12. Let $A B C D$ be a regular triangular pyramid in which the side edges $A D$, $B D$, and $C D$ are pairwise perpendicular. Find all points $X$ in the interior or on the boundary of $A B C D$ for which the volume of the tetrahedron with vertices the orthogonal projections of $X$ on faces of the pyramid $A B C D$ is maximal.
12. Let $X$ be an arbitrary point in $A B C D$. Denote by $M, N, P, Q$ the orthogonal projections of $X$ on the faces $B C D, C A D, A B D, A B C$ respectively, and let

$$
X M=x, \quad X N=y, \quad X P=z, \quad X Q=t
$$

Since $X M, X N$ and $X P$ are pairwise perpendicular, we have

$$
V_{M N P X}=\frac{1}{6} x y z
$$

## Figure 12

It can be easily verified that

$$
V_{P Q M X}=\frac{1}{6} x z t \sin \alpha
$$

where $\alpha$ is the angle between the line $X Q$ and its orthogonal projection on the plane $(X P M)$. Since the planes $(X P M)$ and $(A C D)$ are parallel, $\alpha$ is the angle between the altitude of the pyramid from the vertex $D$ and one of the side faces. We may assume that $A D=B D=C D=1$. Then $A B=B C=C A=\sqrt{2}$ and $\sin \alpha=\sqrt{3} / 3$. So

$$
V_{P Q M X}=\frac{1}{6} \frac{\sqrt{3}}{3} x z t, \quad V_{Q M N X}=\frac{1}{6} \frac{\sqrt{3}}{3} x y t
$$

etc. Therefore

$$
V(X)=V_{M N P Q}=\frac{1}{6}\left(x y z+\frac{\sqrt{3}}{3} t(x y+y z+z x)\right)
$$

Let $x^{\prime}, y^{\prime}, z^{\prime}$ be distances from $Q$ to the faces $B C D, C A D$ and $A B D$ of the pyramid. Then

$$
\begin{aligned}
x^{\prime} & =x+t \sin \alpha=x+\frac{t}{\sqrt{3}} \\
y^{\prime} & =y+\frac{t}{\sqrt{3}} \\
z^{\prime} & =z+\frac{t}{\sqrt{3}}
\end{aligned}
$$

Now we obtain

$$
\begin{aligned}
& V(Q)=\frac{1}{6}\left(x+\frac{t}{\sqrt{3}}\right)\left(y+\frac{t}{\sqrt{3}}\right)\left(z+\frac{t}{\sqrt{3}}\right) \\
& \geq \frac{1}{6}\left(x y z+\frac{\sqrt{3}}{3} t(x y+y z+z x)\right)=V(X)
\end{aligned}
$$

and equality occurs only for $t=0$, i.e. for $X=Q$. Hence we may consider that $X$ lies in the triangle $A B C$. In this case $t=0$ and

$$
\frac{1}{6}=V_{A B C D}=\frac{1}{6}(x+y+z)
$$

i.e.

$$
x+y+z=1
$$

Since $V(X)=\frac{1}{6} x y z$, we obtain by the arithmetic mean-geomteric mean inequality that $V(X)$ is maximal if $x=y=z=\frac{1}{3}$, i.e. when $X$ is the barycenter of the triangle $A B C$.