In what ratio does the point $P$, for which the lines $PB$, $PC$, and $PD$ are pairwise perpendicular to each other, divide the perpendicular segment from vertex $A$ of the regular tetrahedron $ABCD$ to the face $BCD$?
Solution. Let the length of the edges of the tetrahedron be 6 units. Then the length of the medians of the equilateral triangular faces is $\frac{6 \sqrt{3}}{2}=3 \sqrt{3}$ units. Since the tetrahedron is regular, the perpendicular segment from vertex $A$ to the face $BCD$ coincides with the centroid $M$ of the face $BCD$. The height of the tetrahedron can be determined, for example, from the triangle $AMB$ (see Figure 1). The centroid $M$ divides the median in a 2:1 ratio, so $BM=2 \sqrt{3}$, and by the Pythagorean theorem, $AM=\sqrt{AB^{2}-BM^{2}}=2 \sqrt{6}$ units.

![](https://cdn.mathpix.com/cropped/2024_05_02_cf4e23dcffdc033e0f9ag-1.jpg?height=357&width=330&top_left_y=434&top_left_x=884)

Figure 1

The point $M$ is equidistant from all three vertices of the equilateral triangle $BCD$, so the right triangles $PMB$, $PMC$, and $PMD$ are congruent (Figure 2). This means that the right triangle $BPC$ is isosceles. Since the hypotenuse $BC$ has a length of 6, the legs $PB=PC=\frac{6}{\sqrt{2}}=3 \sqrt{2}$ units.

![](https://cdn.mathpix.com/cropped/2024_05_02_cf4e23dcffdc033e0f9ag-1.jpg?height=412&width=336&top_left_y=1098&top_left_x=881)

Applying the Pythagorean theorem in the right triangle $PMB$, we get

$$
PM^{2}=PB^{2}-BM^{2}=18-12
$$

so $PM=\sqrt{6}=\frac{AM}{2}$.

Thus, the point $P$ bisects the segment $AM$.

Remark. A second solution can be constructed based on the cover image of the first solution.