An analytical solution to a geometric problem.

Consider a triangle ABC. Now construct 3 isosceles triangles with the same top angle. Proof that the centre of mass of the triangle formed by the three newly created points is the centre of mass of the original triangle.

This purely geometric problem can be solved very easily if we do not think of points geometrically, but think of them as complex numbers. Assume without loss of generality that the centre of mass of $ABC$ is 0. This means $$A+B+C = 0$$ Call $A'$ the top of the triangle constructed on side $BC$, then is $$A' = B/2 + C/2 + i(B-C)/2 \tan(\alpha/2)$$ with $\alpha$ the top angle. Same works for the other sides: $$B' = C/2 + A/2 + i(C-A)/2 \tan(\alpha/2)$$ $$C' = A/2 + B/2 + i(A-B)/2 \tan(\alpha/2)$$ Summing these three numbers gives obviously zero, so the centres of mass coincide.