C-QAB: F-QDE = QA × QB × QC : QD × QE × QF;

that is, Q-ABC: T-DEF = QA × QB × QC: TD × TE × TF. Therefore, etc.

Q.E.D.

SIMILAR AND REGULAR POLYHEDRONS

569. Polyhedrons which have their corresponding polyhedral angles equal, and have the same number of faces similar each to each, and similarly placed, are called Similar Polyhedrons.

Faces, edges, angles, etc., which are similarly placed in similar polyhedrons are called homologous faces, edges, angles, etc.

570. Since the homologous sides of similar polygons are proportional, the homologous edges of similar polyhedrons are proportional.

571. Since similar polygons are proportional to the squares upon any of their homologous lines, the homologous faces of similar polyhedrons are proportional to the squares upon any of their homologous edges.

572. From § 571 it is evident that the entire surfaces of similar polyhedrons are proportional to the squares upon any of their homologous edges.

Proposition XX

573. 1. Form two similar polyhedrons, and if possible divide them into the same number of tetrahedrons, similar each to each.

2. How does the ratio of any two homologous lines in two similar polyhedrons compare with the ratio of any two homologous edges?

Theorem. Similar polyhedrons may be divided into the same number of tetrahedrons, similar each to each, and similarly placed.

Data: Any two similar polyhedrons, as AJ and A'J'.

To prove that AJ and A'J' may be divided into the same number of tetrahedrons, similar each to each, and similarly placed.

Proof. Select any trihedral angle in AJ, as B, and through the extremities of its edges, as A, G, C, pass a plane. Also through the homologous points, 4', G', c', pass a plane.

Then, § 310, in the tetrahedrons B-AGC and B'-A'G'C', the faces BAC, BAG, BGC are similar to the faces B'A'C', B'A'G', B'G'C', each to each.

..the homologous faces of these tetrahedrons are similar. Also, § 500, the homologous trihedral angles of these tetrahedrons are equal.

Hence, § 569, tetrahedrons B-AGC and B'-A'G'C' are similar. Now, if these similar tetrahedrons are removed from the similar polyhedrons of which they are a part, the polyhedrons which remain will continue to be similar; for the faces and polyhedral angles of the original polyhedrons will be similarly modified.

By continuing to remove similar tetrahedrons from them, the original polyhedrons may be reduced to similar tetrahedrons, and they will then have been divided into the same number of tetrahedrons, similar each to each, and similarly placed.

Therefore, etc.

Q.E.D.

574. Cor. Homologous lines in similar polyhedrons are proportional to their homologous edges.

Proposition XXI

575. Form two similar tetrahedrons. How do the trihedral angles at the vertices compare? Then, how does the ratio of the tetrahedrons compare with the product of the ratios of the homologous edges of the corresponding trihedral angles? (§ 568) How do the ratios of these edges compare with each other? Then, how does the ratio of the tetrahedrons compare with the ratio of the cubes of any two homologous edges?

Theorem. Similar tetrahedrons are to each other as the cubes of their homologous edges.

Data: Any two similar tetrahedrons, as Q-ABC and T-DEF.

To prove

3

Q-ABC: T-DEF = QA3 : TD3 = etc.

Proof. § 569, trihedral Q trihedral T;

=

.. § 568, Q-ABC : T-DEF = QA × QB × QC: TD × TE × TF,

In like manner, the same may be proved for any two homologous edges.

Therefore, etc.

Q.E.D.

576. Cor. Similar polyhedrons are to each other as the cubes of their homologous edges.

1. Into what may two similar polyhedrons be divided? § 573 2. How do the ratios of these portions compare with the ratios of the cubes of their homologous edges?

§ 575

3. How do these ratios compare with the ratios of the cubes of any two homologous edges of the polyhedrons?

§ 574

4. How, then, does the ratio of the sums of these portions compare with the ratio of the cubes of any two homologous edges of the polyhedrons ?

577. A polyhedron whose faces are equal regular polygons, and whose polyhedral angles are equal, is called a Regular Polyhedron.

578. 1. What is the least number of faces that a convex polyhedral angle may have? How does the sum of the face angles of any convex polyhedral angle compare with 360° ? Since each angle of an equilateral triangle is 60°, may a convex polyhedral angle be formed by combining three equilateral triangles? Four? Five? Six? Why? Then, how many regular convex polyhedrons are possible with equilateral triangles for faces?

2. How many degrees are there in the angle of a square? May a convex polyhedral angle be formed by combining three squares? By combining four? Why? Then, how many regular convex polyhedrons are possible with squares for faces?

3. Since each angle of a regular pentagon is 108°, may a convex polyhedral angle be formed by combining three regular pentagons? By combining four? Why? Then, how many regular convex polyhedrons are possible with regular pentagons for faces?

4. Since each angle of a regular hexagon is 120°, may a convex polyhedral angle be formed by combining three regular hexagons? Why? By combining three regular heptagons? Why? What is the limit of the number of sides of a regular polygon that may be used in forming a convex polyhedral angle, and therefore in forming a regular convex polyhedron ?

What, then, is the greatest number of regular convex polyhedrons possible?

579. There are only five regular convex polyhedrons possible, called from the number of their faces the tetrahedron, the hexahedron, the octahedron, the dodecahedron, and the icosahedron.

The tetrahedron, octahedron, and icosahedron are bounded by equilateral triangles; the hexahedron by squares; and the dodecahedron by pentagons.

580. The point within a regular polyhedron that is equidistant from all the faces of the polyhedron is called the center of the polyhedron.

The center is also equidistant from the vertices of all the polyhedral angles of the polyhedron.

Therefore, a sphere may be inscribed in, and a sphere may be circumscribed about, any regular polyhedron.

Proposition XXII

§§ 640, 641

581. Problem. Upon a given edge to construct the regular polyhedrons.

Datum: An edge, as AB.

Required to construct the regular polyhedrons on AB.

Solutions. 1. The regular tetrahedron.

Upon AB construct an equilateral triangle,

as ABC.

A

D

B

At the center of AABC erect a perpendicular, and take a point

D in this perpendicular such that DA = AB.

Draw lines from D to the vertices of ▲ ABC.

Then, the polyhedron D-ABC is a regular tetrahedron.
Proof. By the student. SUGGESTION. Refer to §§ 450, 500.

2. The regular hexahedron.

Upon AB construct the square ABCD, and E upon its sides construct the squares AF, BG, CH, and DE perpendicular to ABCD.

Then, the polyhedron AG is a regular hexahedron.

Proof. By the student. SUGGESTION.

Q.E.F.

A

D

H

B

Refer to § 500.

Q.E.F.

F

C