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Polyhedra
Seminar on Euclidean Geometry
Fabian Frei
May 28, 2015
”Of course, defining something is no guarantee of its existence (think of a
unicorn, for example)” (Hartshorne)
Figure 1: The five Platonic solids identified with the four elements earth, air, fire, water
and the whole universe.
Contents
1 Introduction
2
2 The Five Regular Solids (Chapter 44)
2
3 Euler’s Theorem (Chapter 45)
7
4 Cauchy’s Rigidity Theorem
9
5 Application
13
5.1 Verification of Euler’s theorem for the five regular solids . . . . . . . . . . 13
5.2 Verification of Descarte’s theorem for the five regular solids . . . . . . . . 13
6 Bibliography
14
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1 Introduction
Polyhedra are solid figures bounded by planar polygons. Euclid defined the tetrahedron,
cube, octahedron, icosahedron and dodecahedron by the number and type of faces they
have. He then constructed each one inscribed in a sphere and claims that only these five
are possible. In order to prove this the hypothesis of convexity has to be supplied. In
addition it can be shown then that all dihedral angles are equal.
2 The Five Regular Solids (Chapter 44)
Repetition (Regular Polygons):
A triangle with three equal sides (equilateral), also has three equal angles (I.5). A
triangle with three equal angles (equiangular) has three equal sides (I.6). The same
does not hold for polygons with four or more sides. There, having equal sides does not
imply equal angles, as having equal angles does not imply equal sides.
Definition 2.1 (Regular polygon in the plane). A regular polygon in the plane is a
polygon that is both equilateral and equiangular.
Proposition 2.2 (n-gon). In the real Cartesian plane, for any n ≥ 3,
• there exists a regular polygon of n sides (n-gon) having a given segment as a side,
• any two regular n-gons with a common side are congruent,
• the vertices of the regular n-gon lie on a circle and
• for any two vertices, there is a rotation of the n-gon to itself sending the first vertex
to the second.
Remark 2.3. To be able to divide the angle 2π into n equal pieces, one needs to use
the real numbers. In the Cartesian plane over the field of constructible numbers, or for
those regular polygons that are constructible using compass and marked ruler, not all
values of n are possible (see Sections 29 − 31).
Proof. Mark n equidistant points on the circumference with a circle resulting angles
of 2π/n at the center. Expanding by a scale factor will make the side equal to any
given segment, therefore the existence is proven. For the uniqueness, let there be a
given regular n-gon with side AB. Bisect the equal angles at A and B and let the
angle bisectors meet at O as shown in Figure 2. O is equidistant from A and B. After
repeating this construction at the other vertices, it is obvious that O is equidistant from
all the vertices. Therefore they all lie on a circle with center O. Thus any two regular
n-gons with a common side will be congruent.
A rotation through an angle of 2π/n at the center will send one vertex to the next.
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Figure 2: Construction
Remark 2.4. Multiples of above rotation send the first vertex to any desired vertex.
Remark 2.5. These elementary observations about regular polygons emphasize the
analogy with the three-dimensional case following in the next chapter.
The definition of the regular solids will follow at the end of this section. First there
needs to be done some pre work.
Definition 2.6.
1. A pyramid is a solid figure formed by joining a point to each of the vertices of a
polygon in a plane not containing the point.
2. A tetrahedron is a triangular pyramid formed of four equilateral triangles.
3. A cube is a solid figure contained by six squares.
4. A octahedron and a icosahedron are solid figures bounded by 8 respectively 20
equilateral triangles.
5. A dodecahedron is a figure bounded by 12 regular pentagons.
Convention (Terminology).
• A polyhedron is the surface of a solid figure bounded by plane polygons.
• When two polygons meet, they must have an entire edge in common.
• These plane polygons are the faces of the polyhedron. Their edges are the edges
of the polyhedron, and their vertices are the vertices of the polyhedron.
• Where two faces meet along an edge, there is a so called dihedral angle as shown
in Figure 3.
• At a vertex, the angle in any face passing through that vertex is called a face
angle.
• The collection of all the faces at a vertex makes a solid angle. The solid angle is
not measured by a number, but it can be congruent to another if there is a rigid
motion of the space making one coincide with the other close to the vertex.
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Figure 3: Dihedral angle
• If all the vertices of a polyhedron lie on a sphere, the polyhedron is inscribed in
the sphere.
• If all the faces of a polyhedron are tangent to a sphere, the polyhedron is circumscribed about the sphere.
Proposition 2.7. There exist tetrahedra, cubes, octahedra, icosahedra and dodecahedra
having the following properties:
(a) In each figure, all the dihedral angles are equal.
(b) The vertices of each figure lie on a sphere.
(c) For any two vertices, there is a rigid motion of the figure onto itself sending the first
vertex to the second.
Proof. The proof is shown here for cubes (Exercise 44.1). For all others take a look at
the book (Robin Hartshorne, ”Geometry: Euclid and Beyond”, Springer 2000).
To make a cube, take a square of side 1. At each vertex erect a line perpendicular
to the plane of the square. On those lines, find the points at distance 1 from the
corresponding vertex (all lying on the same side of the plane of the square). Connect
the four new points, resulting in 5 new faces and together with the originally square to
a solid. Clearly all edges have length 1 and the solid is a cube. The dihedral angles
between all the faces are perpendicular by construction. Hence all the dihedral angles
are equal. Erect a line perpendicular to the plane of the original square at its center
find the point whose distance to the plane is 1/2 results in the center of the cube. This
follows from the fact, that this construction can be done with any of the faces, resulting
always the same point. Therefore all vertices lie on a sphere, so the cube is inscribed in
a sphere. A rotation of π/2 about an axis passing through the center of two opposite
squares will send any one vertex to the next.
Remark 2.8. All of the steps in the above construction could be carried out with ruler
and compass in suitable planes. In particular, finding a point on a line at unit distance
from a given point is just a matter of intersecting the line with a circle. Hence these
polyhedra are all constructible with Euclidean tools. In particular, in the real Cartesian
three-space, their coordinates would lie in the field K of constructible number. Put
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otherwise, these five solids exist in the Cartesian three-space over the field K.
Proposition 2.9. In a convex polyhedron all of whose faces are equal regular polygons,
the only possible configurations at a single vertex are 3, 4, or 5 triangles, 3 squares, or
3 pentagons.
Remark 2.10. Note that the condition of convexity is inserted here and necessary
in order to prove the statement. Euclid himself reasoned without this condition, and
therefore did not show an exact proof.
Proof. The key observation for this proof is that the interior angles of the polygons
meeting at a vertex of a polyhedron add up to less than 2π. (Compare also with the
definition of the defect δV in Definition 3.2)
Take an equilateral triangle. The interior angle of an equilateral triangle is π3 . Thus on
regular polyhedron only 3 (Tetrahedron), 4 (Octahedron) or 5 (Icosahedron) triangles
can meet at a vertex. If there were more than 6, their angles would add up to at least
2π, which they cannot. (i.e. 6 triangles at one vertex would lie flat).
Take a square. Since the interior angle of a square is π/2, at most three squares can
meet at a vertex (Cube).
Take a regular pentagon. As it the previous case the only possibility is that three
pentagons meet at a vertex (Dodecahedron).
Polygons with more than six sides cannot form the faces of a regular polyhedron since
their interior angles are at least 2π
3 .
Theorem 2.11. Any polyhedron that is
(a) bounded by equal regular polygons,
(b) convex,
(c) has the same number of faces at each vertex,
is congruent (up to a scale factor) to one of the five constructed in Proposition 2.7.
Furthermore, these five all have the additional properties
(d) all dihedral angles are equal,
(e) the vertices lie on a sphere,
(f) for any two vertices, there is rigid motion of the figure taking one to the other.
Lemma 2.12. If a vertex V of a polyhedron has three faces meeting it, with face angles
α, β, γ, then the three dihedral angles between these faces are uniquely determined by
α, β, γ. In particular, if α, β, γ are equal, the three dihedral angles will also be equal.
Proof. Take Points A, B, C equidistant from V (say distance 1) on the three edges meeting at V . Holding MAV B fixed, let MAV C rotate around the line AV . As it does so, the
point C describes a circle on the unit sphere with center V . Similarly as MBV C rotates
around BV , the point C describes another circle on the unit sphere centred at V . The
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point where these two circles meet is C (or its reflection in the plane of MAV B ). Now it
is clear that all three dihedral angles are uniquely determined by the three face angles
of the triangles at V . (since there are only two (congruent) solutions with the given face
angles.)
Remark 2.13. The conclusion of the Lemma is false for four or more faces meeting at
a vertex.
Proof. (Theorem) Since the figure is convex and has the same number of faces at each
vertex Proposition 2.9 can be applied. There are five cases.
Case 1 : Three equilateral triangles at each vertex. Let A be one of the vertices and
let B, C, D be the adjacent vertices. Since the sides AB, AC, AD, BC, BD, BC are all
equal B, C, D form an equilateral triangle. The vertices B, C, D each have two triangles
already, so BCD makes the third and the whole figure has just these four equilateral
triangles as faces. Because of the previous Lemma 2.12, the dihedral angles are the same
as those in the tetrahedron constructed in Proposition 2.7, so the two tetrahedra are
congruent up to scale factor. Properties (d), (e), (f ) follow from Proposition 2.7.
Case 2 : Three squares at each vertex. By the Lemma 2.12 the dihedral angles are
uniquely determined. In this case they are right angles. Starting at one vertex, the
three square faces fit on a cube. Continuing to adjacent vertices, the faces of the solid
must coincide with those of the cube; hence it is a cube.
Case 3 : Three regular pentagons at a vertex. Similar to Case 2 the dihedral angles are
uniquely determined. They must coincide with the dihedral angles of the dodecahedron
constructed in Proposition 2.7. Starting at one vertex and working the way around, the
figure must coincide with that dodecahedron. The properties (d), (e), (f ) follow from
Proposition 2.7.
Case 4 : Four equilateral triangles at a vertex. Here the Lemma does not apply. Let
A be one vertex. Since there are four faces meeting at A, there are four adjacent
vertices B, C, D, E. At B there are two of the faces (MABC and MABE ) and three of
the edges (AB, BC and BE), so there must be another vertex F s.t. MBCF and MBEF
are equilateral triangles. At C there are three of the required four faces. It follows
that MCDF must be an equilateral triangle forming the fourth face. (This can be seen
from the fact that F has to be the reflection of A at the plane of the square BCDE.
Otherwise F couldn’t be equidistant to B, C and D) Similarly, MDEF is the fourth face
at E. The figure is now complete and bounded by eight equilateral triangles in the same
arrangement as the octahedron of Proposition 2.7. In order to prove the congruence let
A, F be opposite vertices of the figure. Then B, C, D, E are all equidistant from A and
F , so they must lie in the plane that bisects the segment AF . They are also equidistant
from A, so they lie on a sphere with center A that intersects that plane in a circle. So
B, C, D, E lie on a circle. But they are also equidistant from each other, in order, so
they form a square. Let O be the center of the square. Then the segments AF, BD, CE
all pass trough O, and are mutually perpendicular. If an edge of the figure has length
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Figure 4: Octahedron in Case 4
√
1, then O√ is at distance 2/2 from B, C, D, E and it is easy to see that O is also at
distance 2/2 from A and F . Thus A, B, C, D, E, F lie on a sphere with center O. This
was the same construction as in Proposition 2.7. The properties (d), (e), (f ) follow.
Case 5 :(Exercise 44.5) Five equilateral triangles at each vertex. Let A be one vertex.
Since there are five faces meeting at A, there are five adjacent vertices B, C, D, E, F . At
B there are two of the faces and three of the edges, so there must be two other vertices
G and H s.t. MBGH , MBCH , MBF G are equilateral triangles. At C there are three of the
required five faces. There must be another vertex I s.t. MCHI and MCDI are equilateral
triangles. It follows that there are vertices J and K s.t MDIJ , MDEJ , MEJK , MEF K , MCHI
and MCHI are all equilateral triangles. The new vertices G, H, I, J, K all have three of
the required five faces. Therefore the vertex L is added and connected to all these points,
such that the result are five equilateral triangles. The final convex figure has 12 vertices
(A, B, C, D, E, F, G, H, I, J, K, L) and 20 faces. It remains to show that this constructed
figure is congruent to the icosahedron of Proposition 2.7. The properties (d), (e), (f ) will
follow in that case. To complete the proof Cauchy’s rigidity theorem is necessary. This
will follow in chapter 4.
Definition 2.14. A regular polyhedron is a convex polyhedron whose faces are all
equal regular polygons and having the same number of faces meeting at each vertex.
It follows from Theorem 2.11 that the only regular polyhedra are the five Platonic
solids constructed in Proposition 2.7, and that they all have the extra properties (d),
(e), (f) of Theorem 2.11. Alternatively, one could define a regular polyhedron to be a
convex polyhedron all of whose faces are equal regular polygons and that satisfies any
one of the properties (d),(e),(f) of Theorem 2.11.
3 Euler’s Theorem (Chapter 45)
Theorem 3.1 (Euler). Given a convex polyhedron, let ν be the number of vertices, let
e be the number of edges, and let f be the number of faces. Then
ν−e+f =2
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Seminar in Euclidean Geometry
Proof. First remove one face from the polyhedron and do not forget that in the calculation of equation (1). Think of the remaining faces of the polyhedron as made of rubber
and stretched out on a table. It will certainly change the shape of the polygons and
the angles involved, but it will not alter the number of vertices, edges, and faces. Now
perform two kind of operations on this figure.
(a) Choose an edge that separates two faces, or that separates one face from the area
outside the figure, and remove that edge. This decreases the number of edges by one,
and also decreases the number of faces by one, since two faces are now joined together.
So the expression ν − e + f is unchanged.
(b) If at some point in the procedure there is a vertex that has only one edge coming out
of it, remove that vertex and that edge. This decreases both v and e by one, so again
the expression ν − e + f is unchanged.
Repeat step (a) as many times as possible. If step (a) is no longer possible, then there
are no loops in the remaining graph of vertices and edges, so there must be some ends
to the graph (since the graph is finite). Then do step (b). If there are no edges at
all remaining, then the figure must be reduced to points only. But the original figure is
connected, and it remains connected by performing step (a) or (b), so it is just one point.
Then ν = 1, e = 0, f = 1 (the face, that was removed at the beginning), so ν − e + f = 2
Since the expression ν − e + f was unchanged by all the steps, the original ν − e + f is
equal to 2 as required.
In the previous chapter it was shown that the sum of the face angles of the faces at
a vertex must be less than 2π, as it was also used in the proof of Proposition 2.9. Now
take a look at de following angle.
Definition 3.2. The defect δV at the vertex V is defined to be 2π minus the sum of
the face angles at V .
Remark 3.3. The defect δV is always positive.
Corollary 3.4 (Descartes). In a convex polyhedron, the sum of the defects at all the
vertices is equal to 4π.
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Proof.
X
δV =
V
X
X
2π −
(face angles at V )
V
= 2πν −
X
= 2πν −
X
(all face angles)
(n − 2)πfn
n
= 2πν − π
X
nfn + 2π
n
X
fn
n
= 2π(ν − e + f )
(1)
= 4π,
P
where fn is the number of faces having n sides, the total number of faces is f = fn ,
P
the number of edges e is 12 nfn and the sum of the face angles of an n-sided polygon
is (n − 2)π
Remark 3.5. This result is a powerful tool in that it limits the possible vertices with
a particular configuration of faces in a convex polyhedron.
4 Cauchy’s Rigidity Theorem
Problem: Suppose having two convex polyhedra, made of congruent faces, similarly
arranged. You wish to conclude that the two polyhedra are congruent.
Definition 4.1. Let ϕ be a one-to-one correspondence from the set of faces F1 , ..., Ff of
0
0
a first polyhedron to the set of faces F1 , ..., Ff of a second polyhedron, such that for each
i, Fi is congruent to ϕ(Fi ) and furthermore ϕ extents to one-to-one correspondences of
vertices and edges preserving all incidence relations.
Remark 4.2. This formulation allows mirror images, so that congruence means rigid
motion in space followed possibly by reflection in a plane.
Cauchy’s idea was to study how the dihedral angles compare along corresponding
edges. If all the dihedral angles are the same, then the two polyhedra can be build step
by step into congruent figures. On the other hand, if the dihedral angles change, then
one has to track their increase or decrease around each vertex and use Euler’s theorem
to make a contradiction. Intersection of the faces of the polyhedron with a small sphere
around the vertex produces a spherical polygon whose interior angles are precisely the
dihedral angles of the original polyhedron. This is called the vertex figure at the vertex
V . This leads to polygons with changing angles.
Lemma 4.3 (Steinitz). Suppose given in the plane two convex polygons A1 A2 · · · An
and B1 B2 · · · Bn with all sides equal except possibly the last: Ai Ai+1 = Bi Bi+1 for all
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i = 1, 2, ..., n − 1. Suppose also that the angles of the first polygon are less than equal to
the angles of the second, ∠Ai ≤ ∠Bi , for i = 2, ..., n−1 with at least one strict inequality.
Then A1 An < B1 Bn .
Proof. Induction on n
Case 1 For n = 3 it is Euclid (I.24)
Case 2 Suppose that n ≥ 4 and for some i that ∠Ai = ∠Bi . Then MAi−1 Ai Ai+1 is
congruent to MBi−1 Bi Bi+1 . So Ai−1 Ai+1 = Bi−1 Bi+1 and the result follows by applying
the induction hypothesis to the new polygons of n − 1 vertices obtained by omitting Ai
and Bi .
Figure 5: Situation in Case 2
Case 3 Suppose that n ≥ 4 and all the B angles are strictly bigger than the A angles.
0
0
0
Construct a point A1 s.t. A1 A2 = A1 A2 and ∠A1 A2 A3 = ∠B2 . We compare A1 ...An
0
to A1 A2 · · · An and then the latter to B1 · · · Bn . In the first comparison, ∠A3 = ∠A3 ,
0
0
so A1 A2 < A1 A2 by Case 2. In the second comparison ∠A2 = ∠B2 , so A1 A2 < B1 B2
again by Case 2.
0
Case 4 If the new polygon A1 A2 · · · An in Case 3 is not convex, there must be a point
0
A∗1 intermediate between A1 and A1 , for which A∗1 A2 = A1 A2 and A∗1 , An , An−1 are
collinear.
Figure 6: Situation in Case 4
First compare A1 · · · An to A∗1 A2 · · · An and obtain A1 An < A∗1 An by Case 2. Next
compare the polygon A∗1 A2 · · · An−1 to B1 B2 · · · Bn−1 and obtain A∗1 An−1 < B1 Bn−1 by
the induction hypothesis. On the other hand, since A∗1 , An , An−1 are collinear A∗1 An =
A∗1 An−1 − An−1 An . Putting these together:
A1 An < A∗1 An
= A∗1 An−1 − An−1 An
< B1 Bn−1 − Bn−1 Bn
≤ B1 Bn ,
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where the last inequality is just the triangle inequality (I.20). Thus the proof is complete.
The analogous result for polygons on the surface of a sphere:
Lemma 4.4. Let A1 A2 · · · An and B1 B2 · · · Bn be two convex polygons in the plane
or on the sphere, with corresponding sides equal: Ai Ai+1 = Bi Bi+1 for all i = 1, 2, ..., n
(interpreting n + 1 = 1). For each i mark the vertex Ai with + if ∠Ai < ∠Bi , with −
if ∠Ai > ∠Bi , or with no mark if the angles are equal. Then either all corresponding
angles are equal, or in a circuit of the first polygon - ignoring unmarked vertices - the
sign must change at least four times.
Figure 7: Example on how to mark the vertices
Proof. The number of changes is even. Assume there are exactly two changes of sings.
Draw a diagonal Ai Aj as shown in Figure 8. Applying the previous Lemma to the two
resulting convex polygons leads to a contradiction since for the one containing only −
vertices follows Ai Aj > Bi Bj and for the + side: Ai Aj < Bi Bj .
Figure 8: Polygon with only two changes of sign and the diagonal Ai Aj
Theorem 4.5 (Cauchy’s Rigidity Theorem). Suppose there are given two convex polyhedra and a one-to-one map ϕ from the set of faces of one to the other, so that corresponding faces are congruent and are similarly arranged (as described above). Then the
two polyhedra are congruent.
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Proof. For each edge of the first polyhedron, mark it +, −, or no mark, according as its
dihedral angle is less than, greater than, or equal to the corresponding dihedral angle
of the second polyhedron. At each vertex intersect the polyhedron with a small sphere,
and look at the resulting vertex figure. This is a convex spherical polygon, and its
vertices inherit markings + or − from the edges, which by construction correspond to
the increase or decrease of this polygon angles as compared to the vertex figure of the
second polyhedron. Conclude from Lemma 4.5 that for each vertex, if one makes a
circuit of the edges meeting that vertex, either they are all unmarked, or there are at
least 4 changes of sings. Counting the total number of changes of sign in two different
ways will lead to a contradiction
Case 1 Suppose that all edges are marked + or −, in other words, all the dihedral angles
are changing. Let t denote the sum over all the vertices of the number of changes of sign
of edges around that vertex. Then by the Lemma 4.5 clearly t ≥ 4ν.
On the other hand, count by faces. On a triangular face, two adjacent edges must
have the same sign, so that face can contribute at most two changes of sign to its three
vertices. Similarly, a face of n sides can contribute at most n changes of sign if n is even,
or n − 1 if n is odd. In particular,
t ≤ 2f3 +
X
nfn ,
n≥4
Putting together the two inequalities for t gives
4ν ≤ 2f3 +
X
nfn .
n≥4
Now use Euler’s theorem, i.e. ν = e − f + 2 and substitute e =
This gives
X
X
2 (n − 2)fn + 8 ≤ 2f3 +
nfn ,
1
2
P
nfn and f =
P
fn .
n≥4
or
X
(n − 4)fn + 8 ≤ 0,
n≥4
which is impossible, because the terms in the sum are all non-negative.
Case 2 Now suppose there are some marked and some unmarked edges. Of course, if no
edges are marked, all the dihedral angles are equal, and the two polyhedra are congruent.
We will imitate the previous proof using only those vertices and edges that are marked.
This a net. The vertices of the net are those vertices of the polyhedron that have a
marked edge coming out of them. The edges of the net are the marked edges of the
polyhedron. A net-face of the net is any maximal union of faces of the polyhedron that
are not separated by edges of the net. A net-face is no longer a plane polygon, but it is
a connected surface bounded by edges of the net.
Now repeat the previous argument using only the vertices, edges and net-faces of the net.
Denote the numbers of these by ν 0 , e0 , f 0 . The argument is all the same, except for the
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application of Euler’s theorem, which does not apply as stated, because a net may not be
a polyhedron. Nevertheless applying the proof of Euler’s theorem to the net is possible.
The only difference is that the plane figure of points and lines may not be connected. At
the end of the proof, there may be more than one point, i.e. ν 0 − e0 + f 0 ≥ 2. With this
inequality, the argument of Case 1 still works. So there is the desired contradiction. The
only remaining possibility is that all the dihedral angles are equal, so the two polyhedra
are congruent.
Remark 4.6. This result is false without the hypothesis of convexity.
An application of the theorems is the following theorem which is stated here without
proof.
Theorem 4.7. There are exactly eight convex polyhedra all of whose faces are equilateral
triangles. Each one is uniquely determined up to congruence, once the length of an edge
is specified.
5 Application
Exercise 45.1
5.1 Verification of Euler’s theorem for the five regular solids
Tetrahedron: ν = 4, e = 6, f = 4 ⇒ 4 − 6 + 4 = 2
√
Cube: ν = 8, e = 12, f = 6 ⇒ 8 − 12 + 6 = 2
√
Octahedron: ν = 6, e = 12, f = 8 ⇒ 6 − 12 + 8 = 2
√
Icosahedron: ν = 12, e = 30, f = 20 ⇒ 4 − 6 + 4 = 2
√
Dodecahedron: ν = 20, e = 30, f = 12 ⇒ 4 − 6 + 4 = 2
√
5.2 Verification of Descarte’s theorem for the five regular solids
Tetrahedron: 4 vertices and 3 equilateral triangles meeting at each vertex.
√
P
P
δV = 2π − (face angles at V ) = 2π − 3 · π/3 = π ⇒ ν δV = 4 · π
Cube: 8 vertices and 3 equilateral squares meeting at each vertex.
√
P
P
δV = 2π − (face angles at V ) = 2π − 3 · π/2 = π/2 ⇒ ν δV = 8 · π/2 = 4π
Octahedron: 6 vertices and 4 equilateral triangles meeting at each vertex.
√
P
P
2π
δV = 2π − (face angles at V ) = 2π − 4 · π/3 = 2π
ν δV = 6 · 3 = 4π
3 ⇒
Icosahedron: 12 vertices and 5 equilateral triangles meeting at each vertex.
√
P
P
δV = 2π − (face angles at V ) = 2π − 5 · π/3 = π3 ⇒ ν δV = 12 · π3 = 4π
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Dodecahedron: 20 vertices and 3 regular pentagons meeting at each vertex.
√
P
P
π
π
δV = 2π − (face angles at V ) = 2π − 3 · 2π
ν δV = 20 · 5 = 4π
5 = 5 ⇒
6 Bibliography
Robin Hartshorne, ”Geometry: Euclid and Beyond”, Springer 2000
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