$\left[\begin{array}{ll}\text { Coordinate Method in Space } \\ \text { Regular Tetrahedron }\end{array}\right]$

Find the angle between the intersecting medians of two faces of a regular tetrahedron.
## First Method.

Let's find the angle $\alpha$ between the medians $D M$ and $A K$ of the faces $A D B$ and $A B C$ of a regular tetrahedron $A B C D$ (Fig.1). For this, we draw a line through the center $Q$ of the face $A B C$ parallel to $D M$. Let this line intersect the edge $C D$ at point $S$. Then the desired angle is the angle between the intersecting lines $Q S$ and $A K$. Let the edge of the tetrahedron be $a$. From the similarity of triangles $Q S C$ and $M D C$, we find that

$$
Q S = M D \cdot \frac{C Q}{C M} = \frac{a \sqrt{3}}{2} \cdot \frac{2}{3} = \frac{a \sqrt{3}}{3}
$$

Moreover, $C S = \frac{2}{3} C D = \frac{2}{3} a$. From triangle $C K S$ using the cosine theorem, we find that

![](https://cdn.mathpix.com/cropped/2024_05_06_b3d98ec29b07c8345931g-25.jpg?height=100&width=1072&top_left_y=1036&top_left_x=493)

Thus,

![](https://cdn.mathpix.com/cropped/2024_05_06_b3d98ec29b07c8345931g-25.jpg?height=151&width=1063&top_left_y=1199&top_left_x=498)

Let $E$ be the midpoint of the edge $A D$ (Fig.2). We will find the angle $\beta$ between the lines $B E$ and $A K$. For this, we draw a line through point $Q$ parallel to $B E$. This line intersects the median line $E R$ of triangle $A D C$ at point $T$. Then the desired angle is the angle between the intersecting lines $Q T$ and $A K$. From the similarity of triangles $Q T R$ and $B E R$, we find that

$$
Q T = B E \cdot \frac{R Q}{R B} = \frac{a \sqrt{3}}{2} \cdot \frac{1}{3} = \frac{a \sqrt{3}}{6}
$$

Moreover, if point $J$ divides the edge $C D$ in the ratio $\frac{C J}{D J} = \frac{1}{2}$, then $A T = \frac{1}{2} A J$. From triangle $A C J$ using the cosine theorem, we find that

$$
A J^2 = A C^2 + C J^2 - 2 A C \cdot C J \cos 60^{\circ} = a^2 + \frac{1}{9} a^2 - 2 a \cdot \frac{1}{3} a \cdot \frac{1}{2} = \frac{7}{9} a^2
$$

Therefore, $A T^2 = \frac{1}{4} A J^2 = \frac{7}{36} a^2$. Then from triangle $A Q T$ using the cosine theorem, we find that

![](https://cdn.mathpix.com/cropped/2024_05_06_b3d98ec29b07c8345931g-25.jpg?height=154&width=1049&top_left_y=1952&top_left_x=510)

## Second Method.

We introduce a rectangular coordinate system with the origin at point $A$ (Fig.3). The $x$-axis is directed along the ray $A K$. Let $P$ be the projection of point $M$ onto the line passing through vertex $A$ parallel to $B C$. Then the $y$-axis is directed along the ray $A P$. Through point $A$, we draw a line parallel to the height $D Q$ of the tetrahedron. Let $L$ be the projection of point $D$ onto this line. Then the $z$-axis is directed along the ray $A L$. Then the points of interest have the following coordinates:

$$
A(0 ; 0 ; 0), K\left(\frac{a \sqrt{3}}{2} ; 0 ; 0\right), D\left(\frac{a \sqrt{3}}{3} ; 0 ; a \sqrt{\frac{2}{3}}\right), M\left(\frac{a \sqrt{3}}{4} ; \frac{a}{4} ; 0\right)
$$

$$
B\left(\frac{a \sqrt{3}}{2} ; \frac{a}{2} ; 0\right), E\left(\frac{a \sqrt{3}}{6} ; 0 ; \frac{1}{2} a \sqrt{\frac{2}{3}}\right)
$$

Thus,

$$
\overrightarrow{A K} = \left(\frac{a \sqrt{3}}{2} ; 0 ; 0\right), \overrightarrow{D M} = \left(\frac{a \sqrt{3}}{4} - \frac{a \sqrt{3}}{3} ; \frac{a}{4} ; -a \sqrt{\frac{2}{3}}\right) = \left(\frac{-a \sqrt{3}}{12} ; \frac{a}{4} ; -a \sqrt{\frac{2}{3}}\right)
$$

$$
\overrightarrow{B E} = \left(\frac{a \sqrt{3}}{6} - \frac{a \sqrt{3}}{2} ; -\frac{a}{2} ; \frac{1}{2} a \sqrt{\frac{2}{3}}\right) = \left(\frac{-a \sqrt{3}}{3} ; -\frac{a}{2} ; \frac{1}{2} a \sqrt{\frac{2}{3}}\right)
$$

Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_06_b3d98ec29b07c8345931g-26.jpg?height=140&width=801&top_left_y=270&top_left_x=637)

![](https://cdn.mathpix.com/cropped/2024_05_06_b3d98ec29b07c8345931g-26.jpg?height=137&width=784&top_left_y=486&top_left_x=637)

## Answer

 $\arccos \frac{1}{6}, \arccos \frac{2}{3}$.