592. Given an edge, to construct the five regular polyhedrons.

Let A B be the given edge.

D

I. Upon A B to construct a regular tetrahedron.
Upon A B construct the equilateral A

ABC.
$ 232
Find the centre of this A, § 238
and erect OD to the plane A BC.
Take the point D so that A D = A B.
Draw DA, DB, DC.

C

B

ABCD is the regular tetrahedron required.
For, the edges are all equal,

G

§ 450

and hence the faces are equal equilateral A.
and its polyhedral are all equal.

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H

II. To construct a regular hexahedron.

E

D

H

B

Upon the given edge AB construct the square ABCD,

and upon the sides of this square conCstruct the squares EB, FC, GD, HA 1 to the plane A B C D.

Then AG is the regular hexahedron required.

III. To construct a regular octahedron.
Upon the given edge A B construct

the square A B C D.

Through its centre O pass a to

its plane ABCD.

In this take two points E and F, one above and the other below the plane, so that A E and A F are each equal

to A B.

E

Join E and F to each of the vertices of the square.
Then EABC D F is the regular octahedron required.

For, the edges are all equal,

and hence the faces are equal equilateral ▲.

And, since the ADEF and DA C are equal,

A

$450

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DEBF is a square and the pyramid A-D E B F is equal in

all its parts to the pyramid E-A B C D.

Hence, the polyhedral A and E are equal.

In like manner all the polyhedral

of the figure are equal.

IV. To construct a regular dodecahedron.

Upon A B construct the regular pentagon A B C D E. § 395 On each side of this pentagon construct an equal pentagon, so inclined that trihedral shall be formed at A, B, C, D, E. The convex surface thus formed is composed of six regular pentagons.

In like manner, upon an equal pentagon A'B'C' D' E' construct an equal convex surface.

Apply one of these surfaces to the other, with their convexities turned in opposite directions, so that P'O' and P'Q' shall fall upon PO and PQ.

Then every face of the one will, with two consecutive face of the other, form a trihedral Z.

The solid thus formed is the regular dodecahedron required.

V. To construct a regular icosahedron.

Upon A B construct the regular pentagon A B C D E. § 395

At its centre 0 erect a to its plane.

In this take P so that PA

=

= A B.

Join P with each of the vertices of the pentagon;

thus forming a regular pentagonal pyramid whose vertex is P, and whose dihedral s formed on the edges PA, P B, PC, etc. are all equal.

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Taking A and B as vertices, construct two pyramids each equal to the first, and having for bases BPE FG and AG HCP respectively.

There will thus be formed a convex surface consisting of ten equal equilateral A.

In like manner upon an equal pentagon A'B'C' D' E' construct an equal convex surface.

Apply one of these surfaces to the other with their convexities turned in opposite directions, so that every combination of two faces of the one, as P' D' C', P' D' E', shall with a combination of three face of the other, as BCH, BCP, PCD, form a pentahedral Z.

The solid thus formed is the regular icosahedron required. the faces are all equal;

For,

Cons. § 571

Q. E. D.

593. SCHOLIUM. The regular polyhedrons can be formed thus:

Draw the above diagrams upon card-board. Cut through the exterior lines and half through the interior lines. The figures will then readily bend into the regular forms required.

SUPPLEMENTARY PROPOSITIONS.

PROPOSITION XXV. THEOREM. (EULER'S.)

594. In any polyhedron the number of its edges increased by two is equal to the number of its vertices increased by the number of its faces.

Let E denote the number of edges of any polyhedron; V the number of its vertices, F the number of its faces.

B

We are to prove
S

C

Annex a second face SA B by applying one of its edges to an edge of the first face.

There is formed a surface having one edge A B, and two vertices A and D B common to both faces.

.. whatever the number of the sides of the new face, the whole number of edges is now one more than the whole number of vertices.

Annex a third face, SBC, adjacent to each of the former.
The new surface will have two edges SB and B C,

and three vertices S, B and C, in common with the preceding surface.

.. the increase in the number of edges is again one more than the increase in the number of vertices.

According to the same law, for an incomplete surface of F-1 faces

When we add the last face SE A, necessary to complete the surface,

its edges SE, SA and A E, and its vertices S, E and A will be in common with the preceding surface.

.. in a polyhedron of F' faces E = V + F

.. E+ 2 = V+ F.

2.

Q. E. D.

PROPOSITION XXVI. THEOREM.

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

Let E denote the number of edges, V the number of

vertices, F the number of faces, and S the sum of all the angles of the faces of any polyhedron.

We are to prove S = 4 rt. 4 × (V — 2).

Since E denotes the number of

the edges of the polyhedron,

2 E will denote the whole number of sides of all its faces, considered as sides of independent polygons.

S

E

A

D

B

And since the sum of all the interior and exterior of each polygon is equal to 2 rt. s taken as many times as it has sides,

the sum of the interior and exterior of all the faces is equal to 2 rt. ≤ × 2 E.

And since the sum of the exteriors of each face is 4 rt. s,

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the sum of the exterior of all the faces is equal to 4 rt. XF.