(a)r22,1+r22,2≤r2,1(b)r23,1+r23,2+r23,3≤r2,1r3,1+r2,2r3,2(c)r24,1+r24,2+r24,3≤r2,1r4,1+r2,2r4,2(d)1+r2,1+r3,1+r4,1=0(e)r2,2+r3,2+r4,2=0(f)r3,3+r4,3=0⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭
 (25) ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩(a)r22,1+r22,2≤r2,1(b)r23,1+r23,2+r23,3≤r2,1r3,1+r2,2r3,2(c)r24,1+r24,2+r24,3≤r3,1r4,1+r3,2r4,2+r3,3r4,3(d)1+r2,1+r3,1+r4,1=0(e)r2,2+r3,2+r4,2=0(f)r3,3+r4,3=0⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭

Next we show that none of the systems (20)-(25) has a solution.

#### 4.0.1. Systems (20-22)

From part (a) of either of (20-22) we have , and so (d) with (16) imply both . This, however, contradicts (b), so systems (20-22) have no solution.

#### 4.0.2. System (23)

The proof is essentially identical to the proof for systems (20-22) but the contradiction is found at (c) instead of (b).

#### 4.0.3. System (24)

From (24)(a) we have , and this implies, via (24)(d) that . Substituting this into (24)(b) and (c) we get , respectively implying that and have the same sign. This, however, contradicts (24)(e), so system (24) has no solution.

#### 4.0.4. System (25)

The proof is similar to the proof of system (24). From (25)(a) we have , and this implies, via (25)(d) that . This yields, via (25)(b) and further, via (25)(e) . (25)(f) yields . Since we already showed that , substituting into (25)(c) yields a contradiction, so system (25) has no solution.

## 5. Computing the lower bound for Vu0: the (d,h,V)=(3,0,{5,6,7}) problems.

Our goal is to prove Theorem 1, to which we now give an equivalent, more detailed formulations as

###### Corollary 3.

The problems have no solution.

We prove Corollary 3 (and thus also Theorem 1) and Lemma 1) by applying the algorithm laid out in Section 3.

###### Proof.

We show that the system (13)-(14) has no solution for , The case of was proven by elementary considerations in Section 4. The case can also be proven in a similar way, however, only 7 of the 24 systems have such short proofs, the others require more steps: the total length is 22 (hand written) pages. The reason of that it is not included in this paper is that we have found another, optimization-based method which proves the unsolvability for and , too.

Departing from (14), can be expressed as

 (26) rV,k=−V−1∑i=1ri,k,k=1,2,3.

The substitutions (26) convert the system of inequalities (13) and equations (14) into inequalities:

 (27) 3∑k=1r2i,k−3∑k=1ri,krj,k≤0,i=2,…,V−1,
 (28)

where , resulting in systems.

According to (15), , and since , and were eliminated in (26), the number of free variables is . Let us define the parametric function as the weighted sum of the left hand sides of the inequalities (27)-(28) above

 (29) f=V−1∑i=2ci(3∑k=1r2i,k−3∑k=1ri,krj,k)+cV⎡⎣3∑k=1(−V−1∑i=1ri,k)2−3∑k=1(−V−1∑i=1ri,k)rj,k⎤⎦

, where the coefficients are arbitrary positive numbers.

Let us fix the values arbitrarily. If there exist positive coefficients such that is positive everywhere, then the unsolvability of system (27)-(28) follows: assume for contradiction that it has a solution, and substitute this solution in (29), yielding a non-positive value of , which is not possible once is positive.

How to find coefficients and how to check the positivity of ? Positive polynomials [15, 13] have many applications [14, 11].

However, in our case, suprisingly enough a simple randomized search for the coefficients works, at least up to vertices. We benefit from that if is convex, then its minimum value can be calculated simply. The positivity of can be proved by showing its convexity and finding a positive minimum. Since is a multi-variate polynomial of degree two, the first-order conditions of minimality yield a system of linear equations. In order to check convexity, the second-order condition is the positive definiteness of , the Hessian of . Note that depends on coefficients only, following again from that is quadratic.

The algorithm for proving unsolvability of the system of inequalities (27)-(28) is summarized below.

for all systems of inequalities
step 1. generate random positive integer values of c_2,...,c_n
step 2. if Hf is positive definite
then if min f > 0,
then print(’this system is unsolvable’)
go to the next system in the for cycle
else go to step 1
else go to step 1
end for


Appendices 1-4 include the coefficients and the minimum vale of function for all systems written for vertices, respectively. Since the minimum values are positive, all systems are unsolvable.

The computational approach above does not include numerical errors because all calculations deal with integer and rational numbers, resulting in exact rational numbers, too.

###### Remark 4.

In case of vertices, there are some systems of inequalities (e.g. system (27)-(28) written for the choice of for all ), where we could not prove that is positive with appropriate coefficients as all the random trials led to negative minima, and at the same time at least one inequality of the system (27)-(28) was violated – neither unsolvability of all of these systems, or solvability of at least one of these systems follows.

The lower bound for the number of minimal vertices, that a three-dimensional mono-unstable 0-skeleton must have, has now been improved to 8. Upper bounds may come when, for a given , a system of polynomial inequalities (27)-(28) has a solution, and the corresponding polyhedron is convex (see Subsection 3.5). Finding a solution of polynomial inequalities is itself a challenging problem [10, 19].

## 6. Concluding remarks

### 6.1. The dual problem: search for mono-stable polyhedra

So far we demonstrated how Theorem 2 can be converted into an algorithm to establish the lower bound for the number of vertices of a mono-unstable polyhedron. To illustrate the algorithm we computed the case of 0-skeletons and found the lower bound (Theorem 1).

Theorem 3 is the dual of Theorem 2, and it could, in principle, serve as the basis of a dual algorithm, searching for mono-stable polyhedra. However, as we will explain below, the only mass-distribution where this computation would be of comparable difficulty as 0-skeletons for mono-unstable polyhedra is physically rather counter-intuitive. To better understand the background, we point out the connection of our work to Dawson’s research on mono-stable simplices.

Starting on a proof by Conway for the non-existence of a mono-stable tetrahedron, Dawson  investigated the existence of mono-stable simplices in -dimensions. This research was continued in [5, 6] and ultimately led to the proof that for no mono-stable simplex exists and for there exist mono-stable simplices. The case is not yet resolved. The problem of the existence of mono-stable simplices is closely related to our problem and below we will point out the main connections as well as the main differences.

The key idea in Dawson’s arguments is what he calls the projection criterion:

 (30) |xi|<|xj|cosθij,

where is orthogonal to face of the simplex and measures the area of and is the angle between and . If the projection criterion (30) holds, then a - (or )-homogeneous simplex, if placed on face , will tip over to face , so we will call (30) the tipping condition. Next we show

###### Proposition 1.

In the case of homogeneous simplices and 0-skeletal simplices, equation (8) is equivalent to Dawson’s tipping condition (30).

###### Proof.

First we show that

 (31) |xi||qi|=d⋅Vol(s)d+1,

where denotes the volume of the simplex .

Let be a -dimensional simplex with center of mass , let be a -dimensional face of , let the center of mass of be denoted by and let be the vertex opposite . It is known  that (where denotes the length of the line segment ).

Let be a -dimensional simplex defined by the following vertices: face defines vertices and we add as the st vertex. is the height of orthogonal to . From this it follows that for . Since the left hand side of (31) is constant, (30) implies that

 (32) |qj|−|qi|cosθij<0,

and this yields (8) via

 (33) |qj|2−|qj||qi|cosθij<0.

As noted in Proposition 1, Dawson’s tipping condition (30) only applies if the polyhedron is either a homogeneous simplex or a 0-skeleton of a simplex and it is only equivalent to equation (8) under the same condition. Now we show that there exists a material distribution for which Dawson’s tipping condition is reversed: if (30) holds then the simplex will tip from face to face .

###### Definition 9.

We will call a -dimensional polytope with faces a dual 0-skeleton if , with the vectors , () defined in subsection 2.1.

The balance equations for the center of mass of a dual 0-skeleton can be written as

 (34) F∑i=1qi,k=0,k=1,2,…d,

which is analogous to equation (14). Based on Theorem 3 and the balance equations (34) we can construct the dual version of the algorithm presented in section 3.

The name for dual 0-skeletons is motivated by their role in case of simplices where the dual 0-skeleton is uniquely defined and corresponds to a center of mass in the interior of the simplex. In fact, it is straightforward to find the center of mass for the dual 0-skeleton in a simplex: let be a -dimensional simplex and we regard the vectors ( introduced in equation (30). Assume all have identical origin and we denote the endpoints of the vectors by . Now we regard the set of planes each of which is normal to the corresponding vector at the endpoint . It is easy to see that the simplex defined by these planes will be similar to , the set of vectors for can be defined as . If we place unit masses at the points then this mass distribution defines the dual 0-skeleton of . Dawson proved that a homogeneous (or homogeneous 0-skeleton) simplex can tip from face to face if and only if his tipping condition (30) is true. In case of a dual 0-skeleton simplex Dawson’s tipping condition is reversed: it implies the opposite, i.e. that it would tip from face to face .

###### Proposition 2.

For the dual 0-skeleton of any -dimensional simplex the tipping condition (30) is reversed.

###### Proof.

The construction scheme described at Definition 9 proves that the center of mass for the dual-0-skeleton of a -dimensional simplex is an interior point of and with holds for any pair . Expressing the face shadowing condition (8) in terms of yields , which can be obtained from (30) by interchanging subscripts and . ∎

In case of -dimensional polytopes with faces the center of mass corresponding to a dual 0-skeleton may not be in the interior of the polytope. Figure 3 illustrates the 0-skeleton and the dual 0-skeleton of a triangle and also shows a quadrangle where the center of mass for the dual skeleton is not contained in the interior. Figure 3. Illustration of 0-skeletons and dual 0-skeletons in 2D. (a) 0-skeleton of triangle, (b) uniquely defined dual 0-skeleton of the same triangle (c) 0-skeleton of a quadrangle (d) possible dual 0-skeleton of the same quadrangle. Large black dots mark unit masses.

Summarizing, we may say that the utilization of Theorem 3 as the basis of an algorithm to find mono-stable polyhedra is computationally feasible only if we investigate dual 0-skeletons, however, the latter (except for the case of simplices) may not be physically relevant, so the dual algorithm, based on Theorem 3 does not appear to be of practical interest. Since the attention was previously focused on mono-stable polyhedra (and the mono-unstable case was not considered), this observation may explain why this algorithm was not investigated earlier.

### 6.2. Summary

In this paper we improved the previously known [3, 9] lower bound on the minimal number of vertices for a convex, mono-unstable 0-skeleton to . This result also implies the same lower bound for the minimal number of vertices for a convex mono-monostatic 0-skeleton, so we also proved .

On one hand, we think that these lower bounds are not yet close to the actual minimal values. On the other hand, although no mathematical evidence exists, intuitively it looks plausible that the same lower bounds are valid for , i.e., for homogeneous polyhedra.

The algorithm presented in this paper is, in principle, also capable to compute , however, we did not yet attempt to implement the balance equations for this case. Also, the same algorithm is (again, in principle) capable to compute higher dimensional problems. Dawson  investigated the minimal dimension in which a simplex may be mono-stable and in our notation his result can be written for dimensions as .

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## Appendix 1: V=4

 i=2 i=3 j j c2 c3 c4 minf 1 1 94 46 97 1560613/17904 1 2 45 101 39 389290/18397 1 3 70 28 97 696052/63615 2 1 18 91 35 1732/417 2 2 19 53 57 79192/11149 2 3 44 84 26 137704/9511

## Appendix 2: V=5

 i=2 i=3 i=4 j j j c2 c3 c4 c5 minf 1 1 1 99 60 45 101 3607295/47108 1 1 2 39 70 28 97 1181819/48726 1 1 3 18 91 35 19 266291/25140 1 1 4 53 57 44 84 4575347/142162 1 2 1 94 15 51 48 12918697/714980 1 2 2 71 78 40 93 15857847/752306 1 2 3 72 68 79 52 40813288/1367773 1 2 4