62 
UR. H. M. TAYLOR ON A SPECIAL FORM OF THE 
There must, therefore, be 2592/216 = 12 closed pentagons for each duad. 
There are ten lines which do not cut either of the lines of a duad. 
Therefore these twelve pentagons are formed of ten lines, and these ten lines form 
pairs of pentagons in six different ways. 
One such pair of pentagons, none of the sides of which cut either of the lines 
26, 27, is 
18, 5, 14, 2, 12 and 24, 9, 13, 10, 6. 
§ 16. Closed Hexagons. —If the lines a, h, c, d, e,f, taken in order form a closed 
liexagon, it appears that when a is given the number of ways of choosing h, c and d 
is 10.8.4, when a, h, c, d are given, the number of ways of choosing e is 1, and 
that when a, h, c, d and e are given, the number of ways of choosing/ is 1. 
Hence the number of orders of choosing six lines to form a closed hexagon is 
27.10.8,4. 1. 1, 
and each hexagon will appear 6x2 = 12 times. 
The total number of closed hexagons, therefore, is 9. 10.8 = 720. 
Now we have shown that there are only three lines, forming a non-intersecting 
triad, which do not cut one at least of the sides of a closed hexagon. 
There must, therefore be 720/720 = 1 closed hexagon for each triad. 
There are six lines which do not cut any of the lines of a triad. 
Therefore, there is but one closed hexagon formed of the six lines which do not cut 
a triad. 
The hexagon, none of wliose sides cut any of the lines 26, 27, 5 is 1, 2, 12, 
10, 13, 9. 
If a, h, c, d, e, f be the sides of a closed hexagon in order, every line on the 
surface which does not meet a, h. c, d, e or / must meet the lines which meet the pairs 
a,h; h, c; c, d; d, e ; e,f; f, a. 
Now the intersection of the planes a, h and d, c, is a line on the surface ; that is, 
the lines joining a, h ; b, c ; c, d are identical with those joining d, e ; e, f; f, a 
respectively ; and the three form a non-intersecting triad. 
Three other lines, forming a non-intersecting triad, meet them, and they are the 
three lines each of which misses each side of the closed hexagon. 
§ 17. From the closed hexagon, formed of the lines a, b, c, d, e,f, we can form six 
planes, ab, be, cd, de, ef,fa, such that the planes ah, cd, ^intersect the planes be, de,fa, 
in nine of the twenty-seven lines. 
Hence the number of ways of throwing the equation of a cubic surhice into the form 
LMN = PQH, may be found as follows : 
From each such form of the equation we can obtain six closed hexagons, and from 
each closed hexagon we can obtain one such form of equation. 
