50 PROCEEDINGS OP THE AMERICAN ACADEMY. 



they must have one more edge in common ; if, then, we vary continu- 

 ously the fifth arbitrary edge that determines the quadric cone, this cone 

 will always pass through the three double or cuspidal edges and the edge 

 A of the quartic cone, and will cut out all the edges of the quartic cone 

 one at a time. Every quadric cone meets C(") in 2 a points, of which a 

 definite number lie on the three double or cuspidal edges and the edge A, 

 and therefore the same number of points of (7("), say a points, lie on 

 each edge of the quartic cone ; it is evident that, if any other edge be 

 chosen through which the quadric cone is always to pass, the edge A will 

 have a points of C(") on it. If we pass an arbitrary plane through any 

 of the double or cuspidal edges, it will also cut out two edges of the 

 quartic cone, and, if 8 be the number of points of C(") on this double or 

 cuspidal edge, we have 



8+2a:=a — k=4:a, i. e. 8 = 2 a, 



since an arbitrary plane through the vertex cuts out four edges, giving 

 4a = a — k. 



To prove the theorem for group (III), we may employ a method 

 analogous to that used for the Quartic Scroll 'S'(l2, Ig, 4), p. 37; corre- 

 sponding to the planes there, through the double directors, we should 

 use here the planes through the double or cuspidal edges, and instead of 

 the quadric used there we should now employ a quadric cone. 



We shall not consider this proof in detail, but shall now give a proof 

 of the theorem for groups (IV) and (V), — a proof which holds for the 

 other three groups as well, and therefore for all quartic cones. 



If JO is the number of points of C^") on any edge of the cone, it is evi- 

 dent that < jD < a — 1, and p can have, at most, a different values. 

 For convenience, we shall say that all the edges for which p has the 

 same value belong to the same set ; and if there is an infinite number of 

 edges in a set, we shall call it an infinite set, or, if there is a finite num- 

 ber, a finite set. There are, at most, a different sets, and, since a is 

 finite, at least one of these must be an infinite set. Let us suppose, first, 

 that there is only one infinite set, and let a be the value of p for this 

 infinite set ; we can then pass a plane through one of these a-edges 

 (i. e. an edge having a points of C^") on it) and turn it about so that it 

 shall not contain any edge of any finite set ; this plane will then cut out 

 four a-edges, and as it meets C^") in a points, of which k lie at the ver- 

 tex, we have 



4 a -\- k := a, i. e. a — k ^= Aa. 



Now we can pass a plane through any edge of any of the finite sets and 



