12.450 In a triangular pyramid, all faces are equilateral triangles. A plane is drawn through the side of the base, dividing the volume of the pyramid in the ratio 1:3, counting from the base. Find the angle between this plane and the plane of the base.
## Solution.

Let the section $B C D$ of the given pyramid $F A B C$ divide its volume in the ratio 1:3

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-318.jpg?height=302&width=351&top_left_y=1281&top_left_x=84)
(counting from the base). Pyramids $FBCD$ and $ABCD$ have the same base $BCD \Rightarrow$ their volumes are proportional to the heights drawn from $F$ and $A$. And these heights, in turn, are proportional to $FD$ and $AD$, i.e., $\frac{DF}{AD}=3$.

Let $K$ be the midpoint of $BC$. Then $BC \perp AK, BC \perp FK$ (all faces are equilateral triangles) $\Rightarrow$ $\Rightarrow BC \perp AFK$ (Theorem 11) $\Rightarrow BC \perp DK$.

Let $\alpha = \angle FKA, \quad a$ be the edge of the pyramid. Then $AK = FK = \frac{a \sqrt{3}}{2}$, $AF = a$, and by the cosine theorem for $\triangle AFK$, $\cos \alpha = \frac{2 \cdot AK^2 - a^2}{2 \cdot AK^2} = \frac{1}{3}$.

Drop perpendiculars $DN, FM$ to the side $AK$ of $\triangle AFK$. Then $MK = FK \cdot \cos \alpha = \frac{a \sqrt{3}}{6}, AM = AK - MK = \frac{a \sqrt{3}}{3}$. And $AN : NM = AD : DF = 1 : 3 \Rightarrow$ $\Rightarrow NM = \frac{a \sqrt{3}}{4}, DN = \frac{FM}{4} = \frac{a \sqrt{6}}{12} \Rightarrow \operatorname{tg} \angle AKD = \frac{DN}{NK} = \frac{\sqrt{2}}{5}$.

Answer: $\operatorname{arctg} \frac{\sqrt{2}}{5}$.