5. The regular icosaedron. Cons. Upon AB construct a regular pentagon ABCDE.

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At its centre erect a I to its plane, and in this I take a pt. S so that SA : AB.

Join SA, SB, SC, SD, SE.

Then S-ABCDE is a regular pentagonal pyramid (496), and each of its faces is an equilateral A.

Complete the polyedral Z8 at A and B by adding to each three equilateral As each equal to SAB, and making the diedral s around A and B equal.

There is then formed a convex surface CDEF, etc., composed of ten equal equilateral A s.

Construct a second convex surface cdef, etc., equal to the first.

Then the two equal convex surfaces may be combined so as to form a single convex surface, which is the regular icosaedron.

Proof. Suppose the convexity of DG to be up, and the convexity of dg to be down.

Put the two surfaces together so that the pt. D, where two faces meet, falls upon the pt. c, where three faces meet. Then two consecutive face Zs of one surface will unite with three consecutive face s of the other. Thus, the two faces 1 and 9 will unite with the three faces 10, 11, 12, forming a polyedral L of five faces equal to S, without in any way changing the form of either surface, since the three diedral Zs contained by 1 and 9, 10 and 11, 11 and 12 are those which belong to such a polyedral L.

The vertex C will fall at h, and a like polyedral 2 will be formed at that pt.; and so of all the others.

... the polyedral Zs are all equal, and the solid is a regular icosaedron.

Q. E.F. 654. Sch. Models of the regular polyedrons may be easily constructed as follows:

Draw the following diagrams on cardboard, and cut them out. Then cut half way through the board in the dividing lines, and bring the edges together so as to form the respective polyedrons.







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To prove




Proposition 27. Theorem.* 655. In any polyedron the number of edges increased by 2 is equal to the whole number of vertices and faces. I

Hyp. Let E, F, and V denote the number of edges, faces, and vertices respectively of the polyedron S-ABCD.

E + 2 = V + F.
Proof. Beginning with one face
ABCD, we have E= V.

If we annex to this a second face SAB, by applying one of its edges, as AB, to the corresponding edge of the first face, we form a surface having one edge AB, and two vertices A and B common to both faces; therefore, the whole number of edges is now one more than the whole number of vertices.

... for 2 faces ABCD and SAB, E= V +1. If we annex a third face SBC, adjacent to both ABCD and SAB, we form a new surface having two edges SB and BC, and three vertices S, B, C in common with the preceding surface ; therefore the increase in the number of edges is again one more than the increase in the number of vertices.

.. for 3 faces, E= V + 2. In like manner, for 4 faces, E= V + 3. And so on until every face but one has been annexed.

... for (F - 1) faces, E = V+F - 2. But annexing the last face adds no edges nor vertices.

i. for F faces, E= V+F – 2,

E +2 = V +F. Q.F.D. + The proof of this theorem was first published by Euler (1752). $ This proof is due to Cauchy.



1. How many edges has the regular tetraedron ?
2. How many edges has the regular hexaedron ?

3. The frustum of a regular four-sided pyramid is 8 feet high, and the sides of its bases are 3 feet and 5 feet: find the height of an equivalent regular pyramid whose base is 10 feet square.

Proposition 28. Theorem.*

To prove

656. The sum of the face-angles of any polyedron is equal to four right angles taken as many times as the polyedral has vertices less two.

Hyp. Let E, F, and V denote the number of edges, faces, and vertices respectively, and S the sum of the face s of any polyedron.

S = 4 rt. Zs (V 2).
Proof. Since each edge is common to two faces,

... the whole number of sides of the faces considered as independent polygons is 2E.

If we form an exterior / at each vertex of every polygon, the sum of the interior and exterior Zs at each vertex is 2 rt. Zs.

And since the whole number of vertices in the faces is 2E,

... the sum of all the interior and exterior Zs of the faces is 2 rt. 28 X 2E, or 4 rt. _8 X E.

Since the sum of the exterior Zs of each face is 4 rt. Zs,

(151) ... the sum of the ext. Zs of the F faces is 4 rt. Zs X F. ..S+ 4 rt. Zs x F = 4 rt. Ls X E.

:.S= 4 rt. Ls (E – F).
E - F = V - 2.

(655) .. S = 4 rt. Zs (V – 2). Q.E.D.


THEOREMS. 1. Show that a lateral edge of a right prism is equal to the altitude.

2. Show that the lateral faces of right prisms are rectangles.

3. Show that every section of a prism made by a plane parallel to the lateral edges is a parallelogram.

4. The lateral areas of right prisms of equal altitudes are as the perimeters of their bases.

5. The opposite faces of a parallelopiped are equal and parallel.

6. If the four diagonals of a quadrangular prism pass through a common point, the prism is a parallelopiped.

1. If any two non-parallel diagonal planes of a prism are perpendicular to the base, the prism is a right prism. See (546).

8. Any straight line drawn through the centre of a parallelopiped, terminating in a pair of faces, is bisected at that point.

9. The volume of a triangular prism is equal to the product of the area of a lateral face by one-half the perpendicular distance of that face from the opposite edge.

10. In a cube the square of a diagonal is three times the square of an edge.

11. In any parallelopiped, the sum of the squares of the twelve edges is equal to the sum of the squares of its four diagonals.

12. In any quadrangular prism, the sum of the squares of the twelve edges is equal to the sum of the squares of its four diagonals, plus eight times the square of the line joining the common middle points of the diagonals taken two and two.

13. A plane passing through a triangular pyramid,

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