## Task B-3.3.

In the triangular pyramid $A B C D$, the side face $B D C$ is perpendicular to the base $A B C, |B D|=|D C|=1$, and all lateral edges meeting at vertex $D$ form an angle of $60^{\circ}$. Determine the volume of the triangular pyramid.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_1020492bf6e1d2c949fdg-11.jpg?height=597&width=642&top_left_y=2003&top_left_x=707)

According to the problem, $\varangle B D C = \varangle C D A = \varangle A D B = 60^{\circ}$ and $|B D| = |D C| = 1$, so triangle $B C D$ is equilateral, and the height of the triangle $|D K| = \frac{\sqrt{3}}{2}$.

The line $D K$ lies in the plane $B C D$, which is perpendicular to the plane $A B C$, so it is perpendicular to any line lying in the plane $A B C$, i.e., $D K \perp A K$, or $\varangle D K A = 90^{\circ}$.

Notice that $\triangle C K A \cong \triangle B K A (|C K| = |B K|, \varangle C K A = \varangle A K B = 90^{\circ})$, so $|A C| = |A B|$.

Let $|K A| = x$, $|D A| = m$, and $|A B| = n$. From the right triangle $D K A$ we get:

$$
m^{2} = x^{2} + \frac{3}{4} \Longrightarrow x^{2} = m^{2} - \frac{3}{4}
$$

From the right triangle $C K A$ we get:

$$
n^{2} = x^{2} + \frac{1}{4} \Longrightarrow x^{2} = n^{2} - \frac{1}{4}
$$

In triangle $A B D$, apply the cosine rule:

$$
n^{2} = m^{2} + 1 - 2 m \cos 60^{\circ} \Longrightarrow n^{2} = m^{2} - m + 1
$$

From (1) and (2) we get

$$
n^{2} = m^{2} - \frac{1}{2}
$$

From (3) and (4) we get

$$
m^{2} - \frac{1}{2} = m^{2} - m + 1 \Longrightarrow m = \frac{3}{2}
$$

Substituting into (1) we get

$$
x^{2} = \frac{9}{4} - \frac{3}{4} = \frac{6}{4} \Longrightarrow x = \frac{\sqrt{6}}{2}
$$

Finally, the volume of the given triangular pyramid is

$$
V = \frac{1}{3} B \cdot v = \frac{1}{3} \cdot \frac{1 \cdot \frac{\sqrt{6}}{2}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{3 \sqrt{2}}{24} = \frac{\sqrt{2}}{8}
$$