[ Height of a pyramid (tetrahedron).]

The base of the pyramid is a rectangle with a diagonal equal to $b$, and an angle of $60^{\circ}$ between the diagonals. Each of the lateral edges forms an angle of $45^{\circ}$ with the plane of the base. Find the volume of the pyramid.
Prove that the height of the given pyramid passes through the point of intersection of the diagonals of the rectangular base.

## Solution

Let the rectangle $A B C D$ be the base of the pyramid $P A B C D$, and the angle between the diagonals $A C$ and $B D$ is $60^{\circ}$, with $A C = B D = b$. If $P H$ is the height of this pyramid, then from the equality of the right triangles $A P H, B P H, C P H$, and $D P H$ by the leg and the other angle, it follows that point $H$ is equidistant from all vertices of the rectangle $A B C D$, so $H$ is the center of the circle circumscribed around the rectangle $A B C D$, i.e., the point of intersection of its diagonals.

From the right triangle $A P H$, we find that

$$
P H = A H \cdot \operatorname{tg} \angle D A H = \frac{1}{2} b \cdot \operatorname{tg} 45^{\circ} = \frac{1}{2} b
$$

Therefore,

$$
\begin{aligned}
V(P A B C D) & = \frac{1}{3} S(A B C D) \cdot P H = \frac{1}{3} \frac{1}{2} A C \cdot B D \cdot \sin 60^{\circ} \cdot P H = \\
& = \frac{1}{3} \frac{1}{2} b \cdot b \cdot \frac{1}{2} \sqrt{3} \cdot \frac{1}{2} b = b^{3} \sqrt{3} / 24
\end{aligned}
$$

## Answer

$$
b^{3} \sqrt{3} / 24 .
$$