If two of the masses are at (0, 0) and (L, 0), by symmetry the x-coordinate of the third mass must be L/2, and then by the Pythagorean theorem its y-coordinate must be L(âš3)/2. Thus the position of the third mass is (L/2, L(âš3)/2).
Let each mass be m and C be the position of its center of mass.
Then C = (1/(3m)) (m(0, 0) + m(L, 0) + m(L/2, L(âš3)/2)).
= (1/3) (3L/2, L(âš3)/2).
= (L/2, L(âš3)/6).
That is, X_c = L/2 and Y_c = L(âš3)/6.
