An investigation of a triangle in which the distances of the angles from the center of gravity of it may be expressed rationally
"This memoir was made to please a small number of amateurs of Indeterminate Analysis, and contains a very beautiful solution to the problem announced in the title. Here it is in a few words: Let the sides of the desired triangle be 2a, 2b, 2c, and the lengths drawn from their midpoints to the opposite angles be f, g and h, respectively. Take two numbers q and r, and find M = (5qq–rr)/(4qq) and N = (5rr–9qq)/(4rr). Reduce the fraction ((M–N)2–4)/(4(M+N)) to its lowest terms and name the numerator x and the denominator y, and you will have the side 2a = 2qx + (M–N)qy, and the line f = rx–(1/2)(M–N)ry. Make p=x+y and x=x–y, and you will have the sides 2b = pr–qs and 2c = pr+qs, and the lengths g = (3pq+rs)/2 and h = (3pq–rs)/2."
Original Source Citation
Nova Acta Academiae Scientiarum Imperialis Petropolitanae, Volume 12, pp. 101-113.
Opera Omnia Citation
Series 1, Volume 4, pp.290-302.