An elementary treatise on cubic and quartic curves, by A. B. Basset.

250

THEORY OF PROJECTION.

and this line intersects the circumscribing circle at a point B"
such that y = k/3; whence the coordinates of B" are
a=- kc /(1 + k), y = k3...............1..( 13).
Now if in (6) we substitute - (1 + k)/k for k, we shall find that
it is satisfied; hence this quantity is another root of the cubic,
which shows that B" is one of the points of intersection of the
circumscribing circle with the curve; hence B" coincides with B'.
In the same way it can be shown that the coordinates of C' are
a = ky, =- k/(1 + k).................. (14),
and that CC' is parallel to A'B. Whence the three pairs of
straight lines AA', CB'; BB', AC'; CC', BA' are respectively
parallel to one another; also - (1 + k)- is the third root of the
cubic (6). Collecting our results, we find that the coordinates of
A', B', C' are determined by the equations
A', f=ka,, =-ka/(l + k)
B', a=-k/3/(1 + k), 7y = k -............
C', a= ky, /==-ky/(l + ))
The equation of the tangent at A' is
a (32 + + 3) + 3 (332 ) + 3 + (32 + V3) = 0.
Substituting the values of (~, v, f) from the first of (15), this
becomes
a{3ic-(j }~h I - +3[ic}~y {/3 + 0.))2=o.(1,6).
{ (1-+k)3}+_g tI fi}+9 ' (1 k ()2...
To prove that the tangent at A' passes through B', substitute
for (a, f, y) in (16) the coordinates of B' from the second of (15),
and we obtain
k4 3k
(1 + k)4-3k2 + 1 + +(1 k)
which by means of (6) and (11) may be shown to vanish. This
shows that A'B', B'C', C'A' are the tangents at the points of
inflexion A', B', C'.
381. It remains to prove that the triangle A'B'C' is equilateral.
Since BB' is parallel to AC', and ABB'C' lie on a circle,
B'C'A = r -- C'B'B = BA C';
whence A + CAC' = C' + AC'A'.