Problem 1354

3. We shall use the following lemma
Lemma. If an altitude of a tetrahedron passes through the orthocenter of the opposite side, then each of the other altitudes possesses the same property.
Proof. Denote the tetrahedron by $S A B C$ and let $a=B C, b=C A$, $c=A B, m=S A, n=S B, p=S C$. It is enough to prove that an altitude passes through the orthocenter of the opposite side if and only if $a^{2}+m^{2}=b^{2}+n^{2}=c^{2}+p^{2}$.
Suppose that the foot $S^{\prime}$ of the altitude from $S$ is the orthocenter of $A B C$. Then $S S^{\prime} \perp A B C \Rightarrow S B^{2}-S C^{2}=S^{\prime} B^{2}-S^{\prime} C^{2}$. But from $A S^{\prime} \perp B C$ it follows that $A B^{2}-A C^{2}=S^{\prime} B^{2}-S^{\prime} C^{2}$. From these two equalities it can be concluded that $n^{2}-p^{2}=c^{2}-b^{2}$, or equivalently, $n^{2}+b^{2}=c^{2}+p^{2}$. Analogously, $a^{2}+m^{2}=n^{2}+b^{2}$, so we have proved the first part of the equivalence.
Now suppose that $a^{2}+m^{2}=b^{2}+n^{2}=c^{2}+p^{2}$. Defining $S^{\prime}$ as before, we get $n^{2}-p^{2}=S^{\prime} B^{2}-S^{\prime} C^{2}$. From the condition $n^{2}-p^{2}=c^{2}-\ell^{2}$ $\left(\Leftrightarrow b^{2}+n^{2}=c^{2}+p^{2}\right)$ we conclude that $A S^{\prime} \perp B C$. In the same way $C S^{\prime} \perp A B$, which proves that $S^{\prime}$ is the orthocenter of $\triangle A B C$. The lemma is thus proven.
Now using the lemma it is easy to see that if one of the angles at $S$ is right, than so are the others. Indeed, suppose that $\angle A S B=\pi / 2$. From the lemma we have that the altitude from $C$ passes through the orthocenter of $\triangle A S B$, which is $S$, so $C S \perp A S B$ and $\angle C S A=\angle C S B=\pi / 2$.
Therefore $m^{2}+n^{2}=c^{2}, n^{2}+p^{2}=a^{2}$, and $p^{2}+m^{2}=b^{2}$, so it follows that $m^{2}+n^{2}+\mu^{2}=\left(a^{2}+b^{2}+c^{2}\right) / 2$. By the inequality between the arithmetic and quadric means, we have that $\left(a^{2}+b^{2}+c^{2}\right) / 2 \geq 2 s^{2} / 3$, where s denotes the semiperimeter of $\triangle A B C$. It remains to be shown that $2 s^{2} / 3 \geq 18 r^{2}$. Since $S_{\triangle A B C}=s r$, this is equivalent to $2 s^{4} / 3 \geq$ $18 S_{A B C}^{2}=18 s(s-a)(s-b)(s-c)$ by Heron's formula. This reduces to $s^{3} \geq 27(s-a)(s-b)(s-c)$, which is an obvious consequence of the AM-GM mean inequality.
Remark. In the place of the lemma one could prove that the opposite edges of the tetrahedron are mutually perpendicular and proceed in the same way.