If we call the starting point H, then for a given BH and CH, the area of the triangle is maximised when AH is perpendicular to BC. The same is true for the other sides, so BH is perpendicular to AC and CH is perpendicular to AB. This means that H is the orthocentre of the triangle (the point where the three altitudes cross).

Now, for any triangle, the distance between a vertex and the orthocentre is the diameter of the circumcircle multiplied by the cosine of the angle at that vertex, so

25=DcosA

33=DcosB

39=DcosC

This means that the cosines are in the ratio 25:33:39, and because they are vertices of a triangle, the angles sum to 180 degrees.

Let’s use the following identity that is true for angles of a triangle:

(cosA)^2 + (cosB)^2 + (cosC)^2 + 2*cosA*cosB*cosC = 1

Using cosA=25/D etc:

(25^2+33^2+39^2)/D^2 + 2*25*33*39/D^3 = 1

3235D+64350=D^3

Which has the obvious integer root of D=65.

Now we know the circum-diameter is 65 we can calculate the cosines and consequently the side lengths of the triangle:

cosA = 25/65, by Pythagoras the opposite side is 60

cosB = 33/65, opposite side is 56

cosC = 39/65, opposite side is therefore 52.

Now we know the sides of the triangle we can use Heron’s formula to calculate the area.

Area is sqrt(s(s-a)(s-b)(s-c)), where s is the semiperimeter, in our case 84

Area = sqrt(84*24*28*32) = 1344 square miles