Problem 11.7. All vertices of a regular tetrahedron $ABCD$ are on the same side of the plane $\alpha$. It turns out that the projections of the vertices of the tetrahedron onto the plane $\alpha$ form the vertices of a certain square. Find the value of $AB^2$ if it is known that the distances from points $A$ and $B$ to the plane $\alpha$ are 17 and 21, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_06_20f31b6c2b9751326bf1g-6.jpg?height=484&width=438&top_left_y=91&top_left_x=500)
Answer: 32.

![](https://cdn.mathpix.com/cropped/2024_05_06_20f31b6c2b9751326bf1g-6.jpg?height=700&width=601&top_left_y=712&top_left_x=416)

Fig. 9: to the solution of problem 11.7

Solution. Let $A_{1}, B_{1}, C_{1}, D_{1}$ be the projections of points $A, B, C, D$ onto the plane $\alpha$. Let $M$ be the midpoint of segment $B D$, and $K$ be the midpoint of segment $A C$ (Fig. 9). The projections of points $M$ and $K$ onto the plane $\alpha$ are the midpoints of segments $B_{1} D_{1}$ and $A_{1} C_{1}$, respectively, and thus coincide. Therefore, the line $M K$ is perpendicular to the plane $\alpha$. It is also perpendicular to the line $A C$ (the segments $A M$ and $C M$ are equal as medians of equal triangles $A B D$ and $C B D$, so the median $M K$ of the isosceles triangle $A M C$ is also its altitude), which means $A C \| \alpha$. Therefore, $A A_{1} C_{1} C$ is a rectangle and $A_{1} C_{1}=A C$.

Let the side of the square $A_{1} B_{1} C_{1} D_{1}$ be $x$, then by the Pythagorean theorem, $A_{1} C_{1}^{2}=2 x^{2}$. Drop a perpendicular $A H$ to the line $B B_{1}$. Clearly, $A A_{1} B_{1} H$ is a rectangle, so $A A_{1}=H B_{1}=17$ and $A H=A_{1} B_{1}=x$. Also, $B H=B B_{1}-H B_{1}=21-17=4$. By the Pythagorean theorem for triangle $A B H$, we have $A B^{2}=16+x^{2}$. Therefore,

$$
16+x^{2}=A B^{2}=A C^{2}=A_{1} C_{1}^{2}=2 x^{2}
$$

from which we find $x=4$. Then $A B^{2}=2 \cdot 4^{2}=32$.