11 
1901-2.] Prof. Chrystal on Theory of Miller, s Trisector. 
We have obviously 
x 2 + y 2 = a 2 sec 2 2 0 , yjx = tan 30 . 
Hence, if 
we have 
X = J{(x 2 + if - a 2 ) /a 2 } , Y = y/x, 
tan 20 = X , tan 30 = Y . 
Therefore 
tan 0 = tan (30 - 20) , 
= (Y - X)/(l + XY) . 
Hence 
X = tan 20 = 2 tan 0/(1 - tan 2 0) , 
2 (Y -X)(l +XY) 
~~ (1 + XY) 2 - (Y - X) ’ 
The eliminant is therefore 
X{X 2 Y 2 - X 2 - Y 2 + 4XY + 1 j 
= 2 {XY 2 - X 2 Y - X + Y} . 
This may be written 
X(X 2 - 3)(Y 2 - 1) = 2Y(1 - 3X 2 ) ; 
hence 
X 2 (X 2 - 3) 2 (Y 2 - l) 2 = 4Y 2 (3X 2 - l) 2 , 
which gives 
(x 2 + y 2 - a 2 ) (x 2 + y 2 - 4 a 2 ) (x 2 — y 2 ) 2 
= ka 2 x 2 y 2 {3(x 2 + y 2 ) — 4} 2 . . (2). 
The equation (2) may be considerably simplified. Arranging, in 
the first place, according to powers of x 2 + y 2 , we have 
(x 2 - y 2 ) 2 {(x 2 + y 2 ) 3 - 9 a 2 (x 2 + y 2 ) 2 + 24 cfi(x 2 + y 2 ) — 16a 6 } 
= ia 2 x 2 y 2 {2)(x 2 + y 2 ) 2 - 24a 2 (^ 2 + y 2 ) + 16a 4 } . 
Hence we get 
(x 2 — y 2 ) 2 (x 2 + y 2 ) 3 — 9 a 2 (x 2 + y 2 ) 4 + 24 a\x 2 + y 2 ) 3 
- 1 6a 6 (aj 2 + y 2 ) 2 == 0 . 
If we reject the factor (a: 2 + y 2 ) 2 , we get finally for the equation to 
the locus of P 
(x 2 + y 2 )(x 2 - y 2 ) 2 - 9 a 2 (x 2 + y 2 ) 2 + 24a 4 (a; 2 + y 2 ) 
- 16a 6 = 0 .... (3). 
A circular sextic, which we may call Miller'’ s Trisectrix. 
