## Rigid polygons, movable polygons and other polygonsFirst of all, let's have a new look at one of the criteria for the congruence of triangles. The proposition
expresses the following fact:
f1, f2 and f3
which transform AB onto A'B', BC onto B'C' and
CA onto C'A' respectively, then all these transformations
can be performed by a single isometric map f.
An
We can easily see that triangles are the only rigid polygons.
The rigidity of polyhedrons is defined similarly, as the possibility
of replacing a sequence of isometric maps between faces by a single
one. Fig. 1 shows a polyhedron that is not rigid.
Fig.1. A cube with one face replaced by a pyramid
pointing outside and another with the pyramid pointing inside.
Again not much effort is needed to see that a regular tetrahedron
is rigid and so is the cube or the regular octahedron, whereas
the regular icosahedron is not.
Among the non-rigid polygons or polyhedrons those that are Here are two known facts about rigidity and movability of polyhedrons: - convex polyhedrons are rigid (Augustin Louis Cauchy, 1813)
- there are movable polyhedrons (Robert Connelly, 1977)
To make Cauchy's result more precise: both the polyhedron
In the sequel we present a sketch of Cauchy's proof and recall
some examples of movable polyhedrons.
The first step in the proof consists in showing that two convex
polyhedrons with correspondingly congruent sides and correspondingly
congruent dihedral angles are congruent too. This is not hard
to prove, so we leave the details to the readers. The idea is
to choose one face and gluing the other faces to it one by one;
convexity and the congruence of dihedral angles should be sufficient
to make the construction unique. Thus to complete the proof of
the theorem we may just prove that the dihedral angles must be
equal.
For that purpose we will make use of spherical polygons: they
lie on a sphere, their sides are arcs of great circles, which
are those that belong to planes passing through the centre of
the sphere, and their angles are dihedral angles between those
planes. We will also use two observations concerning those polygons.
Let's assume that in an
The proof is inductive.
1. For
Fig. 2 2. Whenever there is equality between the angles, we cut off equal corners from both polygons (see Fig. 2) and thus reduce the problem to the inductive assumption.
3. In the remaining cases we will increase the angle at the vertex
3.1 If this increase does not alter convexity (Fig. 3), the inductive
step consists in applying steps 1 and 2 (you check it for yourself,
can't you?).
Fig. 3. The situation is good on the left and bad
on the right.
3.2. If the polygon thus obtained were not convex, we proceed
differently: turn the side
Fig. 4. This is the case when the angle
By the inductive assumption for (
Similarly, for triangles ,
which completes the proof of Observation 1.
The next observation will be preceded by a definition, too. Given
two spherical
The number of sign changes being pair, we must only exclude 2.
Suppose that this is precisely the number of sign changes for
Let's return to the proof of Cauchy's theorem. Consider a function
transforming a polyhedron
Take a small sphere with the centre in one of the vertices of
the polyhedron (small means "containing only one vertex").
The intersection of the polyhedron with this sphere is a convex
spherical polygon. We repeat the construction at every vertex,
which yields (due to the function
at the corresponding vertices of
The angles of the spherical polygon thus obtained being dihedral
angles of the solid, we may extend the function
to the edges of the polyhedron
We will consider here the case when none of the values of
is zero, thus leaving to the readers the problem of getting rid
- by drawing something, forgetting something - of unnecessary
edges, i.e. those for which the value of
is zero. By Cauchy's lemma we know that the number of sign changes
is in each spherical polygon is not less than 4; thus in the entire
polyhedron the number
Now taking into account the equality ,
where
This, however, implies that ,
which |