Mr. Davies Gilbert on the Regular or Platonic Solids. 163 



The same results may be obtained analytically in a manner 

 exhibiting the extensive generality of algebraic language. 



Let x m the number of regular similar and equal planes 

 bounding the solid body. 

 y = the number of sides in each plane. 

 z = the number of planes, and consequently of spheri- 

 cal angles, meeting at each point. 



Then " gl jnges — the solid angle subtended by each side of 

 the solid from the centre of the sphere. 



But since all the spherical angles meeting at any point of 

 the solid must in every instance equal four right angles to fill 



space; each spherical angle will be equal to — — — -2L- and 



all the spherical angles of a side of the solid = — x 4 right 



angles. But the plane angles of each side of the solid = 

 (2y — 4) right angles. Consequently, the excess of the sphe- 

 rical angles above the plane angles = (4— — 2y + 4) right 



angles. And this equals the solid angle subtended by each 

 side of the solid from the centre of the sphere. Consequently, 



(4 f- — c ly + 4) right angle = nght ang es . 8 z = 4 xy — 



2xyz f 4xz, and x = ^1,),, ' 



Now it is obvious, from the nature of the quantities repre- 

 sented by x, y, and z> that they must in all cases be integral 



and positive. But since x = - — — <- — r— . The denominato 1 " 



2y + (2 — y) must be also affirmative. To ascertain therefore 

 the limit which y cannot exceed without making the expres- 

 sion negative, put 2y + (2 — y) . z = 0. Then 3/= 2 H . 



If z were infinite y = 2 and it attains its next integral 



value; 



when z = 6 y = 3; 



when 2 = 5 y = 3; 



when z = 4 y may be 4 ; 



when z = 3 y may be 6 ; 



when z = 2 y at its maximum infinite: therefore 



any finite number may be taken 

 for it. 

 If z less than 2, the highest value of y is negative. 



Y2 If 



