Another Sadistic Math Challenge?

Since nobody got it, here's the solution:

Make a triangle with legs x1, x2, and x3, and masses m1, m2, m3 on the vertices. Draw the center of mass inside the triangle, then make three radial lines (r1, r2, r3) from the center of mass to each of the three masses. The angle between r1 and r2 is c3, while the angle between r2 and r3 is c1, and the angle between r3 and r1 is c2.

A balance of moments (torques) in the r1, r2, and r3 directions tells us the following:

m1*r1 + m2*r2*Cos[c3] + m3*r3*Cos[c2]

m2*r2 + m1*r1*Cos[c3] + m3*r3*Cos[c1]

m3*r3 + m1*r1*Cos[c2] + m2*r2*Cos[c3]

The law of cosines applied to the three triangles within the main triangle tells us the following:

Cos[c1] = (r2^2 + r3^2 - x1^2)/(2*r2*r3)

Cos[c2] = (r1^2 + r3^2 - x2^2)/(2*r1*r3)

Cos[c3] = (r1^2 + r2^2 - x3^2)/(2*r1*r2)

Substituting these for the cosines in the first three equations, we get:

(2*m1*r1^2 + m3*(r1^2 + r3^2 - x2^2) + m2*(r1^2 + r2^2 - x3^2))/r1 = 0

(2*m2*r2^2 + m3*(r2^2 + r3^2 - x1^2) + m1*(r1^2 + r2^2 - x3^2))/r2 = 0

(2*m3*r3^2 + m2*(r2^2 + r3^2 - x1^2) + m1*(r1^2 + r3^2 - x2^2))/r3 = 0

Solving these three equations simultaneously yields (after much algebra) the following:

r1 = 1/mtotal * sqrt(m2^2*x1^2 + m1^2*x2^2 + m1*m2*(x1^2 + x2^2 - x3^2))

r2 = 1/mtotal * sqrt(m3^2*x2^2 + m2^2*x3^2 + m2*m3*(x2^2 + x3^2 - x1^2))

r3 = 1/mtotal * sqrt(m1^2*x3^2 + m3^2*x1^2 + m3*m1*(x3^2 + x1^2 - x2^2))

where mtotal = m1 + m2 + m3.

And now you know!