In the regular tetrahedron ABCD , AM_1 is the median of the…

In the regular tetrahedron $ABCD$, $AM_1$ is the median of the face $ABC$ and $AM_2$ is the median of the face $ACD$ ($M_1$ is the midpoint of $BC$, $M_2$ the midpoint of $CD$). Find the angle between $AM_1$ and $AM_2$.

Explanation of Method

The angle between two face medians of a regular tetrahedron.

Step 1: Basis vectors

Let $\vec b=\overrightarrow{AB}$, $\vec c=\overrightarrow{AC}$, $\vec d=\overrightarrow{AD}$. All edges equal $a$ and all faces are equilateral, so

\[ \vec b^{\,2}=\vec c^{\,2}=\vec d^{\,2}=a^2,\qquad \vec b\cdot\vec c=\vec b\cdot\vec d=\vec c\cdot\vec d=\frac{a^2}{2}. \]

Step 2: The medians

\[ \overrightarrow{AM_1}=\frac{\vec b+\vec c}{2},\qquad \overrightarrow{AM_2}=\frac{\vec c+\vec d}{2}. \]

\[ |\overrightarrow{AM_1}|^2=\frac{a^2+a^2+a^2}{4}=\frac{3a^2}{4}=|\overrightarrow{AM_2}|^2. \]

Step 3: Scalar product and angle

\[ \overrightarrow{AM_1}\cdot\overrightarrow{AM_2}=\frac14\left(\vec b\cdot\vec c+\vec b\cdot\vec d+\vec c^{\,2}+\vec c\cdot\vec d\right)=\frac14\cdot\frac{5a^2}{2}=\frac{5a^2}{8}, \]

\[ \cos\varphi=\frac{5a^2/8}{3a^2/4}=\frac56. \]

\[\varphi=\arccos\frac56\approx33.6^\circ\]

$\varphi$ $A$ $B$ $C$ $D$ $M_1$ $M_2$

$\overrightarrow{AM_1}\ \text{(median of }\triangle ABC)$
$\overrightarrow{AM_2}\ \text{(median of }\triangle ACD)$

Regular tetrahedron; $M_1$ is the midpoint of $BC$, $M_2$ the midpoint of $CD$. Dashed edges are hidden behind the front faces.

Result

\[\boxed{\,\varphi=\arccos\frac56\approx33.6^\circ\,}\]