738 MR DUNCAN M. Y. SOMMERVILLE ON 
p,=2 4n,—A 4-iA 
anare MV On = 4 +o 
A=2.. (n= [A=6. n,=3=n,, excluded. ] 
n,=5 
A=4. m=3 A> 6. 
nm, =4 
§ 20. Class C. We have here the equation 
ni-d)en-Zen(ie 
r( ~ ny Po Ny Ps a <2, 
or NN.N3(2 — p) + 2(pyNgNz + Pog, + pa) = A >O0. 
We have to solve this equation under the following conditions: 7, + nm, + 1; 5 3, 
3 > p+ 4, therefore p1, p., ps » 2, and we cannot have m1=5, m,=6, n,=7 
together. 
Further, if p,=p,=p3=1, %, %,, and n; must all be even, for the 7,-gon must be 
surrounded alternately with n,.-gons and 1;-gons. Again, if p=4, n;=3, p,;=2, then 
starting with an ,-gon we must have on successive sides an 7,-gon, a double triangle, a 
double triangle, an m»-gon, and so on (fig. 18). Hence ”,, and similarly n,, must be a 
multiple of 3. 
The only possible sets of values of p,, p., 3, 3 are therefore 
Pi =P2,=P3=1, n3=4; 
p,=2, Pi =P.=1, n,=3 or 4. 
We have then 
He nn, p,—- A 
N(p — 2. m3 — 2s) ~ 2pyns 
tata i heh, 16—3A 
nm,=4 1 2n, -8 My -4 * 
Since n, and n, are both even, A must be a multiple of 8. 
NERS Wage (i [A=24. n,=6=n,, excluded.] 
n, =10 
BeaNe. My 6 [A=32. n,=4, excluded. ] 
n= 8 
Pie tim | 6n,-A_, 9-4A 
i 1 — aT ae 
as if 1 Qn, - 6 Ny — 3 
Since ”, and n, are both multiples of 3, A must be a multiple of 18, but A=18 gives 
n,=3 and any higher multiple is excluded, hence there are no developable nodes of this type. 
Dus Pie *) _8m-A . 4-4a 
ie V 4mg-8° " * ny —2 
= Z 
3 
A=4, 1, =3 [A=8. ro" | excluded ] 
n,=5 nm =4 
[A=12. »,=—3=n, , excluded,] 
