62 REV. T. P. KIRKMAN ON THE 
have 3z triangles. If these triangles all vanish, the result 
will have 32’ (2 <x) triangles, and thus we shall at last 
reduce all the figures to-a triply reversible having only 3 
triangles, which is incapable of further reduction in the 
number of its triangles. Any axial plane bisecting the 
base must therefore pass through the vertex of a triangle, 
and bisect the triangle ; it must therefore contain the edge 
k which passes through that vertex. That is, three of the 
edges not meeting the base are axial edges; consequently 
the polyedra considered in the problem have three axial 
edges, and are obtained by the process of ow previous 
problems from (6z + 1)-edra, having 3z triangles, and three 
axial edges that cannot receive points 7. Taking any one 
of these (62+ 1)-edra for the subject of our operations, we 
have 62-6 edges in which to plant m points 7, z.e. ~—-1 
edges in the sixth part of the circuit on which to lay jm 
points 7, as our result is to be (Art. 5) triply reversible. 
We obtain 
w—2|1 
Gt+1) on 
-96 
l r—2|1 
results from every subject of operation, which are to be 
repeated in reversed order in all the six sextants between 
the axial planes so that all shail be triply reversible. Hence 
follows 
Gi) 
R®R3(62 + m+ 1,382) = -2°. R3(627+1,32), 
17 |1 
which vanishes for v=1, unless m=O, showing that no 
triply reversible with three triangles exists except the 
7-edron with three triangles. 
From Art. 15 it is evident that no doubly reversible 
can be thus generated from a triply reversible subject of 
operation. 
21. Prosiem m. To determine the number of singly re- 
