230 
ON THE PROBLEM OF THE IN-AND-CIRCUMSCRIBED TRIANGLE. [514 
Observe this is what the result for Case 1 becomes on writing therein a = D — x, 
viz. the opposite curves a, D may become one and the same curve without any 
alteration in the form of the result. 
Case 5. a = B = x. 
% = (X — 2) cDeFx, % = FeDcX (x — 2), 
where 
(X-2)x + X(x-2) = 2(Xx-X-x); 
therefore 
g = 2 (Xx — X — x) ceDF. 
Case 6. a — c = e = x: perhaps most easily by reciprocation of Case 7 ; or 
Second process, functionally by taking the curve a = c= e to be the aggregate curve 
x + x'. The triangle aBcDeF is here in succession each of the eight triangles: 
x B x JDx F 
x B x' D x' F 
X „ X „ X „ 
rp ry* rp 
tAj yy %Aj yy dj yy 
rp rp rp 
%Aj yy \Aj yy ib yy 
X 
)> 
X 
„ X 
a? 
» 
X 
„ x' 
X 
yy 
X 
„ a! 
where the two top triangles give cf)X and fcc' respectively; the remaining triangles all 
belong to Case 2, and those of the first column give each 2 (a? — x) x'BDF, and those 
of the second column each 2 (x 2 — x') xBDF. We have thus 
(f> (x + x ) — (f>x — (fix' = {6 (x 2 x' + xx' 2 ) — 12xx'} BDF. 
Hence obtaining a particular solution and adding the constants, we have 
(px — (2x 3 — 6x 2 + olx + /3X + yf) BDF\ 
it is easy to see that a, /3, y are independent of the curves B, D, F\ and taking 
each of these to be a point, and the curve a = c = e to be a conic, then it is known 
that <f)x= 2 ; we have therefore 2 = 16 — 24 + 2a + 2^8 + 6y, that is a. + /3 + 3y = 5. 
The case where the curve a = c — e is a line gives 0=2 — 6 + a + 3y, that is a + 3y = 4 ; 
but it is not easy to find another condition; assuming however y = 0, we have a = 4, 
/3 = 1, and thence 
cpx = (2a? — 6a? + 4x + X) BDF, 
or say 
g = [2x (x — 1) (x — 2) + X] BDF: 
this is a good easy example of the functional process, the use of which begins to 
exhibit itself; and I have therefore given it, notwithstanding the difficulty as to the 
complete determination of the constants.