[Height of a pyramid (tetrahedron).]

The base of the pyramid is a right triangle with a hypotenuse equal to \( c \) and an angle of \( 30^{\circ} \). The lateral edges of the pyramid are inclined to the base plane at an angle of \( 45^{\circ} \). Find the volume of the pyramid.
Prove that the height of the given pyramid passes through the center of the circle circumscribed around the base, i.e., through the midpoint of the hypotenuse of the base triangle.

## Solution

Let $D H$ be the height of the triangular pyramid $A B C D$, where $A B C$ is a right triangle with $\angle C=90^{\circ}$, $A B=c$, and $\angle A=30^{\circ}$. Since $D H$ is perpendicular to the plane $A B C$, the segments $A H$, $B H$, and $C H$ are the projections of the oblique lines $A D$, $B D$, and $C D$ onto the plane $A B C$. By the given condition,

$$
\angle D A H=\angle D B H=\angle D C H=45^{\circ}.
$$

The right triangles $D A H$, $D B H$, and $D C H$ are equal by the leg and acute angle, so $A H=B H=C H$ and $H$ is the center of the circle circumscribed around triangle $A B C$. Since this triangle is a right triangle, $H$ is the midpoint of the hypotenuse $A B$. We then find

$$
\begin{gathered}
D H=A H \cdot \operatorname{tg} \angle D A H=\frac{1}{2} c \cdot \operatorname{tg} 45^{\circ}=\frac{1}{2} c, B C=A B \cdot \sin 30^{\circ}=\frac{1}{2} c \\
A C=A B \cdot \cos 30^{\circ}=\frac{1}{2} c \sqrt{3}
\end{gathered}
$$

Therefore,

$$
\begin{gathered}
V(A B C D)=\frac{1}{3} S(A B C D) \cdot D H=\frac{1}{3} \frac{1}{2} B C \cdot A C \cdot D H= \\
=\frac{1}{3} \frac{1}{2} \frac{1}{2} c \cdot \frac{1}{2} c \sqrt{3} \cdot \frac{1}{2} c=c^{3} \sqrt{3} / 48
\end{gathered}
$$

## Answer

$c^{3} \sqrt{3} / 48$