The collected mathematical papers of Arthur Cayley.

401] A NOTATION OF THE POINTS AND LINES IN PASCAL'S THEOREM.

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viz. HABNJ, is the pentad which contains HA, the arrangement of the last three
letters B, N, J thereof being arbitrary; HEFLD is the pentad that contains HE, but
the last three letters are so arranged that the columns HBF, HNL, HJD are each of
them a triad, IMG is then the residue of the pentad AEIMIG, and KCO is the
complementary triad to AEH, but the arrangement of the letters IMG, and of the
letters KGCO, are each of them determinate; viz. these are such that we have BFICO,
NLMKO, JDGCK, each of them a pentad.
And this being so we derive from the arrangement
2g AH, EH;
3 m KC, KO, CO;
6h AI, AM, AG; El, EM, EG;
12z IB, IF, MN, ML, GJ, GD; HB, HF, HN, HL, HJ, HD;
9p AB, AN, AJ; EF, EL, ED; BF, NL, J D;
12r CB, CF, CJ, CD; OB, OF, ON, OL; KN, KL, KJ, KD;
12t FL, FD, LD; BN, BJ, NJ; IC, IO; MK, MO; GK, GC;
3w IM, IG, MG;
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viz. the line AE in question meets AH, EH each of them in a point g; KG, KO, CO
each in a point m; and so on. By constructing in the same way an arrangement for
each of the lines AH, &c., we find the nature of the point of intersection of any
two of the lines AB, AE, AH, &c.; and we may then present the results in a table
(see Plate), which shows at a glance what is the point of intersection (whether a
point g, m, h, z, p, r, t, or w) of any two of the Pascalian lines.
I further remark that representing the 45 Pascalian points as follows:
12.34=a 13.24=g 14.23=m 15.23 = s 16.23=y
12.35 = b 13. 25= h 14. 25 = n 15.24 = t 16.24 =
12.36 = c 13.26 = i 14. 26 = o 15. 26 = u 16. 25= a
12.45=d 13.45=j 14.35=p 15. 34 = v 16.34==/,
12. 46 = e 13.46 = k 14.36 = 15.36=w 16.35 = y
12.56=f 13.56 = 1 14.56 = r 15.46 = x 16.45=8
23.45 = 25.34 = X 34. 56 = p
23.46 = 25.36 = 35.46 =
23.56 = 25. 46 = v 36.45 = r
24.35 = 26.34 =
24. 36= 26.35 = w
24.56 = 26.45 = r