Rigid polygons, movable polygons and other polygons
First of all, let's have a new look at one of the criteria for the congruence of triangles. The proposition
if , then
expresses the following fact:
if there are isometric maps 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 n-sided polygon is said to be rigid if it satisfies the following condition:
whenever there are isometric maps f1, f2, ..., fn transforming the sides of the polygon onto the corresponding sides of another one, then there is a single such map f that does all these transformations.
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 movable
(the so-called flexors) can be distinguished. In this class
not only are there figures that have congruent sides or faces
without being congruent themselves, but there is also a continuous
transition from one of two such figures to the other which preserves
congruent sides or faces at every step. It can be easily deduced
that each non-rigid polygon is movable, and there certainly are
non-rigid polyhedrons that are not movable, as is the case with
the solid of Fig. 1.
Here are two known facts about rigidity and movability of polyhedrons:
To make Cauchy's result more precise: both the polyhedron and
its image are expected to be convex (see Fig. 1).
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
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 n-sided polygon the side XiXi+1
has length ai (for i=1,
2, ..., n-1), the length of XnX1
is an, and the angle at the i-th
vertex is .
If two convex spherical n-sided polygons satisfy the equalities for i=1, ..., (n-1) and for i=2, ..., (n-1), then . Moreover, if at least one of the inequalities between the angles is strict, the inequality between the sides is strict, too.
The proof is inductive.
1. For n=3 the proof is the same as for plane triangles
(how does this proof go, dear reader?)
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 Xn-1 up to .
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,
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 Xn-1 Xn
around Xn-1 until the figure becomes
convex again (Fig. 4).
Fig. 4. This is the case when the angle Xn-2
is strictly less than
and the points X2,
lie on the same arc.
By the inductive assumption for (n-1)-sided polygons X2, ..., Xn-1, X'n and we have .
Similarly, for triangles X1, Xn-1, Xn and X1, Xn-1, X'n we have . Altogether this yields
which completes the proof of Observation 1.
The next observation will be preceded by a definition, too. Given
two spherical n-sided polygons S and ,
the function assigns
to i the value +1 when ,
the value 0 when , and
the value -1 otherwise. We say that a sign change occurs, if
Observation 2 (Cauchy's lemma)
For any S and such that for i=1,..., n, either no sign change occurs or there are at least 4 such changes.
The number of sign changes being pair, we must only exclude 2.
Suppose that this is precisely the number of sign changes for
S and . Let the
vertices Xi and Xj
separate that part of the polygon which has
from that with and
call the two parts P1 and P2, correspondingly. Applying
Observation 1 to P1 and ,
we get , while an application
of the same observation to P2 and
yields - a contradiction.
Let's return to the proof of Cauchy's theorem. Consider a function
transforming a polyhedron
H onto a polyhedron
so that each face is congruent to its image. We must prove that
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 Hand convex
spherical polygons with the corresponding sides equal. Thus both
Observations can be applied to the polyhedrons.
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 H by defining (still on
the fixed sphere) for an edge k the function ,
if at the corresponding vertex Xi of
the corresponding spherical polygon ,
where = -1, 0, +1.
Now we say that two edges are adjacent if they belong to the same
face and have a common endpoint. Define a function
on adjacent edges k and l so that .
Our aim now is to prove that the function
is constant with value zero.
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 N of such changes is not less than
4V, where V is the number of vertices in the solid.
On the other hand, if Si is the number
of i-sided faces of the polyhedron, we have (why?).
Now taking into account the equality ,
where E is the number of edges in the polyhedron, we arrive
This, however, implies that ,
which contradicts Euler's formula! And this is the end.