The second circle serves to tell us that the resulting traingle FHG is isoscelese, with one of its equal sides lying coincident with the xaxis and therefore with the diameter of the first circle.
For purposes of finding the area consider the base to be segment HF, and altitude measured from point G to that line.
The altitude intersects HF at some point (x,0), and G has coordinates (x,h) where h is the height of the triangle.
Consider point G' on the circle, with coordinates (x,h). Chord GG' intersects diameter EF, dividing GG' into two segments of length h and EF into segments of length 2+x and 2x.
Then h*h = (2x)(2+x).
So h = sqrt(4x^2), but we knew that already from G being on the circle.
Since the triangle is isosceles with the base HF being equal to side GF, the length of the base is the same as that of GF: sqrt(h^2 + (2x)^2).
Substituting for h we get the base equals sqrt(84x) = 2 sqrt(2x).
Thus the area of the triangle is sqrt((4x^2)(2x)) = sqrt(82x^24x+x^3).
Differentiating the above, the numerator, which has to be set equal to zero, is 3x^24x4. Applying the quadratic formula gives solutions as x = 2/3 +/ 4/3.
x=6/3 = 2 results in a zero area "triangle".
x=2/3 results in
h = sqrt((8/3)(4/3) = 4sqrt(2)/3
base = sqrt(32/3) = 4sqrt(2)/sqrt(3)
Area = 32/(3sqrt(3))/2 = 16sqrt(3)/9 ~= 3.07920143567801

Posted by Charlie
on 20080111 13:22:04 