84 ON HEXAGONAL FACETS 1N A CIRCLE. 
IV. But this does not represent the number of entire facets 
in the circle, for many of these facets will be cut through by 
the circle, and the fractional parts must be rejected (by the 
question). We must therefore seek a more accurate method 
of calculation. 
V. Take now, as Casr I, that of N an odd number, the 
centre of the circle containing the facets, comciding with the 
centre of one of the facets. 
It is evident from fig. 1 that hexagonal facets must be 
arranged on any surface, plane or curved, in the following 
order : 
Ist. A facet in the centre. 
2d. Six facets round this central one. 
3d. Twelve round those. 
Ath. Eighteen round the last ; and so on, increasing by six 
in each term of the series. 
Hence the whole number of facets in the hexagonal ar- 
rangement, whose diameter N = 2n + 1, may be thus found. 
Let H be the number required, then H, = 3x.» +1+1. 
EXAMPLE. 
N = 35 
2n = 34 
n=17 
~. H,=51 x 18+1=919. 
Case II. 
VI. If N be even, the centre of a circumscribing circle 
will fall in the bisection (A) of a side of a facet (fig. 2). 
Here, therefore, we must find, by the rule for Case I, the 
number of the facets in the half hexagon whose centre is the 
centre of the facet next to A (observing that if n be the 
number of facets in AB, n—1 will be the number of terms 
in the arithmetical series), twice the number so found will be 
the number in the whole hexagonal arrangement, less by the 
row of facets on the diameter AC(= 2n). 
“. Hy, =3n.n—1+2n +1. 
It has now been shown how to find the exact number of 
