4. In a regular truncated quadrilateral pyramid, inscribe a pyramid by taking the upper square as the base and the center of the lower square as the apex. The sides of the squares are: $a$ for the lower square, and $b$ for the upper square. What is the height of the pyramid if the lateral areas of the two pyramids are equal? (Determine under what conditions the problem has a solution.)
Solution. Let $M, N, M_{1}$ and $N_{1}$ be the midpoints of the edges $A B, C D, A_{1} B_{1}$ and $C_{1} D_{1}$ respectively (see diagram). Then the quadrilateral $M N M_{1} N_{1}$ is a trapezoid with bases $\overline{M N}=a$ and $\overline{M_{1} N_{1}}=a$. We are looking for the height $H=\overline{S S_{1}}$ of this trapezoid. Let $\overline{M M_{1}}=h_{1}$, $\overline{M_{1} S}=h_{2}$. For the lateral areas $P_{1}$ and $P_{2}$ of the given pyramid and the inscribed one respectively, we have:
![](https://cdn.mathpix.com/cropped/2024_06_05_13ae78702b3aed5f03b4g-3.jpg?height=376&width=1220&top_left_y=1588&top_left_x=260)

$$
P_{1}=\frac{4(a+b)}{2} h_{1}=2(a+b) h_{1}, \quad P_{2}=\frac{4 b h_{2}}{2}=2 b h_{2}
$$

From $P_{1}=P_{2}$ we get

$$
2(a+b) h_{1}=b h_{2}
$$

From the right triangle $M E M_{1}$ (see diagram) we have

$$
H^{2}=h_{1}^{2}-\frac{(a-b)^{2}}{4}
$$

and from the right triangle $S S_{1} M_{1}$ we have

$$
H^{2}=h_{2}^{2}-\frac{b^{2}}{4}
$$

Thus,

$$
h_{1}^{2}-\frac{(a-b)^{2}}{4}=h_{2}^{2}-\frac{b^{2}}{4}
$$

i.e.

$$
4 h_{2}^{2}=4 h_{1}^{2}-a^{2}+2 a b
$$

From (1) and (2) we get

$$
h_{1}^{2}=\frac{b^{2}(2 b-a)}{4(2 b+a)}
$$

so

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

The problem has a solution if and only if $2 b^{2}-a^{2}>0$, i.e., $a<b \sqrt{2}$.

## III year