311. On the plane, there is a square $A B C D$. On the sides $B C$ and $C D$, points $P$ and $Q$ are taken such that $|C P|+|C Q|=$ $=|A B|$. Let $M$ be a point in space such that in the tetrahedron $A P Q M$, all faces are equal triangles. Determine the geometric locus of the projections of points $M$ onto the plane perpendicular to the plane of the square and passing through the diagonal $A C$.

The orthocenter of the tetrahedron.
311. Consider a cube $A E F G A_{1} E_{1} F_{1} G_{1}$ with an edge equal to the side of the square $A B C D$. Take points $P$ and $Q$ on the edges $A_{1} E_{1}$ and $A_{1} G_{1}$ such that $\left|A_{1} P\right|=|B P|=|C Q|$, $\left|A_{1} Q\right|=|Q D|=|P C|$ (Fig. $\left.62, a\right)$. Consider the rectangle $A_{1} P M_{1} Q$. The point $M_{1}$, due to the condition $\quad\left|A_{1} P\right|+\left|A_{1} Q\right|=\left|A_{1} E_{1}\right|$, lies on the diagonal $E_{1} G_{1}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-150.jpg?height=428&width=415&top_left_y=333&top_left_x=671)

Fig. 61.

Therefore, if $M$ is the projection of $M_{1}$ onto $E G$, then the tetrahedron $A P Q M$ has all faces equal to $\triangle A P Q$. The square $A B C D$, in which the triangle $A P Q$ lies, is obtained from the square $A E E_{1} A_{1}$ by rotating around the diagonal $A F_{1}$ by some angle $\alpha$ (Fig. 62, b). Since the plane $E G A_{1}$ is perpendicular to the diagonal $A F_{1}$, $B D$ lies in this plane. But the planes $A E E_{1} A_{1}$, $A B C D$, and the lines $E G$, $E A_{1}$, $A_{1} G$, and $B D$ are tangent to the sphere inscribed in the cube. Therefore, the angle between the plane $A B C D$ and $A_{1} E G$ is constant, it is equal to the angle $\varphi$ between the planes $\boldsymbol{A} E E_{1} A_{1}$ and $A_{1} E G$, for which

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-150.jpg?height=392&width=423&top_left_y=1178&top_left_x=148)
a)

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-150.jpg?height=426&width=508&top_left_y=1164&top_left_x=578)

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-150.jpg?height=161&width=252&top_left_y=1565&top_left_x=360)

Fig. 62 b
$\cos \varphi=\frac{1}{\sqrt{3}}$. But the planes $A_{1} E G$ and $A B C D$ intersect along the diagonal $B D$. Therefore, the point $M$ lies in the plane passing through $B D$ and forming an angle $\varphi$ with the plane $A B C D$, and the geometric locus of the projections of the points $M$ will be two segments emanating from the midpoint of $A C$ at an angle $\varphi$ to $A C$ such that $\cos \varphi=\frac{1}{\sqrt{3}}$, and having a length of $a \frac{\sqrt{2}}{2}$ (Fig. $62, c$).