578 
PROBLEMS AND SOLUTIONS. 
[705 
Equating the discriminant to zero, this is 
Or finally 
0 = A 2 V = 4 (AC — B 2 ) 3 - (3ABC - A 2 D - 2B 3 ) 2 . 
{3a 2 x 2 + 3 b 2 y 2 — 4a 4 + 4a 2 ò 2 — 45 4 ) 3 
+ {9 (a 2 — 2b 2 ) a 2 x 2 + 9 (6 2 — 2a 2 ) b 2 y 2 — 8a 6 + 12a 4 Z> 2 + 12a 2 6 4 — 6 6 } 2 = 0, 
which is the required equation. 
[Vol. xxi., January to June, 1874, pp. 29, 30.] 
4298. (Proposed by J. W. L. Glaisher, B.A.)—With four given straight lines 
to form a quadrilateral inscribable in a circle. 
Solution by Professor Cayley. 
Let the sides of the quadrilateral taken in order be a, b, c, d\ and let its 
diagonals be x, y\ viz. x the diagonal joining the intersection of the sides a, b 
with that of the sides c, d; y the diagonal joining the intersection of the sides 
a, d with that of the sides b, c; then, the quadrilateral being inscribed in a circle, 
the opposite angles are supplementary to each other. Suppose for a moment that 
the angles subtended by the diagonal x are 6, tt — 6, we have 
and thence 
that is, 
and similarly 
x 2 = b 2 + c 2 + 2 be cos 6, x 2 = a 2 + d 2 — 2 ad cos 6 ; 
{ad + be) x 2 = ad (b 2 + c 2 ) + be (a 2 + d 2 ) = (ac + bd) {ab + cd), 
x 2 = {ac + bd) 
ab + cd 
ad + be ’ 
y 2 = {ac + bd) 
ad + be 
ab + cd’ 
agreeing as they should do with the known relation xy = ac+ bd: the quadrilateral 
is thus determined by means of either of its diagonals. It is however interesting 
to treat the question in a different manner. 
Considering a, b, c, d, x, y as the sides and diagonals of a quadrilateral, we have 
between these quantities a given relation, say 
F {a, b, c, d, x, y) = 0, 
and the quadrilateral being inscribed in a circle, we have also the relation xy = ac + bd; 
which two equations determine x, y; and thus give the solution of the problem.