DERIVED FROM THE REGULA tt POLYTOPES. 57 



investigate how far we can proceed in this way by using the first 

 net symbol only. This much more complicated symbol is 



[4(2 + |/2> 1 +4 + K2, 4(2+ K2>2 +2 + ^2,4(2 +K2> 3 +2 + ^2,4(2 + K2K + 1/2], 

 2#, ; being even. We abridge it into the following form, clear by itself: 



[4+V/2, 2-f-V2, 2--J-V2, V 7 2], (8 + 4V/2)fl 1 ,fl 2 ,fl 8> fl 4 ,2a i even, 



where %, a 2 ><fe> a 4 preceded by the common factor 8~j-4\/2 repre- 

 sents the immovable part. 



Face gap prismotope. By extension the centre ^ , % , ^ , Ö of the 

 face (2,0,0) of [2,0,0,0] passes into the new origin O' with 

 the coordinates f (2 + V%) , f (2 + V2) , -f (2 -(- V"2) , 0. By 

 supposing the four ^ of the net symbol to disappear we get inter 

 alia the set of vertices (4 -f V'2, 2 -f V2, 2 -)- V^2) [\/2], i.e. 

 a P 3 . These are the only vertices contained in the net symbol above 

 mentioned lying at minimum distance -| VS from O' , but as we 

 shall see immediately the two other net symbols contain other ver- 

 tices partaking of this property. However, in order to sharpen our 

 analytic tools, we leave these other net symbols alone for a moment 

 and try to deduce these lacking vertices from the simple properties 

 of the prismotope with two regular generating polygons in planes 

 perfectly normal to each other. By means of the P s just found 

 we know that one of these polygons is a triangle, and the 

 character of the other polygon can be deduced from its circum- 

 radius. For the relation p ± 2 -\- p% = p 2 between the circumradii p i} 

 p 2 , p of the two generating polygons and the prismotope itself 

 gives, as we have p = ±\/% and ^ = -| Y^G , ^ = -f V6, i.e. 

 the second polygon is also a triangle and the prismotope a (3 ; 3). 

 We have therefore only to find a third position of the first triangle, 

 the two end planes of P d containing already two positions , and 

 this third position can be found by remarking that the centres of 

 these three equipollent triangles are the vertices of an equilateral 

 triangle with O' as centre. So, if p, q,r, s are the coordinates 

 of the centre of this third position we have that the triangle with 

 the three vertices 



f + V/2, 

 f + V/2, 



f + V/2, 

 t + V'2 , 



f + V/2, 



f + V/2,- 



V/2 

 — VI 



P 

 must admit 



S 



r 



s 



| (2 + V/2), f (2 + V/2), |(2 + V/2), 

 as centre. From this it ensues that we have 



p=z S = r = % + 2V2,s= 0, 



