Brazilian Mathematical Olympiad

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 $T'$ whose vertices are the centroids of the faces of $T$. Then $T'$ is similar to $T$, with ratio $1 : 3$ (the faces of $T'$ are parallel to the faces of $T$ and the ratio of the altitudes is $1 : 3$). The circumradius of $T'$, 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 $R'$ 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 $2R = \frac{a}{\sin \angle A} < \frac{l}{\sin 60^{\circ}} = 2R'$. 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 $O'$, which is the circumcenter of $ABC$, lies inside $ABC$, so $ABC$ is not obtusangle. Thus its circumradius $R'$ is less than the circumradius $R''$ 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 $OO'$ 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
$$3V = d_1S_1 + d_2S_2 + d_3S_3 + d_4S_4 = r(S_1 + S_2 + S_3 + S_4) \\ \implies r = \frac{S_1d_1 + S_2d_2 + S_3d_3 + S_4d_4}{S_1 + S_2 + S_3 + S_4},$$
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 $O'$ and $A$ is right-angled in $O'$. So $d_1^2 = R^2 - R'^2$ and then $R^2 - R'^2 \le r^2 \iff R'^2 \ge R^2 - r^2$. But, by looking at $U$, $a = \sqrt{3}R'' = \frac{2}{3}\sqrt{6}R \implies R''^2 = \frac{8}{9}R^2$. But $R'' > R'$, so $\frac{8}{9}R^2 > R^2 - r^2 \iff R < 3r$, a contradiction. So one of the sides of $ABC$ does not exceed $a$.