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PROBLEMS AND SOLUTIONS.
[485
Solution by the Proposer.
The equation y 3 = x?z, is satisfied by the values x : y : z = 1 : 0 : 6 3 ; and con
versely, to any given value of the parameter 0 there corresponds a point on the cubic
y 3 = x 2 z. Consider the five points corresponding to the values (0 1 , 0,, 0 3 , 0 i} 0 5 ) respec
tively ; the equation of the conic through these five points is
x 2 , 2/ ) ) yz, zx, xy
l, 0s, 0 X \ e x \ 0 1
where the remaining four lines of the determinant are obtained from the second line
by writing therein 0 2 , 0 3 , 0 i} 0 5 successively in place of 6 1 . Writing for shortness
r*(ft, 0», &3, 0a) 0 5 ) to denote the product of the differences of the quantities
(0 l , 0 2 , 0 3 , 0 i} 0 5 ), the equation contains the factor ^ (0 2 , 0 2 , 0 3 , 0 4 , 0 5 ), and we may
therefore write it in the simplified form
?*(ft, ft, ft, ft, ft)
x>, y 2 , z 2 , yz, zx, xy
1, 0?, 0', 0‘, 0 3 , 0\
= 0.
Hence putting in this equation 0 1 = 0 2 = 0 3 = 0+ = 0 5 = <f>, we have the equation of the
conic of five-pointic intersection at the point The result in its reduced form
may be obtained directly without much difficulty, but it is obtained most easily as
follows: let the function on the left hand of the foregoing equation be represented by
(a, b, c, f g, h\x, y, zf,
then writing x : y : z = 1 : 0 : 0 3 , we have
(a, b, c, f g, h][ 1, 0, 0 3 ) 2
1
0., 0 3 > 04» 0.)
1, 0 2 , 0 6 , 0 i , 0 3 , 0
1, 0 2 , 0A 0 2 \ 0 3 , 0,
(0-0 1 )(0~0,)(0-0 3 )(0-0 i )(0 - 0 5 )
r*(ft ft, ft, ft, ft, ft)
l, ft, ft, ft, ft, a
l, <v, ft*, ft*, ft*, ft
= (0-ft)(0-ft)(0-ft)(0-ft)(0-ft)(0 + ft+ft + ft + ft + ft);
for the determinant, which is a function of the order 16 in the quantities (0, 0 1} 0. 2 , 0 3 , # 4 , 0 5 )
conjointly, divides by (0, 0 lf 0 2) 0 3 , 0 it 0 5 ), which is a function of the order 15; and
as the quotient is a symmetrical function of 0, 0 1} 0. 2 , 0 3 , 0 it 0 5 , it must be equal,
save to a numerical factor which may be disregarded, to 0 + 0 2 + 0 2 + 0 3 + 0 i + 0 5 .
