11.5. Given a regular triangular pyramid $S A B C$, the edge of the base of which is equal to 1. Medians of the lateral faces are drawn from vertices $A$ and $B$ of the base $A B C$, which do not have common points. It is known that the edges of a certain cube lie on the lines containing these medians. Find the length of the lateral edge of the pyramid.
Answer: $\frac{\sqrt{6}}{2}$.

Solution. The specified medians $A D$ and $B E$ of the lateral faces $A S B$ and $B S C$ lie on skew lines, and the skew edges of the cube are mutually perpendicular. Therefore, the condition for the existence of a cube with edges on the specified lines is equivalent to the perpendicularity of these lines. Thus, we need to find the length $b$ of the lateral edge of the pyramid, where the angle between the skew medians of the lateral faces is $90^{\circ}$. There are various methods to solve this.

The first method. Using the formula for calculating the median of a triangle, $A D = B E = \frac{\sqrt{2 \cdot 1^{2} + 2 b^{2} - b^{2}}}{2} = \frac{\sqrt{b^{2} + 2}}{2}$. By parallel translation by the vector $\overrightarrow{E D}$, the image of the median $B E$ is the segment $F D$ (see Fig. 11.5a). From triangle $A B F$ using the cosine rule, $A F = \sqrt{A B^{2} + B F^{2} - 2 A B \cdot B F \cdot \cos 120^{\circ}} = \frac{\sqrt{7}}{2}$. Triangle $A D F$ is a right-angled and isosceles triangle, so $A F = A P \sqrt{2}$. We obtain the equation: $\frac{\sqrt{b^{2} + 2}}{2} \cdot \sqrt{2} = \frac{\sqrt{7}}{2}$. The solution to this equation is $b = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}$.

Another similar method of calculation is possible if segments parallel to the given medians are drawn from vertex $S$, with lengths equal to twice the median.

![](https://cdn.mathpix.com/cropped/2024_05_06_c9ae9b97d8e763e17998g-2.jpg?height=460&width=394&top_left_y=1451&top_left_x=527)

Fig. 11.5a

![](https://cdn.mathpix.com/cropped/2024_05_06_c9ae9b97d8e763e17998g-2.jpg?height=406&width=397&top_left_y=1502&top_left_x=1249)

Fig. 11.5b

The second method. Let $\angle S B A = \angle S B C = \alpha$. Consider the vectors $\overrightarrow{B A}, \overrightarrow{B C}$, and $\overrightarrow{B S}$ and express $\overrightarrow{A D}$ and $\overrightarrow{B E}$ in terms of them: $\overrightarrow{A D} = \frac{1}{2} \overrightarrow{B S} - \overrightarrow{B A}, \overrightarrow{B E} = \frac{1}{2} \overrightarrow{B S} + \frac{1}{2} \overrightarrow{B C}$ (see Fig. 11.5b). Then, $\overrightarrow{A D} \cdot \overrightarrow{B E} = \left(\frac{1}{2} \overrightarrow{B S} - \overrightarrow{B A}\right) \cdot \left(\frac{1}{2} \overrightarrow{B S} + \frac{1}{2} \overrightarrow{B C}\right) = \frac{1}{4}(\overrightarrow{B S} - 2 \overrightarrow{B A}) \times (\overrightarrow{B S} + \overrightarrow{B C}) = \frac{1}{4}\left(\overrightarrow{B S}^{2} + \overrightarrow{B S} \cdot \overrightarrow{B C} - 2 \overrightarrow{B A} \cdot \overrightarrow{B S} - 2 \overrightarrow{B A} \cdot \overrightarrow{B C}\right)$.

Considering that $|\overrightarrow{B A}| = |\overrightarrow{B C}| = 1$ and $|\overrightarrow{B S}| = b$, we get:

$$
\overrightarrow{A D} \cdot \overrightarrow{B E} = \frac{1}{4}\left(b^{2} + b \cdot 1 \cdot \cos \alpha - 2 b \cdot 1 \cdot \cos \alpha - 2 \cdot 1 \cdot 1 \cdot \cos 60^{\circ}\right) = \frac{1}{4}\left(b^{2} - b \cos \alpha - 1\right)
$$

The non-zero vectors $\overrightarrow{A D}$ and $\overrightarrow{B E}$ will be perpendicular if and only if their dot product is zero. Thus, $\frac{1}{4}\left(b^{2} - b \cos \alpha - 1\right) = 0$.

From triangle $A B S: b \cos \alpha = \frac{1}{2} A B = \frac{1}{2}$, so $b^{2} = \frac{3}{2}$, which means $b = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}$.

The vector basis and auxiliary angle can also be introduced differently, for example, $\overline{S A}, \overline{S B}, \overline{S C}$ and the planar angle at vertex $S$.

A method of solution using an auxiliary volume is also possible.

## Grading Criteria:

+ The correct answer and a fully justified solution are provided

$\pm$ The reasoning is correct, but there is an arithmetic error in the final calculation

干 The problem is explained, why it reduces to the condition of perpendicularity of the medians, but the further solution is missing or contains errors

- Only the answer is provided