XI OBM

There are two main steps in the proof: first, we'll prove a tridimensional analogous to the famous $R\ge 2r$ inequality; then, we'll prove the bidimensional analogous to the problem. Combining these two facts and other arguments will lead us to a solution.

The first step is proving that if $R$ and $r$ are the circumradius and the inradius of the circumscribed and inscribed spheres of a tetrahedron $T$ then $R\ge 3r$.

To do this, consider the tetrahedron $T^{\prime }$ whose vertices are the centroids of the faces of $T$. Then $T^{\prime }$ is similar to $T$, with ratio $1:3$ (the faces of $T^{\prime }$ are parallel to the faces of $T$ and the ratio of the altitudes is $1:3$). The circumradius of $T^{\prime }$, which is $\frac{R}{3}$, must be bigger than $r$, because the smallest sphere through four points, one in each face of $T$, is the inscribed sphere. So $R\ge 3r$.

Now we prove that if $ABC$ is not obtusangle and has all sides smaller than the side $l$ of an equilateral triangle then the circumradius $R$ of $ABC$ is less than the circumradius $R^{\prime }$ of the equilateral triangle. Indeed, by the sine law, if $a$ is the greater side of $ABC$ then $60^{\circ }\le \angle A\le 90^{\circ }$ and $2R=\frac{a}{\sin \angle A}<\frac{l}{\sin 60^{\circ }}=2R^{\prime }$. Let $T$ be a tetrahedron and $U$ be the regular tetrahedron with edge length $a$ and same circumradius $R$ as $T$. Let $O$ be the circumcenter of both $T$ and $U$ and suppose $ABC$ is the face nearer to $O$. Since $O$ is inside $T$, by problem 2 of 1987 its projection $O^{\prime }$, which is the circumcenter of $ABC$, lies inside $ABC$, so $ABC$ is not obtusangle. Thus its circumradius $R^{\prime }$ is less than the circumradius $R^{\prime \prime }$ of the equilateral triangle with side $a$.

If $d_1\le d_2\le d_3\le d_4$ are the distances from $O$ to each face ($d_1$ being the distance $O{O}^{\prime }$ from $O$ to $ABC$) and $S_1,S_2,S_3,S_4$ the corresponding areas of the faces, then the volume $V$ of $T$ is such that $$\begin{gathered} 3V=d_1{S}_1+d_2{S}_2+d_3{S}_3+d_4{S}_4=r(S_1+S_2+S_3+S_4) \\ ⟹r=\frac{S_1{d}_1+S_2{d}_2+S_3{d}_3+S_4{d}_4}{S_1+S_2+S_3+S_4}, \end{gathered}$$ that is, $r$ is the weighted mean of $d_1,d_2,d_3,d_4$ (the weights are the $S_i$'s). This means that the smallest value $d_1$ is not bigger than $r$, that is, $d_1\le r$.

The triangle formed by $O$, its projection $O^{\prime }$ and $A$ is right-angled in $O^{\prime }$. So $d_1^2=R^2-R^{\prime 2}$ and then $R^2-R^{\prime 2}\le r^2⟺R^{\prime 2}\ge R^2-r^2$. But, by looking at $U$, $a=\sqrt{3}{R}^{\prime \prime }=\frac{2}{3}\sqrt{6}R⟹R^{\prime \prime 2}=\frac{8}{9}{R}^2$. But $R^{\prime \prime }>R^{\prime }$, so $\frac{8}{9}{R}^2>R^2-r^2⟺R<3r$, a contradiction. So one of the sides of $ABC$ does not exceed $a$.