PROBLEMS AND SOLUTIONS. 
567 
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Then the points 0, O', 0" lie on a conic circumscribed about the triangle ABC. 
In fact, if in the subsidiary theorem the inscribed conic be a circle, and the 
points I, J be the circular points at infinity, the point 0 will be the centre of the 
circle, that is, the point of intersection of the interior bisectors of the angles ; O' will 
be the point of intersection of the lines to the points of contact of the inscribed 
circle; and 0" the point of intersection of the perpendiculars on the sides of the 
triangle; and, these three points being on a conic circumscribed about the triangle, 
the general theorem will apply to the three points in question. 
I first prove the subsidiary theorem. Taking x = 0, y = 0, z — 0 for the equations 
of the sides of the triangle and (a, ¡3, 7), (a', /3', 7') for the coordinates of the points 
I, J respectively; the equation of the inscribed conic is 
V® > \!y , \/z = 0, 
Va, V/3, V7 
Va, VÆ', V7 
or say 
a \]x + b \]y + c \fz = 0, 
where 
a = \/ßy — V/3'7 =p — p 1} b = V^yof — t/ya = q — q 1} c— Va/3' — VV/3 = r — r lt suppose. 
The coordinates of the point of intersection of the lines from the vertices to the 
points of contact on the opposite sides are 
1 
¥ 
1 
that is, 
111 
(p-pi? ‘ (q-qi) 2 ’ ( r - n) 2 ' 
The equation of the line IJ is 
((3y — /3'y) x + (7a! — y'a.) y + (a/3' — a'¡3) z = 0 ; 
or, what is the same thing, 
(p 2 —pi*)x+ (q 2 — qi 2 )y+ (r 2 — rf) z — 0. 
Representing this for a moment by \x + ¡xy + vz = 0, the coordinates of the pole of 
this line, in regard to the inscribed conic a s/x + b \Jy + c \/z — are as 
c 2 ya + ¥v : a?v + c 2 A. : ¥\ + a?fx. 
Now 
cyi + ¥v = (r — r^f (q 2 — qf) + (q — q^f (r 2 — rf), 
= (r- n) (q ~ ?i) [(r - r0 (q + ft) + (? - ft) (r + r a )], 
= 2 (r - r a ) (q - ft) (qr - ftn),