4. The base of a pyramid is a rectangle with sides $a$ and $b$, and the lateral edges are equal to $c$. The pyramid is intersected by a plane parallel to the base, dividing it into two parts with equal volumes.

Find the distance from the plane to the vertex of the pyramid.
Solution. Let $A_{1} B_{1} C_{1} D_{1}$ be the rectangle obtained as the intersection of the plane with the pyramid (see diagram). Let $V$ be the volume of the pyramid $A B C D S$, and $V_{1}$ be the volume of the pyramid $A_{1} B_{1} C_{1} D_{1} S$. From the problem statement, we have

$$
V=2 V_{1}
$$

The two pyramids are similar, so we have $P: P_{1}=H^{2}: x^{2}$, i.e., $P_{1}=\frac{x^{2} P}{H^{2}}$. From (1) we get $P_{1}=\frac{P \cdot H}{2 x}$; therefore, $x=\frac{H}{2} \sqrt[3]{4}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_d61831f2f852c6e23b39g-2.jpg?height=577&width=572&top_left_y=726&top_left_x=913)
Let's find the height $H$ of the given pyramid.

From triangle $S O C$ we get

$$
H^{2}=c^{2}-\frac{a^{2}+b^{2}}{4}=\frac{1}{4}\left(4 c^{2}-a^{2}-b^{2}\right)
$$

i.e., $H=\frac{1}{2} \sqrt{4 c^{2}-a^{2}-b^{2}}$. Consequently,

$$
x=\frac{H \sqrt[4]{4}}{2}=\frac{\sqrt[3]{4}}{4} \sqrt{4 c^{2}-a^{2}-b^{2}}
$$

## 2nd year