[ Properties of sections ] $[$ Regular tetrahedron $]$

Through the vertex $D$ of a regular tetrahedron $A B C D$ with edge $a$ and the midpoints of edges $A B$ and $A C$, a plane is drawn. Find the area of the resulting section.
Let $M$ and $N$ be the midpoints of edges $AB$ and $AC$ respectively. The resulting section is an isosceles triangle $DMN$ (the equality $DM = DN$ follows from the congruence of triangles $ADM$ and $ADN$). Let $O$ be the center of the equilateral triangle $ABC$ (then $DO$ is the height of the tetrahedron $ABCD$), and $K$ be the intersection point of segment $MN$ with the median $AL$ of triangle $ABC$. Then $K$ is the midpoint of segments $MN$ and $AL$, $DK$ is the height of triangle $DMN$, and since

\[
\begin{gathered}
AO = \frac{2}{3} AL = \frac{2}{3} \cdot \frac{1}{2} a \sqrt{3} = \frac{1}{3} a \sqrt{3} \\
OK = AO - AK = \frac{1}{3} a \sqrt{3} - \frac{1}{4} a \sqrt{3} = \frac{a \sqrt{3}}{12} \\
DO = \sqrt{AD^2 - AO^2} = \sqrt{a^2 - \frac{a^2}{3}} = a \sqrt{\frac{2}{3}}
\end{gathered}
\]

we find from the right triangle $DOK$ that

\[
DK = \sqrt{DO^2 + OK^2} = \sqrt{\frac{2a^2}{3} + \frac{a^2}{48}} = \frac{a \sqrt{11}}{4}
\]

Therefore,

\[
S_{\triangle DMN} = \frac{1}{2} MN \cdot KD = \frac{1}{2} a \cdot \frac{a \sqrt{11}}{4} = \frac{a^2 \sqrt{11}}{16}
\]

## Answer

$\frac{\mathbf{a}^{2} \sqrt{11}}{16}$