( 501 ) 



a square or a regular pentagon, with the assumed point always as 



centre, according to p having the value 

 3, 4 or 5, the section of the system of 

 the p + 2 bounding spaces of the new 

 cell passing through K^ with a space 

 normal to UK^ — e.g. with a space 

 which according to the normal in to OK^ 

 is perpendicular to the plane of projec- 

 tion — is a right /^-lateral prism, of 

 which the segment LL' of m enclosed 

 between /, /' is the axis and the per- 

 pendicular endplanes project themselves 

 in L and L' . From this ensues: 



"Through a vertex pass p -\- 2 bounding 

 spaces, i.e. Q:=z p -\-2'\ 

 "Through an edge pass three bounding spaces, i.e. P=3"- 

 "Through a \ertex pass 'Ip edges, so ph is the number of edges, 

 i.e. K=:pir 



"The system of the bounding spaces consists of two groups, namely 

 of e regular polyhedra with q faces, and r semi-regular polyhedra 

 {e/,k',/') with equivalent vertices truncated at the vertices as far 

 as the centres of the edges, i.e. it! = (? -|- r". 



"As the polyhedra of the second group have e' -{- /' := k' -\- 2 

 faces and a face is common to two bounding spaces, the number of 

 faces is half the sum of qe and r{k' -f- 2) or qe and pk -\- 2r, i.e. 

 2F=qe-{-pk^2r". 

 Thus the result is : 



"The first of the two cells, {E, K, F, R, F, Q), deduced out of the 

 regular cell {e, k,f, r, p, q) has the characteristic numbers 



E=k, K = pk, F=.\{qe-Ypk)^ 



R 



e -f- r, 



Here the law of Eüler^-|- F = K -\- R may serve as verification. 

 In reality the difference of the two members of this equation 



E-\-F-{K-^R)=zk+ kiqe+ p/^) -\-r-ipk [-e^ r) 

 z=: k — e ^ ^ (qe — pk) 

 = ll{q-2)e-{p-2)k\ 



is equal to zero in consequence of the relation (1). 



3. The second of the new cells deduced out of the regular cell 

 is enclosed by the polar spaces of the centres K^ of the edges with 

 respect to the spherical space through those points, i. e. by the 



34* 



