148 
Proceedings of the Poyal Society of Edinburgh. [Sess. 
SECTION Y. 
On the Description of Geometrical Curves through Given Points 
§ 64. Lemma III. 
A curve C n meets a conic in 2 n points and a cubic in Sn points. 
The proof is analytical. 
F rom it is suggested : 
Cor. 1. Two curves C m and C n seem to cut in mn points. 
This is easily proved when one of the curves is y = x m ; but the general 
demonstration is beyond Maclaurin’s powers. The truth of the statement 
is assumed in what follows. 
Cor. 2. Two curves of degree n cut in n 2 points. Thus we may find 
two curves of degree n through the same n 2 points. Now the equation to 
C n involves \(n 2 + Sn) conditions, and .’. \(n 2 + 3ri) points may not be 
sufficient to determine a curve uniquely when \(n 2 -\-3ri) is not greater 
than n 2 . 
Thus nine points may not uniquely determine a cubic, and yet ten points 
are too many. 
[This is the source of the so-called Cramer’s Paradox. Cramer, who 
simply repeats what Maclaurin gives with the additional application 
to quartics, quotes Maclaurin as his authority (vide Cramer, Courbes 
algebriques). 
The paradox is therefore Maclaurin’s and not Cramer’s.] 
Cor. 3. If, of the points given to determine a C n , nr+1 lie on C r , where 
n >r, then either the problem is impossible or the C n degenerates into C 
along with C n _ r . 
Cor. 4. A curve C cannot have more than %(n — l)(n— 2) double points. 
Cor. 5. If, on a curve C m , three points are multiple of order m/2 and 
one of order — — 1, all the other points will be simple. 
§ 65. Projp. XXV 
shows how to draw a curve C n through 2^ + 1 given points one of 
which is an (n — l)-ple point. 
Prop. XXVI 
shows how to draw a C 2n through as many points as suffice to determine 
a C n , and other three points each of which is an ?i-ple point. 
