IX] OF EULER'S LAW 739 



We may add yet two more formulae, both related to the last, and 

 all derivable, ultimately, from Euler's Law * : 



3/3+2/4+/5-^12 + 0.C3+2c4+4c5+... + 0./e+/7 + 2/8 (6) 



and 



3C3+ 2c4 + C5 = 12 + 0,/3+ 2/4+ 4/5 + ... + O.Ce + C7 + 2c, ...(7). 



These imply that in every polyhedron the triangular, quadrangular 

 and pentagonal faces (or corners) must, taken together and multi- 

 plied as above, be at least twelve in number. Therefore no poly- 

 hedron can exist which has not a certain number of triangles, 

 squares or pentagons in its composition; and the impossibihty of 

 a polyhedron consisting only of hexagons is demonstrated once 

 again. 



Formulae (5), (6) and (7) further shew us that not only is a three- 

 way polyhedron of hexagons impossible, but also a four- way 

 polyhedron of quadrangles, or one of six- way corners and triangular 

 facets ; all of which become the more obvious when we reflect that 

 the plane angles meeting in each point or node must be, on the 

 average, in the first case 3 x 120°, in the second 4 x 90°, and in the 

 third 6 X 60°. 



Lastly, having now considered the case of other than trihedral 

 corners, we may learn a simple but very curious relation between 

 the number of faces and corners, arising (like so much else) out of 

 Euler's Law. In a polyhedron whose corners are all n-hedral, 

 nC=2E; therefore (by Euler) 7iC/2 + 2 = F + C; therefore 

 2F ={n-2)C + 4. 



Therefore, if 



n=3, 2F=4:+ C] 



= 4, =4 + 2cl (8). 



= 5, = 4 + 3Cj 



Let us look again at the microscopic skeleton of Dorataspis 

 (Fig. 341). We have seen that some of its facets are hexagonal, 

 the rest pentagonal; there happen to be eight of the former, and 

 therefore (as we now know) there must be twelve of the latter. 



* Derivable from Euler together with the formulae for the "edge-counts," 

 viz. llnF^ = 2E, and SnC„ = 2^; which merely mean that each edge separates 

 two faces, and joins two corners. 



