A sphere of radius 10 is inscribed in the frame of the tetrahedron A sphere of radius $10$ is inscribed in the frame of the tetrahedron(i.e. touches all the edges). The sum of the lengths of edges of a tetrahedron is $180$. Prove that the volume of the tetrahedron does not exceed $3000$.
I would be glad to receive ideas/hints. My main concern here is that we cannot assume the tetrahedron is regular: it would have been a trivial problem otherwise.
 A: If a triangle has a fixed perimeter and a fixed inradius, it has a fixed area.

In our case, let us consider the six (red) points on the edges in which the given sphere meets the tetrahedron frame. These points have the same distance from the center of the sphere and they split the tetrahedron frame in such a way that there are $3$ segments with equal lengths meeting at every vertex of the tetrahedron. Since the given sphere has radius $10$, the distance between every edge of the tetrahedron and its opposide edge is $\leq 20$. Additionally, the sum of the lengths of an edge and the opposite edge is constant and it equals $\color{red}{\ell_1}+\color{orange}{\ell_2}+\color{purple}{\ell_3}+\color{green}{\ell_4}=60$. In particular, the product of the lengths of two opposite edges is at most $900$ by the AM-GM inequality. 
Since (see here)

Given two opposite edges $AB$ and $CD$ and the shortest (perpendicular) distance $EF(=d)$, between them, we can calculate the volume of the tetrahedron as follows: let $\varphi$ be the dihedral angle between the planes $ABEF$ and $CDEF$. Then the volume of tetrahedron $ABCD$ equals $V= \frac{1}{6}(AB) (CD) (EF)\sin\varphi$.

It follows that the volume of our tetrahedron is 
$$ V \color{red}{\leq} \frac{1}{6}\cdot 900\cdot 20 = \color{red}{3000}.$$
Equality is achieved only in the regular case.
