10 Mr Blythe, On the forms of [Nov. 11, 



Take areal coordinates, the tetrahedron of reference being 

 that formed by the planes a, /S, y, 8 ; and let 

 fx = (7ja + mS — y) ih + 8, 

 v = (l 2 a + m 2 @ — 7) n 2 + 8. 



Let the lines 4, 9, 13 form an isosceles right-angled triangle, 

 hypotenuse in 9. The planes 4, 12, 2 and 13, 14, 15 are per- 

 pendicular to 4, 9, 13. 



15 cuts the edge 13 in the ratio l 2 : 1, 



14 „ 13 „ k : 1, 



2 „ 4 „ m x : 1, 



12 „ 4 „ m 2 \ 1. 



If 7 = cut a = 0, /3 = at a distance (^ above 8 = 0, and 14, 2 

 and 12, 15 meet a = 0, ft = at distances d, and c? 3 above S = 0, then 



7b 1 {d. 2 — ^i) = c£ 2 , 



n 2 (d 3 — o\) = d s . 



We see therefore that by taking arbitrary positions for nine of 

 the straight lines we determine all the constants in the equation 

 a{3y = K8/j,v, except K. 



Take a point 0, {3 1} y 1} on the straight line 4. Denote the 

 ratio 7! : & by p, we shall find that 



X x a = 8, \. 2 fi = fi, \y=v 

 will reduce to two independent equations both satisfied by 0, /3 X , 

 713 if 



\ 2 =n 1 {m l -p), \ 



X 3 p = ?i 2 (ra 2 - p), \ (A). 



and a, x (m 2 ?i 2 — ihp) = — >V l 2 (^Wa — hp) J 



These two equations will represent a straight line on the 

 surface if XiX 2 \ 3 (x/3y = 8p,v coincide with a/3y = K8fiv, that is if 



A-jA^A^-fi = J.. 



If K be known it is evident that using this equation with (A) 

 we get a cubic equation for p. 



When p is determined we get but one value of \, \ 2) X 3 for 

 each value of p, and thus obtain three definite straight lines on the 

 surface. 



If, however, we consider K as unknown, and take an arbitrary 

 value for p, say p x , we find K from the equation 



p 2 (m 2 n 2 - ^pj) = - Kn?n? (raj - p x ) (m, - p x ) (l^h - kpi), 

 and the other values of p from the quadratic 



(Iji^ - l. 2 p) (m 2 — p) (nh — p) _ (m. 2 n. 2 — i^p) p 

 (l 1 m. 2 -l 2 p 1 )(m 2 -p 1 )(;m 1 -p 1 ) (m. 2 n 2 -')i 1 p 1 ) p ± ' 

 obtained after simplification and division by p — p 1 . 



