Solving for a Length given the Medians

Let $a=BC$ , $b=CA$ , and $c=AB$ . By Apollonius Theorem, we find that $m_c=CD=\sqrt{3}=\sqrt{\frac{2(a^2+b^2)-c^2}{4}}$ . Analogously we find $m_a=AE=4=\sqrt{\frac{2(b^2+c^2)-a^2}{4}}$ and $m_b=BF=\sqrt{10}=\sqrt{\frac{2(a^2+c^2)-b^2}{4}}$ . By using these three results we can get $c=\frac{2}{3}\sqrt{2(m_a^2+m_b^2)-m_c^2}$ . By substituting lengths of the medians as given in the problem, we get $c=\frac{14}{3}$ .

Note that $EF$ is a midsegment of $\triangle ABC$ hence by similar triangles, $c=2(EF)$ which means that $EF=\frac{1}{2}c=\frac{7}{3}$ . Therefore $j=7$ , $k=3$ which gives us an answer of $7+3=10$ .