## Task B-3.7.

Given is a triangular pyramid $S A B C$ where the side $S B C$ is perpendicular to the base $A B C$, and $\varangle A S B=\varangle B S C=\varangle C S A=60^{\circ}$ and $|S B|=|S C|=1$. Determine the volume of the pyramid $S A B C$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_5ec97620b8cc23e4b297g-25.jpg?height=560&width=600&top_left_y=311&top_left_x=634)

1 point

From $|S B|=|S C|=1$ and $\angle B S C=60^{\circ}$, it follows that triangle $B S C$ is an equilateral triangle with side length 1.

The lateral face $B S C$ is perpendicular to the plane $A B C$, so the height of triangle $B S C$ is the height of the pyramid, and we have

$$
|S N|=\frac{a \sqrt{3}}{2}=\frac{\sqrt{3}}{2}
$$

Triangles $A B S$ and $A C S$ are congruent, so $|A B|=|A C|$, which means triangle $A B C$ is isosceles, and its median $A N$ is also the height. Therefore, triangle $A N C$ is a right triangle. Then,

$$
|A C|^{2}=|A N|^{2}+|N C|^{2}=|A N|^{2}+\frac{1}{4}
$$

Applying the cosine rule in triangle $A S C$, we get

$$
|A C|^{2}=|A S|^{2}+|S C|^{2}-2|A S| \cdot|S C| \cdot \cos 60^{\circ}=|A S|^{2}+1-|A S|. \quad 1 \text { point }
$$

Thus,

$$
|A N|^{2}+\frac{1}{4}=|A S|^{2}+1-|A S|
$$

Applying the Pythagorean theorem in triangle $A N S$, we get

$$
|A N|^{2}=|A S|^{2}-|S N|^{2}=|A S|^{2}-\frac{3}{4}
$$

so,

$$
|A S|^{2}-\frac{3}{4}+\frac{1}{4}=|A S|^{2}+1-|A S|
$$

which gives $|A S|=\frac{3}{2}$. Then,

$$
|A N|^{2}=\frac{9}{4}-\frac{3}{4}=\frac{6}{4}
$$

so, $|A N|=\frac{\sqrt{6}}{2}$. Finally, the desired volume is

$$
V=\frac{1}{3} \frac{|A N| \cdot|B C|}{2} \cdot|S N|=\frac{\sqrt{2}}{8}
$$

## COUNTY COMPETITION IN MATHEMATICS

## 4th grade - high school - B variant

1. March 2023.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE TASK, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.