In other words, to make the center of mass of all 3 masses be at the intersection of the medians, we now only have to concentrate on making the center of mass of the lower 2 masses be at their midpoint. Using Archimedes’ Law of the Lever, we know this happens when the unknown mass is also m.
Thus, the center of mass of 3 equal masses placed at the vertices of a triangle is located at the intersection of the triangle’s medians.
Now, we can compute the ratio of the lengths that the red median is split into by the other medians. We go back to computing the center of mass piecemeal by first computing the center of mass of the 2 leftmost masses and imagining a single point mass of mass 2mplaced there. We’ve purposefully assigned the masses so that the center of mass of this 2m-mass and the m-mass in the lower right corner will be at the intersection of the medians, hence the ratio of the lengths that the red median is split into, by Archimedes’ Law of the Lever, must be 1 : 2. We deduce that the length we are looking for (indicated by the double-headed arrow) is 1/3 the length of the red median.
Now let’s turn our attention to the other length: