( 285 ) 



and the group II, to a point .I/,-; if Q, is chosen as Q,-, .)/, must 

 coincide either with .1/, or with J/ 5 . 



The chosen line c, furnishes one point of intersection more, on 

 J\I\\ so there must be one group of icosahedra more with an edge 

 situated in the intersecting space. Indeed, we tind only one group 

 VI,, IV, belonging again to Q x . 



Case e A . On plate VIII 6 under e we have to deal with a central 

 section of C 800 , from which ensues that the line e i passes through 

 the centre .1/ of the pentagon. Moreover the groups 1IÏ 4 , IV 4 , \ \ 

 furnish successively a point Pi, a point Q,-, a point M%. So r K is a 

 diameter through a vertex of the pentagon, e.g. P l M l Q l . Hero no 

 other point of intersection appears. 



11. It would be possible to go on in this manner and to treat 

 successively, proceeding from the easier cases to the more difficult 

 ones, the remaining lines through two remarkable points, the lines 

 through only one remarkable point, the lines parallel to one of the 

 sides. We prefer however to explain now, for an arbitrary case, 

 how the ratio of division of the side of the starpentagon corre- 

 sponding to a determined group of icosahedra can be found by means 

 of Fig. 3 of the quoted memoir, which is repeated here with slight 

 modification as fig. 4. 



We therefore consider the group IV, of plate II 6 mentioned above 

 under d x , and remember that the icosahedral sections corresponding 

 to this group are determined, according to the quoted memoir, by 

 planes normal to the plane of fig. 4 in a line parallel to pp 1 . If 

 the edge normal in q to the plane of that diagram is once more the 

 edge Ai> and the chord pp' situated in that plane the side of the 

 starpentagon, then the point S on that side determining the diagonal 

 plane in question is found by drawing through q the line qS paral- 

 lel to pp 1 . Now pin is the smaller segment of the line pp' divided 

 internally in medial section and the same relation holds for. p II r=wq 

 with respect to the segments p'r = p u s = sq. So if the ratio of the 



side of the regular pentagon to its diagonal is indicated by -, we 



d 



deduce from similar triangles 



phv : irq = pi i' :w<S, 



which may be transformed into 



p*q : phc = pS : pw . 

 This leads to 



pS _ 3d + 2s pw _ 3d + 2s d — s (2-f e) (3— e) 1 -f e 1 



pp ~ ~ 3'/ + * ' pp ~~ M 4- » 'I ~T4~*~ ~ <~-4 5 " T e ' 



19* 



