744 MR DUNCAN M. Y. SOMMERVILLE ON 
Now if an n,-gon occurs in the same network in a different combination pj’, 5’, p4; 
N,, N_, %,, then a,, the angle of the n,-gon, must satisfy the two equations 
T 
5 cos — 
= n , ’ = 
Ors = 4, Py 41+ Po O_ + Pyay= 27. 
- @& 
sin —4 
2 
If we substitute in the first equation a value of a, corresponding to a possible set of 
values of p,', p>, p, obtained from the second, we must get an integral value for n,. 
This will not in general happen. 
The following negative results may also be obtained at once : 
1. No composite types exist with only one kind of polygon, for the angles are all 
equal, and there must therefore be the same number of polygons at each point. 
2. No composite types exist with p=3 at each point, for the angle and the side 
determine n. 
§ 28. A certain number of composite types may be obtained by the following 
TuHrorEeM.— Whenever a simple type contains a group of polygons bounded by a 
regular polygon, the replacement of that group by a single polygon will in general gue 
a composite type. 
When such a group occurs it may be replaced by a single polygon, for in the 
corresponding polyhedron the boundary of the group lies in one plane. Further, the 
replacement of the group removes at least one line from the nodes at the boundary, and 
the resulting network contains some nodes with p lines and some with at most p—1, 
2.€. it contains at least two different species of nodes, and is therefore composite. 
It follows that, in order that the simple type may give a composite type in this 
way, p must be >3. It may happen that the angle of the boundary polygon is >180°. __ 
If we exclude this case we get the following composite types: 
1. From 3,”° by replacing five covertical triangles by a pentagon. 
This may be done in three ways, replacing 1, 2 or 3 sets by pentagons.* 
The three networks are of different types. They may be denoted as follows, the 
symbol within brackets denoting the species of node and the coefficient the number of 
times it occurs in the network. 
(a2) 6(8;) + 5(8,5)) (fig. 41) 
(6) 2(8;) + 6(8,5,) + 2(8,5,) (fig. 42) 
(c) 3(335,) + 6(3,5,) (fig. 43) 
2. From 3,°4.'° by replacing one of the octagonal groups (fig. 27 deleting the part 
within the heavy lines). 
12(3;4,) + 8(4,8,) 
3. From 3,°°5,”4,*° (fie. 36) by replacing the decagonal group. 
* Tf two opposite groups of five triangles are replaced by pentagons we get the simple type 3,1°5,? (like fig, 25). 
+ If both groups are replaced we get the simple type 4,°8,? (like fig, 24), 
