CHAPTER II.

THE FIVE REGULAR SOLIDS.

702. Def. A regular polyhedron is one of which all the faces are identically equal regular polygons and all the polyhedral angles are identically equal.

REMARK. A regular polyhedron is familiarly called a regular solid.

703. PROBLEM. To find how many regular solids are possible.

1. Let us consider any vertex of a regular polyhedron. Since at least three faces must meet to form the polyhedral angle at each vertex, and since the sum of all the plane angles which make up the polyhedral angle must be less than 360° (§ 661), we conclude:

Each angle of the faces of a regular solid must be less than 120°.

2. Since the angles of a regular hexagon are each 120°, and the angles of every polygon of more than six sides yet greater, we conclude:

Each face of a regular solid must have less than six sides. Such faces must therefore be either triangles, squares, or pentagons.

3. If we choose equilateral triangles, each polyhedral angle may have either 3, 4, or 5 faces, because 3 x 60°, 4 x 60°, and 5 × 60° are all less than 360°. It cannot have 6 faces, because each angle of the triangle being 60°, six angles would make 360°, reaching the limit. Therefore no more than three regular solids may be constructed with triangles.

4. Because each angle of a square is 90°, three is the only number of squares which can form a polyhedral angle. Therefore only one regular solid can have square faces.

5. Because each angle of a regular pentagon is 108°, a

polyhedral angle cannot be formed of more than three pentagons. Therefore only one regular solid can be formed of pentagons.

6. We therefore conclude that not more than five regular solids are possible. That five can really be constructed we show in the following way.

704. The Tetrahedron. If we take three equal equilateral triangles, AOB, BOC, and

COA, and join them at the common vertex, 0, ABC will be an identically equal equilateral triangle. Therefore a polyhedron will be formed having as faces four identical equilateral triangles.

This solid is a regular tetrahedron.

705. The cube, or regular hexahedron, is a regular parallelopiped of which the faces are all squares. Its construction is clear from §§ 685–693.

706. The Octahedron. Let ABCD be a square; O, its centre; PQ, a line passing through O perpendicular to the plane of the square.

On this line take the points P and Q, such that PA, PB, PC, and PD, also QA, QB, QC, and QD, are each equal to a side of the square. The figure P-ABCD-Q will be a regular

B

P

solid, having for its sides eight equilateral triangles.

Proof. 1. Because all the lines from Por Q to A, B, C, and D are equal, all the eight triangles which form the faces are equal and equilateral.

2. Let a diagonal be drawn from A to D.

Because O is the centre of the square ABCD, this diagonal will pass through O, and the lines PQ and AD, which intersect in O, are in the same plane. Therefore

P, A, Q, and D are in one plane.

3. In the triangles APD, ABD, AQD we have

Side AD common,

All the other sides equal.

Therefore the triangles are identically equal; and because ABD is a right angle, APD and AQD are also right angles, and the quadrilateral APDQ is a square equal to the square ABCD.

4. Because the polyhedral angle at B has its four faces and its base APDQ equal to the four faces and the base ABCD of the polyhedral angle at P,

Polyhedral angle B = polyhedral angle P.

5. In the same way it may be shown that any other two polyhedral angles are equal. Therefore the figure P-ABCD-Q is a regular solid having eight equilateral triangles for its faces. This solid is called the regular octahedron.

Р

707. The Dodecahedron. Taking the regular pentagon ABCDE as a base, join to its sides those of five other equal pentagons, so as to form five trihedral angles at A, B, C, D, and E, respectively. G Because the face angles of these trihedral angles are equal, the angles themselves are identically equal.

(§ 660)

Therefore the dihedral angles

formed along the edges AK, BL,

M

B

C

D

CM, etc., are equal to the dihedral angles AB, BC, etc.

Therefore the face angles EAK, LBC, etc., are identically equal to the angles of the original regular pentagons.

Pass planes through JLF and GKF, etc., and let FP be their line of intersection. Then, continuing the same course of reasoning, it may be shown that the face angles GKF, FLJ, etc., are all angles of 108°, or those of a regular pentagon. Completing this second series of five pentagons, we shall have left a pentagonal opening, which being closed, the surface of the polyhedron will be completed as shown in the diagram. The solid thus formed has 12 pentagons for its sides, and is called a regular dodecahedron.

708. The Icosahedron. Let five equilateral triangles form

a polyhedral angle at P, such that the dihedral angles along PA, PB, etc., shall all be equal.

The base ABCDE will then form a regular pentagon.

Complete the polyhedral angles at A, B, C, D, and E by adding to each three other equilateral triangles, and making the dihedral angles around A, B, C, etc., all equal.

H

It can be then shown, as in the case of the dodecahedron, that the lines F, G, H, I, J will form a second regular pentagon.

If upon this pentagon we construct five equilateral triangles, FQG, GQH, etc., having their vertices at Q, the solid will be completed, and its faces will be formed of 20 equilateral triangles.

This solid is called the icosahedron.

709. The five regular solids are therefore:
The tetrahedron, formed of

3 triangles.

The cube, or hexahedron, formed of 6 squares.
The octahedron, formed of

The dodecahedron, formed of

8 triangles.

12 pentagons.

The icosahedron, formed of

THEOREM X.

20 triangles.

710. The perpendiculars through the centres of the faces of a regular solid meet in a point which is equally distant from all the faces, from all the edges, and from all the vertices.

Hypothesis. ABCDE and ABKLM, two faces of a regular polyhedron, intersecting along the edge AB; 0, Q, the centres of these faces; OR, QR, perpendiculars to the faces.

Conclusion. These perpendiculars meet all the perpendiculars through the centres of the other faces in a point R equally distant from all the faces, edges, and vertices.

Proof. From O and Q drop perpendiculars upon the edge AB. Then

1. Because these perpendiculars are dropped from the centres of regular polygons, they will fall upon the middle point P of the common side AB.

2. Because PO and PQ are perpendicular to AB at the same L point P, and OR and QR are perpendiculars to the intersecting

planes, they will meet in a point (§ 627).

Let R be their point of meeting. Join PR. Then3. In the right-angled triangles POR and PQR we have Side PR common,

POPQ (being apothegms of equal polygons). Therefore these triangles are identically equal, and

4. If S be the centre of any other face adjacent to ABCDE, it can be shown in the same way that the perpendicular from S will meet QR in a point.

5. Because the angles between the faces Q and S are the same as between 0 and Q, it may be shown that the perpendicular from S will meet QR in the point R.

6. Continuing the reasoning, it will appear that all the perpendiculars from the centres of the faces meet in the same point R.

7. If from R perpendiculars be dropped upon all the edges and all the vertices, these perpendiculars, together with those upon the corresponding faces and the lines like QP and QB from the centres of the faces to the edges and vertices, will form identically equal triangles. Hence will follow the conclusion to be demonstrated.

NOTE. We have given but a brief outline of the demonstration, which the student may complete as an exercise. The conclusions may also be considered as following immediately from the symmetry of the polyhedron.