Schol. 1. The preceding theorem comprehends, as a particular case, figures symmetrical with reference to a plane.

Schol. 2. Every triangular prism, right or oblique, has four diametral planes, one of which is the plane half way between the bases parallel to them; and the three others are the planes passing through the lateral edges and through the diameters of the bases.

CENTER OF MEAN DISTANCES.

The point which has been named center of mean distances in a polygon in a previous appendix (II., def. 4), has a property with reference to a plane which we have shown it to have with reference to a line.

THEOREM 8. The perpendicular let fall from the center of mean dis‐ tances upon a plane drawn at pleasure in space, is equal to the quotient of the algebraic sum of the perpendiculars let fall from the different vertices upon this plane divided by the number of vertices. This sum is zero when the plane passes through the center of mean distances, and vice versa.

The demonstration will be similar to that in the corresponding one in a previous appendix (II., def. 4 et seq.).

It is to be observed, that the vees of which the point in question is the ce. of mean distances need not be in the same plane as they were supposed to be in the previous appendix.

Schol. To determine the center of mean distances for any nuinber of points not in the same plane, draw three planes at pleasure which cut each other (suppose, for the sake of simplicity, at right angles). Let fall, from the different vertices upon each of these planes, perpendiculars; find afterward for each plane the algebraic sum of its

diculars, and divide this sum by the number of vertices. Finaay, at distances equal to the three quotients, draw planes parallel to the first three, and their common intersection will be the point sought.

When four points are not in the same plane, these points, combined three and three, determine a tetrahedron. This being observed: THEOREM 9. In every tetrahedron the lines which join the middle ints of the edges not adjacent, all meet in a point which is the center of man distances of the four vertices.

Schol. This point is also found in three planes parallel to the faces, and at a distance equal to one quarter the distance of the opposite vertex from each face.

THEOREM 10. The four lines mining the vertices with the centers of mean distances of the opposite fazes, meet in a point which is the cen

ter of mean distances of the vertices. This point is one quarter the distance from the center of mean distances in each face to the opposite

vertex.

OF CENTERS OF SIMILITUDE.

THEOREM 11. If all the vertices of a polyhedron be joined with a point in space by lines, and upon these lines, or three prolongations, portions be taken proportional to the lines themselves, the vertices of a new polyhedron will be thus obtained, which is directly or inversely similar to the first.

This point is called a center of similitude, EXTERNAL in the first case, INTERNAL in the second.

The proof of the above is in all respects similar to that in a previous appendix (App. II.).

CENTERS OF SIMILITUDE OF SPHERES.

If lines be drawn tangent to two circles meeting each other, one pair internally and the other pair externally; and if these circles and tangents be set in revolution about the line joining the centers, the circles will generate spheres, and the tangents, cones enveloping the spheres, and the points of contact will generate circles which will be the curves of contact of the cones and spheres; the planes of these circles of contact will be perpendicular to the axis. The vertices of these cones, at the points in which the tangents intersect, are called centers of similitude of the two spheres, the one internal, the other

external.

Prove that every plane tangent to one of these conic surfaces is tangent to the two spheres.

And, conversely, that every plane tangent to the two spheres is tangent to one of the conic surfaces.

Two spheres in space would have an infinite number of common tangent planes. One of these would be determined by another condition, as, that it should pass through a given point, or be parallel to a given line, or tangent to a third sphere, &c.; and there would be two planes which would fulfill the required condition in the first two cases; in the last there might be four systems of two planes tangent to the three spheres, to wit: two planes comprehending the three spheres between them, and six placed two and two between one of the spheres and the two others.

This second case gives rise to a remarkable theorem analogous to one in a previous appendix (App. II.), for three circumferences of a circle.

THEOREM 12. The six centers of similitude of three spheres exterior to one another are situated THREE AND THREE upon a same line, to wit, the three external centers of similitude, then one of the external and two internal, giving in all four lines.

For, first, let us consider the two tangent planes which embrace the three spheres between them. These planes being tangent to the three cones which envelop the spheres, must both pass through the vertices of these cones, and, consequently, their intersection must. The other tangent planes will, in a similar manner, serve to demonstrate the other part of the theorem.

This theorem serves to prove the correctness of the theorem for the case of three circumferences (App. II.), because the centers of similitude of these circles are the same as the centers of similitude of three spheres, of which these circles are great circles.

It is thus that sometimes propositions in Plane Geometry may be demonstrated in a more simple manner by the aid of truths relating to geometry in space.

REGULAR POLYHEDRONS.

A regular polyhedron is one in which the faces are equal regular polygons, and the diedral angles equal. From this definition it will follow that the polyhedral angles will also be equal.

THEOREM.

There can be but five regular polyhedrons.

This follows from Prop. 2, of Polyhedral Angles, that a polyhedral angle can not be formed unless the sum of the plane angles which form it is less than four right angles.

If we take equilateral triangles, each angle of which is two thirds of a right angle, to form a polyhedral angle, we may combine these in threes, fours, and fives, but not more, because 6 X

angles.

2 12

== 4 right

3

If we take squares, each angle of which is one right angle, to form a polyhedral angle, we can combine them in threes alone, for 4 X 1= 4 right angles.

If regular pentagons, each angle of which is

can be combined but in threes.

5 × 2-4

= 1}, they

If hexagons, each angle of which is 13, they can not be combined

even in threes to form a polyhedral angle, and three is the least number of planes that can be employed for this purpose.

It is evident that still less can regular polygons of a greater number of sides be employed.

There can, therefore, be formed but three regular polyhedrons of triangular faces, but one of square faces, and but one of pentagonal fates, in all five, which is the greatest number that can possibly exist. Schol. It remains to be shown that five regular polyhedrons can be formed.

CONSTRUCTION OF REGULAR POLYHEDRONS.

10. TO CONSTRUCT A REGULAR TETRAHEDRON.

Take an equilateral triangle; erect at the center of its inscribed circle a perpendicular to its plane; with one of its vertices as a center, and a radius equal in length to one of its edges, cut this perpendicular in a point; join this point with the vertices of the triangle, and the regular tetrahedron will be formed.

20. TO CONSTRUCT A REGULAR HEXAHEDRON OR CUBE.

We leave this to the student, being too easy to require explanation.

30. A REGULAR OCTAHEDRON.

Upon a line equal to one of the sides of the equilateral triangle, which is to be a face, construct a square; erect at the center of this square a perpendicular to its plane, and take upon this perpendicular, on each side of the plane, a distance equal to one half the diagonal of the square; joining the points thus determined with the vertices of the square, the polyhedron required is formed.

N.B.-The center of the square is a center of symmetry. It is also the center of figure.

40. A REGULAR ICOSAHEDRON.

Construct first a pentagon upon the side of the given equilateral triangle; at the center of this figure erect a perpendicular to its plane; with a radius equal to the side of the triangle, cut this perpendicular in a point; this point being joined with the vertices of the pentagon, will furnish five equilateral triangles formed about it; form now a second pentahedral angle, with one of the angles of the pentagon as a vertex, and two of its faces will be the same as those of the first pentahedral angle formed; with a third vertex of the same triangle, to which the other two already employed belonged, form a third

pentahedral angle; for this purpose two new faces will be required. There will thus be united ten triangles, forming a sort of polyhedral cap, such that the angles at the border are formed by alternately two and three triangles. This polygonal line, which terminates the surface, has its sides equal, but its vertices not in the same plane. If now a second polyhedral cap be constructed equal to the first, its diedral angles will have the same value as those in the other. Then, without breaking the continuity, we can unite the double angles of the border of the first cap with the triple angles of the border of the second, and vice versâ; whence will result a figure of twenty equal faces equally inclined.

50. A REGULAR DODECAHEDRON.

Suppose that with three regular pentagons a trihedral be formed, which is possible (see last th.). The three diedral angles of this trihedral angle are equal. Now with new pentagons, equal to the preceding, can be formed in the same manner, successively, at the vertices of one of these pentagons, other trihedral angles, all of the same magnitude. There will result six regular pentagons, composing a polygonal cap, such that the angles of the border are formed alternately of one and of two plane angles.

[The same remark as above applies to this border.]

If a second cap be imagined, equal to the first, they can be united, border to border, so that the single angles of the one accord to the double angles of the other; and thus will be formed a figure of twelve faces, equal, and equally inclined to one another.

Schol. 1. To construct a regular polyhedron mechanically, taking one of the faces as a base of construction, upon a sheet of pasteboard make the development of all the faces, then fold these different faces upon their edges in a suitable manner.

Schol. 2. All the regular polyhedrons except the tetrahedron have a center of symmetry which is identical with the center of figure. All have, also, planes of symmetry. These are, in general, planes perpendicular upon the middle of the edges, or upon the middle points of lines joining opposite vertices, taken two and two, or else planes passing through the opposite edges, two and two.

Schol. 3. The regular tetrahedron has 4 vertices, 4 faces, and 6 edges.

General Scholium upon Polyhedrons. These expressions, which can