SOLUTION OF THE PROBLEM OF THE POLYEDRA. 95 
edges, showing the two faces and the two summits bound- 
ing each edge; for if the edges 
_E and F make an angle on the 
solid, they will have then both a 
common face and a common 
summit. 
Thus to form a paradigm of 
the 9-acral 8-edron here figured, 
which is one of those described 
by the equation, 
5, +5pt+4e4+4e+3p4 304 324+3u 
=5,+4,+3,+3,4+3,4+8,4+3,+3,+3,=2-15, 
we mark the faces in any way A BC.-.-.-, and the summits 
abc--- Wenext form a table of eight columns, headed 
ABC..-, making nine rows headed by abc:- 
ay 
ye ae se A oe oh Om 
eel. Many te Pao ta, eyes) Ee, 
eee ob tot: teen oe, 
ie aaa Begone! 
Co) aa ag A wee 
3 D,, F, E, 
3 D, F, G 
Bak AG Down uiiGe arate Ses 
a gt idee pale ara gata aes 5 
3 Rife Cadet wr hans Site 
putting M,, for the point common to the column M and 
the row n, and entering M,=1, or M,=0, according as 
the face marked M has or has not the summit marked n 
for one of those about it, 7.e. according as the summit has 
or has not the face for one of those about it. 
Thus we have entered A,=1=F,=B,, because the 
pentace a is in the pentagon A, the tetragon F and the 
triangle B; and D,=0=G,, because that pentace is not 
