Do the medians (or other cevians) form all the triangles? I want to know whether set of medians of all triangles, or some other class of cevians, can form the set of all the triangles? 
For example, in the case of altitudes, $(4,7,10)$ is an counterexample. While $(4,7,10)$ can form a triangle, there is no triangle with altitudes $(4,7,10)$. 
 A: From the Wikipedia article on Median, the lengths of the medians in terms of the sidelengths are: 
$m_a = \dfrac{1}{2}\sqrt{2b^2+2c^2-a^2}$
$m_b = \dfrac{1}{2}\sqrt{2c^2+2a^2-b^2}$
$m_c = \dfrac{1}{2}\sqrt{2a^2+2b^2-c^2}$
Solving for $(m_a,m_b,m_c)$ in terms of $(a,b,c)$ yields: 
$a = \dfrac{2}{3}\sqrt{2m_b^2+2m_c^2-m_a^2}$
$b = \dfrac{2}{3}\sqrt{2m_c^2+2m_a^2-m_b^2}$
$c = \dfrac{2}{3}\sqrt{2m_a^2+2m_b^2-m_c^2}$
So for any desired median lengths $(m_a,m_b,m_c)$, we can find sidelengths $(a,b,c)$ such that the triangle with side lengths $(a,b,c)$ has medians of length $(m_a,m_b,m_c)$. 
The area of a triangle is given by: 
$T = \sqrt{s(s-a)(s-b)(s-c)} = \dfrac{4}{3} \sqrt{\sigma (\sigma - m_a)(\sigma - m_b)(\sigma - m_c)}$, 
where $s = \dfrac{a+b+c}{2}$ and $\sigma = \dfrac{m_a+m_b+m_c}{2}$. 
From this, we can see that $(a,b,c)$ satisfies the triangle inequality iff $(m_a,m_b,m_c)$ satisfies the triangle inequality. (If either triple did not satisfy the inequality, the corresponding quantity under the radical would be negative).
