In every face of a polyhedron take any point, Which for abbreviation we may call the centre of that face. Join the centre of each face with the centres of the adjoining faces ; we have thus a new polyhedron, and the points may be so taken, that those lying in the faces which meet at any corner, shall all be in the same plane. The new polyhedron has obviously as many corners as the old one had faces; and as many faces as the old one has °timers: the umber of edges being the game in both : and if we call a corner triangular, quadrangular, &c., according as three, four, &c., angles meet there, the new solid has as many triangular, &c., faces, as the old solid has triangular, &c., corners, and rice rural. These polyhedrons may be called conjugate to one another.
Thus there is a triangular tetrahedron (four-faced solid) with four triangular corners: consequently the conjugate solid is another tetra hedron of the same kind. The quadrangular hexahedron (of six four :tided faces) has S triangular corners : the conjugate solid has therefore S triangular faces, and six quadrangular corners (the triangular octahedron). The pentagonal dodecahedron (having 12 five-sided faces) luis 20 triangular corners : the conjugate solid has therefore 20 triangular faces and 12 pentagonal corners (the triangular icosa hedron). The solids mentioned in this paragraph are those which may be mule of equilateral and equiangular faces. [Reout.An Swans.] Again, a solid can be formed with 14 quadrangular faces, having S triangular corners and S quadrangular ones; its conjugate solid has therefore 8 triangular and 8 quadrangular faces, with 14 quadrangnlar corners; the number of edges in both being 8+8+14-2, or 28.
Let F,. fie., be the number of triangular, quadrangular, agonal, fie, faces in a solid, and ;, c„ c„ fie., the number of triangular, quadrangular, pentagonal, corners. Let r,o,r, be the total number of faces, corners, and edges; then we have (1) (2) Again, since 3 + 4 F.+ . ... is the total number of sides of all the faces, before they are joined, and since the junction joins each with another, we have half the preceding for the number of edges, or 2 e=3;+4r,4-5r,+...(8) 2 E--.3 4 0,1-5 (4) But F+C=E +2, whence we deduce 2 c=4+e,+2r,+3r,+...(5) 2 c.+3 c,+... (6) Hence r,+ r, +.... and + c, + must be even numbers ; for if these be subtracted from the even numbers 2 c and 2 r, it will be seen that even numbers are left : or the number of odd-sided figures must be even, and also the number of odd-angled corners. Moreover
the number of corners must be made up of (1) a couple ; (2) half as many as there are odd-sided faces; (3) 1 for every quadrangle and pen tagon, 2 for every hexagon and heptagon, 3 for every octagon and nonagon, fie.; and the same will be true if we write faces for corners, and corners for faces.
Since every face bats at least three sides, and every corner at least three angles, 2 E cannot fall short of 8 F, nor of 8 c. Hence, neither 4 r —6 r, nor 4 F.-6 c can be negative, that is, neither of the following can be negative : 3 c,+2 ;+c,-12—c,-2 c,— (7) 3 (8) Hence it appears that there must be either triangular, quadrangular, nr pentagonal faces, and either three-angled, four-angled, or live-angled corners. Call these the essential faces and corners. Bence the fol• lowing readily follows : If the essential faces be all triangles, there must be 4 at least ; if all quadrangles, 6 at least ; if all pentagons, 12 at least : and the same of the corners. If the non-essential faces be all hexagons, or the non essential corners six-angled, it would appear • that the minimum number of essential faces and corners need not be increased, how many hexagons soever, or six-angled corners, there may be.
Wo easily show that the formula; in (7) and (8) are 2 r,+ 4 re-F G r, + and 2 o, + 4 c, + 6 ;+ . from which, by addition, we find r,+ceefia-r,+c,+2(r„+c,)+3(r,+;)+....
That is, calling both triangles and three-angled corners by the name of triplets, quadrangles and four-angled cornefa by that of quadmaicts, &c., we have the following theorem :—Space cannot be inclosed without tripleta and the triplets are in number 8, and one for each quintuplet, and two for each sextuplet, And three for each septuplet, &e.
If all the corners be three-angled, we have 2 x=8 c, or (8) vanishes. If then all the faces be of sides not exceeding six, we have 3r,+2r,a-r,=12.
Similarly, if all the fates be triangular, and the corners nowhere more than six-angled, we must have 3 c,+2 o,+c,=12.
Bence it follows that when all the corners are three-angled, and all the faces either pentagons or hexagons, the number of pentagons can be neither more nor less than 12 : also that when all the faces are trianglea, and all the corners five-angled or six-angled, the number of five-angled corners can be neither more nor less than 12.