( 532 ) 



doublerectangular hyperspherical tetrahedron can be arranged in 

 three cycles, two of 6 and one of 3 elements, namely 



1. (h n — o la , a 14 , i n — a 84 , « S4 , « a8 , « 13 ), 



2. (è n — yl H , ^ 4J , a 18 , a ï4 , A lt , hn — A 41 ), 



3. (4„ , £ x — a„ , .4 18 ), 



so that it is possible to allow the elements of each cycle to undergo 

 all simultaneously a cyclic transformation, if only afterwards those 

 of the second cycle are replaced by their complements, 1 ) without 

 these elements ceasing to be the elements of a possible doublerect- 

 angular hyperspherical tetrahedron. 



These same simultaneous transformations may thus be applied to 

 each formula holding in general for elements of the doublerectan- 

 gular hyperspherical tetrahedron. 



5. If we again apply the construction described in § I to the 

 newly formed spherical triangle, etc. we find a closed range of five 

 spherical triangles of which the hypothenusae form a spherical 

 pentagon. 



The sides of these five spherical triangles are parts of live great 

 circles on the sphere, namely the circles part of which is formed by 

 the three sides of the original spherical triangle and the two polar 

 circles of the vertices of its oblique angles. These live great circles 

 form, however, another second similar range of five spherical triangles, 

 namely that of the opposite triangles of the former range. 



6. We can likewise deduce in a manner indicated in § 3 out of 

 a doublerectangular hyperspherical tetrahedron a range of such 

 tetrahedra of which the faces all belong to six spheres, namely the 

 spheres part of which is formed by the" faces of the original tetra- 

 hedron and the polar spheres of the points A x and A A . 



Let us call B x the polar sphere of A lt B 2 that of A 4 , B s the 

 sphere A x A> A 3 , B 4 the sphere A A A x A t , B b the sphere A s A t A, 

 and B 9 the sphere A, A s A 4 . 



Each of these spheres divides the hypersphere into two halves of 

 which I shall designate the one to which the original tetrahedron 



*) If we write the second cycle » 



(J 3t — A l4 , \Jt — A 4i , a lt , i st — a 34 , A l% , .4 41 ) 

 or 



then no replacement of the elements by their complements is necessary, but the 

 cycle has lost its symmetry with respect to the tetrahedron. 



