PARTITION OF A PARALLELEPIPED INTO TETRAHEDRA. 713 



is, by applying them face to face, we see that the line joining the corners opposite to the 

 united faces is longer than any line in the cube. There can therefore be only one in 

 this complex. The rest of it, that is, two-thirds of the cube, must be made up of other 

 tetrahedra, each of which has a volume equal to one-sixth of the cube, and there must 

 therefore be four of them. Now, as Q has no part of the surface of the cube, these four 

 must have the whole of it. This can only be by each of them having three half-faces of 

 the cube. This is the case only with A ; accordingly, this complex consists of one & and 

 four A's (fig. 1). This quinquepartite division of the cube may be noted as Q, 4A. # As 

 this is the only division containing Q, all the rest must be sexpartite divisions, containing 

 six equal-volume tetrahedra. 



Let i, 8, I, y represent the numbers respectively of tetrahedra of the forms I, A, L, T 

 in a combination forming a cube. We have (1) i+S+l+y=6 because of the volume, 

 and (2) i + 3$ + 2l + 2y = 12 because of the surface. From these we see at once that 

 i = 8, and that I + y is always an even number. 



Not only is i = 8, but I and A are always necessarily connected together, the equi- 

 lateral triangle of the one being applied to that of the other so as to form a figure which 

 we shall call (IA). 



That this is so can easily be shown. The equilateral triangle which forms a face of 

 the I and of the A is an " internal " face, that is, a face in the interior of the cube, and 

 must therefore, in the complex, be covered by another internal face, or by other internal 

 faces, or by portions of other internal faces. It cannot be covered by other internal 

 faces or by parts of such, because no other triangle in the system has an angle of 60°, 

 and there are no two angles in the system which together make up an angle of 60° (of 

 the two angles of the scalene triangle, one is greater than 60° and the other less than 30°), 

 so that the internal equilateral triangle must be covered by another of the same. But 

 two A's cannot go together, they would form a double triangular pyramid, entirely sur- 

 rounded by six half-faces of the cube ; four of the five corners of this double triangular 

 pyramid would indeed coincide with corners of the cube, but the fifth corner, while in a 

 body-diagonal of the cube, would be far from the position of any corner of the cube. 

 And no more can two I's go together, for they would form a figure in which two half- 

 faces of the cube would meet at a re-entrant angle. The internal equilateral triangle of 

 a A must therefore be covered by the internal equilateral triangle of an I, and therefore 

 neither I nor A can occur except in combination as an (IA). 



(IA) is an oblique square pyramid, the base of which is a face of the cube, and the 

 apex of which is one of the four corners of the cube not in the base. Of its corners , 

 three are of one sign and two of the other. It can be divided into an I and a A by a 

 plane passing through the three cosignal corners. But it can be divided into two tetra- 

 hedra in another way. A plane through the apex and the two cosignal corners (the 



* If we leave out the condition forbidding new corners, we can obtain, from this quinquepartite division, a case 

 of division into six equal-volume tetrahedra, by cutting Ci into two equal tetrahedra by a plane containing one edge and 

 bisecting the opposite edge. 



