DEFINITION. A polyhedral angle is said to be convex when any polygon, formed by a plane cutting every face, is convex; otherwise it is said to be concave.

Theorem 27. In any convex polyhedral angle the sum of the face angles is less than a perigon. Given any convex polyhedral

angle, V-ABC.....

To prove

that AVB+BVC

.....

+< CVD + < perigon.

Proof. 1. Let the faces of the angle

be cut by a plane. This
will form a convex poly-
gon of n sides (n = 5 in

the figure), abc

.....

Def. convex polyh. Z

E a

A

=

3. Then Σp (n-2) st. 4, or Σp + 2 st. n st. £.

I, th. 21

I, th. 19

4. And Σv + Σb = =n st., since there is a st. for

..Σv+b=Σp + perigon. Steps 3 and 4; ax. 1

each A.

5.

6.

.. Σv < perigon, ·.· Σb> Σp.

Th. 26

EXERCISES. 626. The three planes which bisect the three dihedral angles of a trihedral angle intersect in a common line whose points are equidistant from the three faces. (See th. 20, cor., and I, th. 32.)

627. Suppose a polyhedral angle formed by three, four, five equilateral triangles. What is the sum of the face angles at the vertex ?

Section 5. - Problems.

POSTULATES OF CONSTRUCTION. As in Plane Geometry it was postulated that a line could be drawn and produced, and a circle drawn, so in Solid Geometry it is postulated that a plane may be drawn and revolved about a line as an axis, and, later, that a sphere, cylinder, and cone may be drawn.

Problem 1. Through a given point to pass a plane perpendicular to a given line: (1) the point being without the line, (2) the point being on the line.

1. Given the line YY' and point P

Proof.

X

3. Then MN, the plane of OP, OX, is the required plane.

2. Given the line YY' and the point O upon it.

Required
Construction.

Proof.

through O to pass a plane

YY'.

Problem 2. Through a given point to pass a plane parallel to a given plane.

Solution. Draw two intersecting lines in the given plane. Through the given point draw two lines parallel to these lines, thus determining the required plane.

Problem 3. Through a given point to draw a line perpendicular to a given plane: (1) the point being without the plane, (2) the point being in the plane.

1. Given the plane MN and the

4. Then PP' is the required perpendicular.

Proof. 1. CA plane CPP'.

Th. 6, cor.

2. Given Required

the plane MN and the point R within it.
through R to draw a perpendicular to MN.

Construction. 1. From any external point S draw ST

2. From R draw RQ || TS.

Proof. Then RQ is the required perpendicular.

MN.

Case 1

I, pr. 6
Why?

EXERCISES.

628. From the point of intersection of two lines to draw

a line perpendicular to each of them.

629. To determine the point whose distances from the three faces of a given trihedral angle are given. Is it unique ?

630. From the vertex of a trihedral angle to draw a line making equal angles with the three edges.

631. The three planes, through the bisectors of the face-angles of a trihedral angle, perpendicular to those faces, intersect in a common line whose points are equidistant from the edges. (See I, th. 31.)

632. In how many ways can a polyhedral angle be formed with equilateral triangles and squares?

BOOK VII. POLYHEDRA.

Section 1. General and Regular Polyhedra.

DEFINITIONS. A solid whose bounding surface consists entirely of planes is called a polyhedron; the polygons which bound it are called its faces; the sides of those polygons, its edges; and the points where the edges meet, its vertices.

If a polyhedron is such that no straight line can be drawn to cut its surface more than twice, it is said to be convex; otherwise it is said to be concave.

Unless the contrary is stated, the word polyhedron means convex polyhedron. The word convex will, however, be used wherever necessary for special emphasis.

If the faces of a polyhedron are congruent and regular polygons, the polyhedron is said to be regular.

Count

EXERCISES. 633. Draw a figure of a polyhedron of four faces. the edges, faces, and vertices and show that the number of edges plus two equals the number of faces plus the number of vertices.

634. Do the same for a polyhedron of five faces; also for one of six faces.

635. Take a piece of chalk, apple, or potato, and see if a seven-edged polyhedron can be cut from it.

636. What is the locus of points on the surface of a polyhedron equidistant from two given vertices? (The distances are to be taken as usual on a straight line, and not necessarily on the surface.)

637. What is the locus of points equidistant from two given nonparallel faces of a given polyhedron ?

638. To find a point equidistant from two given vertices of a polyhedron, and from two given non-parallel faces.

Theorem 1. If a convex polyhedron has e edges, v vertices, and f faces, then e +2=f+v.

Proof. 1. Imagine ABC..... Z formed by adding adjacent faces, begin

ning with any face as ABCD.....

of a sides, then adding face

M, of b sides, and so on.

2. Let e,= the number of edges, and

of vertices, after r faces have been put together.

3. Then e1 = a, and v1 = a, since the first face had

a sides.

4. . adding an adjacent face M of b sides, gives only (b-1) new edges, and (b-2) new vertices (Why?), 5... ea+b-1, v2 = a+b-2, so that e-V2-1. 6. Similarly, es va 2 (Why? Write it out like step 5), and, in general, e,v,r-1.

7. But the addition of the last, or fth face, as XYZ, after all the others have been put together, gives no new edges or vertices.

9. That is, e-vf-2, so that e+2=f+v; for ef=e, and v+= v.

COROLLARY. For every polyhedron there is another which, with the same number of edges, has as many faces as the first has vertices, and as many vertices as the first has faces.

For in the equation e + 2 = f + v, the f and v may be interchanged without affecting the e.

NOTE. This theorem is known as Euler's, although Descartes knew and employed it. The above proof is, however, essentially De Morgan's. The theorem is very useful in the study of crystals.