I assume that the "diameter of a triangle" means the diameter of the smallest circle containing it.
Assume the triangle has sides a, b and c and angles A, B and C (where angle A is opposite side a etc.). Calculate angle A from the law of cosines. Then:
a^2 = b^2 +c^2 - 2*b*c*cos(A) so:
A = arccos((b^2 + c^2 - a^2)/(2*b*c)) -------------------(1)
Then, using the usual construction for finding the center, O, of the circumscribed circle (i.e. a perpendicular bisectors of each side is constructed and all three meet at point O) we can draw the radius from O to A. This will divide angle A into two smaller angles (in some cases one angle will be larger than A and the other will be negative, but they will still add up to angle A). Using simple trigonometric formulas we can calculate angle A by adding the two together. The result then is:
A = arccos(b/(2*r)) + arccos(c/(2*r)) -----------------------------(2)
If we then equate the right hand sides of (1) and (2) we get an equation involving a, b, c and r. These can be solved to find r in terms of a,b and c. I used the easy way using Maple and got the result shown below.
r = abs[sqr(-b^4+2*b^2*c^2+2*b^2*a^2-c^4+2*c^2*a^2-a^4)*a*c*b]/
[ b^4-2*b^2*c^2-2*b^2*a^2+c^4-2*c^2*a^2+a^4]
The diameter is, of course, 2*r.
This answer checks the cases for an equilateral triangle and for a 3,4,5 right triangle, each of which is easy to do by much simpler methods.
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