38 PROCEEDINGS OP THE AMERICAN ACADEMY. 



orator </'. As there is a finite number of pairs of values of ./• and y, we 

 may arrange them in order of magnitude, calling// the greatest, i. e. we 

 shall say that g is the greatest cumber of points of C"" 1 on any generator; 

 correspondingly, g is the smallest number of poiuts of C" 1 on any gen- 

 erator, since g + g' = k. Now a plane through // cuts out two genera- 

 tors, and as we revolve the plane about D' the number of points of C [,,) on 

 the two generators in the plane is constant and equal to a — 8', where 

 8' is the number of points of G' (a) ou JJ . If then we pass a plane through 

 J)' and a generator g' , having the least number of points of C 1 " 1 on it, 

 the other generator in this plane must have the greatest number g of 

 points of < '" on it ; therefore g + g — a — 5', and since g + g' = a — 8, 

 we have 8 = 8', i. e. C'"' meets each of the double director.-. /) and 1)', 

 in the same Dumber of points, 8. If now a plane be passed through a 

 generator g and the director D it will cut out a generator g\ and if 

 through this generator //' and L/ we pass a plaue it will cutout another 

 generator </, so that there are at least t.vo generators g. Through these 

 two generators g, the directors D and />' (which four lines form a 

 gauch-quadrilateral), and an arbitrary point we can pass a quadric ; 

 I), D' , and the two generators counting for 2 + 2 + 1 + 1 = ('. lines in 

 the intersection of the quadric and scroll. Each generator meets the 

 quadric in two points, one on D and one ou D\ and it we take the arbi- 

 trary point on any generator, this generator will lie on the quadric, and 

 the remaining intersection of the quadric and scroll will be another gen- 

 erator. Thus by varying this arbitrary point all the generators of the 

 scroll, two at a time, will be successively cut out by a variable quadric 

 that always contains D, /)', and the two chosen generators^. This 

 quadric always meets C"" in 2a points, of which 28 lie on D and I >' 

 and 2g lie on the two chosen generators </, so that the remaining 

 2 " — 2 8 — 2 g points lie on the other two generators cut out b\ the 

 quadric , bul a — 8 = g + </' and therefore 2 a — 2 8 — 2 g = 2 </' . 

 Now there is no generator thai has fewer than g points of ('• mi it and 

 the number of points of <'" on the two generators together Is 2 g / j 

 therefore each must have g f points of C?"' ou it. Therefore everj gen- 

 erator of the scroll, excepl the two chosen generators g, has g* points of 

 ( on it. In like manner, if wechoose two generators g*, through which 

 (he variable quadric is always to pass, we have the Bum oi the number 

 of points of '"""on the other two generators cul onl \>\ the quadric 

 always equal to 2 a - 2 8 — 2 ;/ 2 g, and, since do generator hat 



; •<• than g points of CW on it, everj generator on the scroll, excepl 



the tw( has g points of on it. fherefore <i </ and 



