176 Mr. W. M. Hicks on some Effects of Dissociation 

 Zy y, or y > 1 ; that is, the roots must be such that 



X and y > < i, 



z, 1-Z-2.V, l-z-2i/>0<l. 



If we regard x, y, z as coordinates of a point in space, this 

 condition asserts that the point may lie anywhere within a 

 pyramid whose base is a square having the axes of a; and ?/ for 

 adjacent sides and the length of a side = ^, and whose vertex 

 is on the axis of ^ at a height 1 from the origin. It can be 

 shown, as follows, that we always have one, and only one, 

 solution of the above equations satisfying these conditions. 



Let A, B, (Plate III. hg. 1) be the axes of coordi- 

 nates, A D B C the pyramid within which Qv, y, z) must 

 lie. The three equations (1) represent three quadrics ; and we 

 have to show that they have one, and only one, intersection 

 within this pyramid ; and this we shall do by proving that 

 they divide the pyramid in such a way that they must cut one 

 another at some one point within, and at only one. Taking 

 the first, or 



asz{^—z — 2xy — x\2aiiX-^ai2?/ + aioZ-\-ai^.i/-^auy^\-0, 



we see that it has a generating line through C parallel to B ; 

 also it cuts both A and B I) between 0, A and B, D. For, 

 putting z=y = 0, we have, to find x, 



as^(l — 2xy—a;^2aii.v — 2ai-sx + ai^-hau\ =0: 



when ^ = this is positive, and when x = ^ it is negative ; thus 

 there is one, and only one, root between and A. Similarly 

 there is one, and only one, between B and D. Thus the sur- 

 fiice divides the pyramid into two portions by a diaphragm 

 extending from C, and intersecting the faces G A, C B D, 

 and the base. 



Similarly the second surface divides the pyramid by a dia- 

 phragm extending from C, and intersecting the faces COB, 

 C A D, and the base. Therefore they must intersect on a 

 simple curve extending from C to a point E in the base, and 

 lying between wholly within the pyramid. 



The third quadric has B D, D A for generating lines, and 

 also cuts C in one, and only one, point. For putting 

 .r = z/ = 0, we have, to find z, 



a,^-zy-z { 2aooZ + (^oa + ao4)(l -^)\=^ 0, 



which is positive when z = 0, and negative when^=l. Hence 

 this again divides the pyramid by a single continuous dia- 

 phragm extending from B D, D A across to a point in C ; 



