Cor. 1. It is readily seen that only one set of face angles can be obtained when a set of dihedral angles is given; so that when the dihedral angles of a three-faced corner are given, the face angles are given also; and the face angles can be expressed in terms of the dihedral angles.

Cor. 2. A three-faced corner is given when the dihedral angles, and their order, are given.

To construct a three-faced corner when its face angles are given is analogous to constructing a triangle when its sides are given; and to construct the corner when its dihedral angles are given is analogous to constructing the triangle when its angles are given. And this latter is a definite problem with respect to the corner, but an indefinite one with respect to the triangle.

45. Problem. To find the locus of a point equidistant from three given points not in line.

Let A, B, C be the points, and let U be the rightbisector plane of AB, and V be the right-bisector plane of AC (Art. 24. Def.).

Every point equidistant from A and B is on U (Art. 25. conv.), and every point equidistant from A and C is on V. And the required locus is the common line of U and V. But this line evidently passes through the circumcentre of the triangle ABC and is normal to its plane.

Hence the locus of a point equidistant from three given points, not in line, is the axis of vertices of the circumcircle of the three points considered as a cone-circle.

Cor. The three right-bisector planes, of the joins of three points, taken two and two, form an axial pencil.

46. Problem. To find a point equidistant from four given points which are not complanar, and no three of which are in line.

Let A, B, C, D be the four points, and let PO be the locus of a point equidistant from

A, B, and C. Join D, the fourth point, to any one of the other three, as C, and draw the rightbisector plane, X, of CD.

As D is not complanar with A, B, and C, the plane X is not parallel to PO, and therefore A meets PO at some point O. But O is equidistant from A, B, and

C, and it is also equidistant from C and D.

Therefore O is equidistant from A, B, C, and D.

Cor. 1. The line OP is the common line to three bisector planes, namely, those of AB, BC, and CA (Art. 45. Cor.), and X is a fourth plane which goes through the point O. The two remaining bisector planes, those of AD and CD, must pass through the same point 0.

Therefore the six right-bisector planes of the joins of four non-complanar points, of which no three are in line, pass through a common point and form a sheaf of planes.

Cor. 2. The four points can be combined to form four different triangles, and the lines, such as PO, which pass through their circumcentres and are normal to their planes, all pass through O and form a sheaf of lines.

Cor. 3. As the line PO can meet the plane X in only one point, there can be only one point equidistant from A, B, C, and D.

EXERCISES C.

1. Any face angle of a three-faced corner is greater than the difference between the other two.

2. Show how to construct a corner symmetrical with a given

corner.

3. Show that the three bisectors of the dihedral angles of a three-faced corner have a common line, and that this line is an isoclinal to the three faces.

4. There are four isoclinal lines through the vertex to the three faces of a three-faced corner.

5. In the figure (2) of Art. 41 denote O'A' by p.

C'D2 = C'B''2 = DA2 + A'C'2 - 2 DA' A'C' cos A (P. Art. 217.)

whence by reduction and dividing by cos b cos c,

cos a = cos b cos c + sin b sin c cos A;

or,

cos A = (cos a

cos b cos c)/sin b sin c;

which expresses a dihedral angle in terms of the face angles.

6. Express a face angle in terms of the dihedral angles. (Employ the property of the reciprocal corner.)

7. If the face angles of a three-faced corner are each 60°, show that the cosine of a dihedral angle is .

8. In 46 where is the locus if A, B, C are in line?

9. In 47 where is the point O if the four points be complanar ? where if three points be in line?

SECTION 4.

POLYHEDRA.

47. Def. A spatial figure formed of four or more planes so disposed as to completely enclose a portion of space is a polyhedron. It is analogous to the polygon in plane geometry, and its plane section is always some form of polygon.

The faces of the polyhedron are those portions of planes which are concerned in forming the closed figure, but for generality the term is sometimes extended to outlying parts of these planes.

The adjacent faces meet by twos to form edges, and the edges are concurrent in groups of three or more to form corners.

When a polyhedron is such that no line can meet more than two of its faces, it is convex.

48. Theorem. In any polyhedron the sum of the number of faces and the number of corners is greater by two than the number of edges.

Proof. Any polyhedron may be supposed to be built up by beginning with one face, and to it adding a second face, and then a third, and so on until the figure is completed.

Denote, in general, the number of corners by C, the number of faces by F, and the number of edges by E.

B

1. Let us start with a single face, U. The number of edges is the same as the number of corners, and we have one face. Therefore the equation C+FE+1 is satisfied. 2. To U add the face V. In so doing V loses one of its edges, BC, and two of its corners, B and C, by union with similar parts of U. So that in adding V we increase F by 1, and we increase E by one more than the increase of C; and hence the equation C+F=E÷1 is still satisfied.

U

W

3. To U and V add W. This new face loses two of its edges, DC and CG, and three of its corners, D, C, and G. Here again we add one face and one more edge than corner, so that C+FE+1 is still satisfied.

4. It is readily seen that in adding any face whatever, that face loses one more corner than edge by union with other faces, until we come to the last face necessary to complete the polyhedron.

This face loses all its edges and all its corners, so that by adding this face we increase the number of faces by 1 without interfering with the numbers of edges or corAnd hence in the completed polyhedron we have

ners.

C+FE+2.

This beautiful theorem is usually attributed to Euler, and is known as Euler's theorem on Polyhedra, but it appears to have been known before his time.