514] ON THE PROBLEM OF THE IN-AND-CIRCUMSCRIBED TRIANGLE, 
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which is 
= X 2 ( # + 1) 
+ X (# 2 — 2# — 19) 
+ # 2 — 19# 
+ £(- 2X — 2x+ 18). 
And then, by means of the equations 
we verify that 
(x — 4) 2t = (x — 4) (X 2 — X + 8x — 3f), 
(X - 4) 28 = (X - 4) (# 2 - # + 8X - 3£), 
(# — 3) t = (# — 3) ( — 3# + f), 
(X-3)* = (X— 3) ( — 3X + a 
d = (# — 4) 2t + (X — 4) 28 + (# — 3) t + (X — 3) k. 
S3. Correspondence (a, e). 
From the values of d — 8 — S', c — 7 — y we have 
e — e — e =(— X 2 + 13X + 4# — 54) A, 
and then 
e = e' = (X — 2) (a: — 3) (X — 3) (« — 3); 
that is 
e = 2 (# — 3) 2 (X — 2) (X — 3) 
+ (— X 2 + 13X + 4# — 54) (— 2X — 2# + 2 + P), 
which is 
= X 3 ( 2) 
+ X 2 ( 2# 2 —10# -10) 
+ X (— 10# 2 + 26# + 44) 
+ 4# 2 + 44# 
+ | (-X 2 + 13X + 4# - 54), 
and then 
(# — 4) (# — 5) 2t = (# — 4) (# — 5) (X 2 — X + 8# — 3a 
{(X - 4) (X - 5) + # - 3} 28 = {(X - 4) (X - 5) + # - 3} (# 2 - # + 8X - 3a 
{ 3 (# — 3) (# — 4) + # — 3} i =(#-3)(3#-ll)(- 8# + a 
2(X — 3) (X — 4)« = 2 (X — 3)(X- 4) (- 3X+f); 
and summing these values and comparing, 
c = (x - 4) (# - 5) 2t + 2 (X - 3) (X - 4)« 
+ [(X - 4) (X — 5) + # — 3] 28 + [3 (# — 3) (# - 4) + # - 3] i. 
The united points (a, e) are in fact, 1°, each of the # —4 intersections of a double 
tangent with the curve, in respect of the two contacts and of the remaining # — 5 
intersections; 2°, each double point in respect of the two branches and of the pairs 
of tangents from it to the curve; 3°, each of the # — 3 intersections of each of the