383] PROBLEMS AND SOLUTIONS, 
ature, 
nation 
cos 6, 
for 
iving 
the 
iding 
nurse 
< 4 • 
^ 2 ’ 
rertex 
ration 
of a 
The 
see 
for 
tively 
and 
nine 
rn to 
are situate on a conic, appears at once by writing down its equation : viz., 
(be + ca + ab) (x 2 + y 2 + z 2 ) — (a 2 a- b 2 + c 2 ) (yz + zx + xy) = 0, 
which is satisfied by the coordinates of each of the six points. 
2. It is interesting to remark that the six points on the conic form, not 
a general inscribed hexagon, but a hexagon such as is mentioned in Prob. 1512 
(vol. II. p. 51), viz., one in which the three diagonals pass respectively through the 
Pascalian points (intersections of opposite sides): in fact, in the hexagon 143526, forming 
the equations of the sides and diagonals, these are 
14. (b + c) x— ay— a z = 0, 
15. —cx— c y + (a + b)z = 0, 
16. —bx + (c + a)y— b z — 0, 
36. (a + b) x — 
34. — b x — 
25. (c + a) x — by— b z = 0, 
26. —ax— a y+ (b + c) z = 0, 
24. — c x + (a + b) y — c z — 0, 
Gy— c z = 0, 
b y + (c + a) z = 0, 
35. —a x +(b + c) y— a z = 0 ; 
so that the lines 14, 25, 36 meet in the point x = 0, y + 2 = 0, 
„ 16, 24, 35 „ y = 0, z +x = 0, 
„ 15, 26, 34 „ z = 0, x + y — 0. 
3. It is further to be remarked that the six points lie on the cubic curve 
a? + y 2 + z z xyz 
a 3 + b 3 + c 3 abc ’ 
and are consequently the six points of intersection of this cubic by the above mentioned 
conic. 
4. The points (x = 0, y + z = 0), (y = 0, z + x — 0), (z = 0, x + y = 0) are the three 
real inflexions of the cubic; hence, attending only to the cubic, and starting from the 
arbitrary point (a, b, c) on this curve, it appears by what precedes, that we may, 
by means of the three real inflexions of the cubic, construct the system of six points, 
(the construction is, in fact, identical with that given in my Solution of Problem 1744, 
vol. iv. pp. 32—37, [ante p. 597] the six points being six out of the therein 
mentioned eighteen points); and it further appears, that these six points lie on a conic. 
5. As regards the second part of the proposed Problem, consider the cubic curve 
x? + y 3 + z 3 + Qlxyz = 0 ; the three real lines containing the nine points of inflexion are 
the lines x = 0, y = 0, z=0; and the points A, B, C are therefore (y = 0, z= 0), 
(z =0, x = 0), (x = 0, y = 0) respectively. From each of these points we may draw to 
the curve six tangents, and we have thus on the curve eighteen points, which are a 
particular case of the system in the Solution of Prob. 1744. Or if from each of the 
points we draw two properly selected tangents, (when the cubic has an oval these 
c. y. 77