12.465 The height of a regular triangular pyramid SABC is $\boldsymbol{H}$. Through the vertex $A$ of the base $ABC$, a plane is drawn non-perpendicular to the opposite lateral edge SC. This plane forms an angle $\alpha$ with the base plane. Find the volume of the part of the pyramid enclosed between the base plane and the section plane.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-322.jpg?height=339&width=332&top_left_y=1196&top_left_x=102)

Let SO be the height of the given regular pyramid $SABC$. Drop perpendiculars from vertices $A$ and $B$ to the edge $SC$: they will intersect at one point $D$ (since the lateral faces are equal) $\Rightarrow SC \perp BDA$ (Theorem 11). Since through a point only one plane can be drawn perpendicular to a given line, the plane $BDA$ is the one mentioned in the problem's condition. Let $K$ be the midpoint of $AB$. Then $AB \perp CK, AB \perp SK \Rightarrow AB \perp CSK$ (Theorem 11) $\Rightarrow$ $\Rightarrow AB \perp DK \Rightarrow \angle DKC = \alpha. CO = H \cdot \tan \angle CSO = H \cdot \tan \alpha$, $AB = \sqrt{3} \cdot CO = \sqrt{3} H \cdot \tan \alpha, CD = CK \cdot \sin \alpha = \frac{3}{2} CO \cdot \sin \alpha = \frac{3}{2} H \cdot \tan \alpha \cdot \sin \alpha$,
$DK = CK \cdot \cos \alpha = \frac{3}{2} H \cdot \tan \alpha \cdot \cos \alpha \Rightarrow$

$V_{ABCD} = \frac{1}{3} S_{ABD} \cdot CD = \frac{1}{3} \cdot \frac{AB \cdot DK}{2} \cdot CD = \frac{3 \sqrt{3}}{8} H^3 \tan^2 \alpha \cdot \sin^2 \alpha$

Answer: $\frac{3 \sqrt{3}}{8} H^3 \tan^2 \alpha \cdot \sin^2 \alpha$