Form nine other pyramids equal to F-ABCDE.

Now one of these can be made to coincide with F-ABCDE in five different ways. For it makes no difference which side of its base coincides with AB.

Hence all of the diedral angles FA, FB, etc., are equal to any one of the diedral angles of the second pyramid, and are therefore all equal to each other.

Now place one of the seven pyramids, say A'-B'F' E' KG, so that the diedral angle A'F' shall coincide with its equal AF and the faces A'F'B' and A'F'E' with their equals AFB and AFE; thus adding the new faces EAK, KAG, and GAB.

Place a second pyramid, B'-C'F'A'G'H, so that the diedral angles B'F' and B'A' shall coincide with their equals BF and BA, and the faces B'C'F', B'F'A', and B'A'G' with their equals BCF, BFA, and BAG; thus adding the new faces GBH and HBC.

Similarly place two others, C'-D' F' B' H'I and D'. E' F' C'I'J, with their vertices at C and D; thus adding the new faces HCI, ICD and IDJ, JDE.

Place a fifth, E'-K'A'F'D'J', so that the diedral angles E'A', E'F', and E'D' shall coincide with their equals EA, EF, and ED, and the faces E'A'K', E'F'A', E'D'F', and E'J'D' with their equals EAK, EFA, EDF, and EJD; thus adding the new face JEK.

The four other pyramids can be similarly placed with their vertices at G, H, I, and J; thus adding the new faces OGK and OGH, OHI, OIJ, and OJK.

The polyedron thus completed, having twenty equal equilateral triangles for faces and having its diedral angles all equal, is a regular icosaedron.

Q. E. F.

738. Remark.-The five regular polyedrons may be made from cardboard as follows: Draw on cardboard the figures given below, and on the inner lines cut the cardboard half through with a penknife. Cut the figures out entire and fold the cardboard as shown for the icosaedron in the accompanying plate.

GENERAL THEOREMS ON POLYEDRONS

PROPOSITION XXXVI.

THEOREM

739. The number of the edges of any polyedron increased by two is equal to the number of its vertices increased by the number of its faces.*

Denote the number of its edges by E; the number of its vertices by V; and the number of its faces by F.

Let us put together the surface of the polyedron face by face and compare the number of edges with the number of vertices at each step.

If we take one face, as ABCD, the number of edges is obviously equal to the number of vertices.

That is, for one face, E = V.

Now let us add a second face, say a quadrilateral ABGF, to the first by placing the edges AB together. The new surface, consisting of ABCD and ABGF, will have three new edges, AF, FG, and GB, and two new vertices, F and G.

*This theorem was discovered by Euler (1707-1783).

The whole number of edges will be then one greater than the whole number of vertices.

However many sides the second face may have, it is easily seen that the number of new edges added will be one more than the number of new vertices.

Therefore for two faces, E=V+1.

Next add a third face ADEF by placing an edge of it in coincidence with an edge of each of the first two faces.

We thus add two new edges, DE and FE, and one new vertex, E.

However many sides the third face may have, the increase in the number of edges is one more than the increase in the number of vertices.

Hence for three faces, E= V + 2.

We can in this way prove the following table:

When the number of faces is F-1, the surface is not closed.

To close it we add the last face.

In so doing we place each edge and each vertex of the

last face in coincidence with an edge and vertex of the open surface.

Adding the last face then increases neither the number of edges nor the number of vertices.

That is, for F faces, E=V+F−2.

or

E+2=V+F.

Q. E. D.

PROPOSITION XXXVII.

THEOREM

740. The sum of the angles of all the faces of any convex polyedron is equal to four right angles taken as many times as the polyedron has vertices less two.

Let S denote the sum of the angles of all the faces, and V the number of vertices of any convex polyedron. Also let R denote a right angle.

Any one face is a convex polygon.

Let the number of its sides be n.

Produce the sides in succession as in $ 69.

The sum of the exterior angles thus formed is 4R. The sum of the interior and exterior angles is 2Rx 1. § 22 Do the same for all the faces of the polyedron considered as independent polygons of n, n', n'', etc., sides.

Then the sum of the exterior angles of the F faces is 4RX F.

The sum S of their interior angles plus the sum of their exterior angles is 2R(n+n'+n" + etc.).

That is,

S+4R×F=2R(n+n'+n" + etc.).

Now, if E denotes the number of edges of the polyedron, n+n'+n'' + etc. =2E,

since each edge is a side of two polygons.