AustriaMO2011

The tetrahedron has two equilateral faces whose sides are of length $2$, and two isosceles faces with two sides of length $2$ and one of length $1$. Let $F$ be the area of each equilateral face and $G$ the area of each isosceles face. It is obvious that $F>G$ holds. Further, let $a$ and $b$ be the distances of $P$ from the equilateral faces and $c$ and $d$ the distances from the isosceles faces.

If $V$ is the volume of the tetrahedron, we have $$\begin{array}{rl}3V& =F(a+b)+G(c+d)=F(a+b+c+d)-(F-G)(c+d)\\ & ⟺a+b+c+d=\frac{3V+(F-G)(c+d)}{F}.\end{array}$$ Since $F-G>0$, the value of $a+b+c+d$ is minimal for $c+d=0$, which is the case for $c=d=0$. The minimum value is therefore assumed for points $P$ on the common edge of the isosceles faces, i.e. on the edge with length $1$.

On the other hand, we also have $$\begin{array}{rl}3V& =F(a+b)+G(c+d)=G(a+b+c+d)+(F-G)(a+b)\\ & ⟺a+b+c+d=\frac{3V-(F-G)(a+b)}{G}.\end{array}$$ Since $F-G>0$, the value of $a+b+c+d$ is maximal for $a+b=0$, which is the case for $a=b=0$. The maximum value is therefore assumed for points $P$ on the common edge of the equilateral faces.