Given three concentric (having the same center) circles.
Their radii are 1,2 and 3.
Place three points A,B,C - each on a different circle to get a triangle ABC
with a maximal area.

Without loss of generality, let the side BC be parallel to the x axis and at a distance d below it with OB = 2 and OC = 3 so that 0 < d < 2. For the triangle to have maximum area, the point A must then be at the highest point of the unit circle and the three points must have the following coordinates:

Just as the line AO is perpendicular to side BC for maximum area, by similar arguments BO is perpendicular to CA and CO is perpendicular to AB, making O the orthocentre of triangle ABC.

Using the gradients of AB and CO:(d + 1)/sqrt(4 - d^{2}) = sqrt(9 - d^{2})/d

which can be squared to gived^{2}(d + 1)^{2} = (4 - d^{2})(9 - d^{2})

then simplified tod^{3} + 7d^{2} - 18 = 0

whose only solution between 0 and 2 is:d = 1.458757..

The area is given by 0.5(d + 1)[sqrt(4 - d^{2}) + sqrt(9 - d^{2})]and when

d = 1.4587.., the maximum area works out to be 4.904822..

[It does seem surprising that, despite knowing the lengths OA, OB and OC, and knowing that O is the orthocentre, we still need to solve a cubic to find the area. Am I missing something?]