570 
PROBLEMS AND SOLUTIONS. 
[383 
and if X, Y, Z are the coordinates of the point il, then we have 
X Y Z A 
^ ~o "h = 0, 
a ß y 
X Y t Z 
T7 + ö> d / = 0, 
a P 7 
X Y Z 
ol" + ß" + y" 
0. 
The equations of the sides of the inscribed triangle obtained by means of the 
point 0 are 
x y z _ 
-- + ! + - = °, 
a ß y 
X y z 
- - I + - =0, 
a ß y 
*+y_£=0 
« + /8 y 
and the like for the triangles obtained by means of the points 0' and 0" respectively. 
Hence, for a set of corresponding sides of the three triangles, we have, e.g., 
x , y z _ x y z n x y z 
-- + ^ + - = 0, - - + 4 + - = 0, - -, + L + -, 
cl ß y a ß y a ß y 
0, 
and it is clear that these equations are simultaneously satisfied by the values 
x : y : z = -X : Y : Z, 
and we have the like expressions for the other sets of corresponding sides; that is, 
we have for the coordinates of the vertices of the resulting triangle 
(-X : Y : Z), (X : - Y : Z), (X : Y : -Z); 
and hence also the equations of the sides of the triangle in question are 
X 
Y 
+|-°, 
z x . 
Z + X~ 0, 
X + F -0 ’ 
that is, it is a triangle circumscribed about the triangle ABO. The equations of the 
lines joining the corresponding vertices of the two triangles are 
y _ Z Z _ X X __ y 
Ÿ~Z’ Z~X’ X Y ’ 
and these lines meet in the point (X : Y : Z), which is the point H, the intersection 
of the polars of 0, O', 0"; the demonstration of the theorem is thus completed. 
{The expression Polar of a point in regard to a triangle denotes a line constructed 
as follows :—viz., 0 being the point and ABO the triangle, then, taking on BO a point 
a, the harmonic in regard to the points B and G of the intersection of BO by AO] 
and in like manner on OA and AB the points b and c respectively, the three points 
a, b, c lie on a line which is the polar of the point O. If the equations of the 
sides are x = 0, y — 0, z = 0, and the coordinates of the point are (a, /3, y), then