Does three medians determine a triangle? If given three medians, is there only one triangle that has these three medians? How do I prove that?
 A: Apologies about the lack of a diagram. It is important that you supply it.
Let $ABC$ be a triangle, and let $M_a$ be the midpoint of side $BC$. Let $\theta=\angle AM_aB$. Then $\angle AM_aC=180^\circ -\theta$. Let $m_a$ be the length of the median $AM_a$.
By the Cosine Law applied to $\triangle AM_aB$ we have
$$c^2=(a/2)^2+m_a^2-2(a/2)(m_a)\cos\theta.$$
Similarly, we have
$$b^2=(a/2)^2+m_a^2-2(a/2)(m_a)\cos(180^\circ -\theta).$$
Adding, we get, since $\cos(180^\circ-\theta)=-\cos\theta$, that
$$b^2+c^2=a^2/2+2m_a^2,$$
so 
$$2b^2+2c^2-a^2=4m_a^2.$$
We get similar equations for the other medians. Now given $m_a$, $m_b$, and $m_c$ we can solve the system of linear equations for $a^2$, $b^2$, and $c^2$.
Remark: Note that given three lengths, these lengths are not necessarily the lengths of the medians of a triangle. In terms of our equations, there may not be positive solutions for $a^2,b^2,c^2$. This is geometrically obvious, since it is clear that there is no triangle with medians $1,1,1000$. In fact the medians of any triangle themselves form a triangle, so satisfy the triangle inequality.
