$\left.\begin{array}{c}\text { Sphere inscribed in a pyramid } \\ \text { Regular pyramid }\end{array}\right]$

In a regular quadrilateral pyramid with a height not less than $h$, a hemisphere of radius 1 is placed such that it touches all the lateral faces of the pyramid, and the center of the hemisphere lies on the base of the pyramid. Find the smallest possible value of the total surface area of such a pyramid.
Let $O$ be the center of the base $ABCD$ of a regular quadrilateral pyramid $PABCD$. The center of the hemisphere coincides with point $O$, and the hemisphere touches the lateral faces at points lying on the apothems. Let $M$ be the midpoint of side $BC$ of the square $ABCD$, and $K$ be the point of tangency of the hemisphere with the lateral face $PBC$. Then point $K$ lies on the segment $PM$, $OK \perp PM$, and $OK=1$. Denote the height $PO$ of the pyramid by $x$, the side of the base by $a$, and the angle between the lateral face and the base plane by $\alpha$. Then

$$
\angle POK = \angle PMO - \alpha, \quad OM = \frac{1}{2}a
$$

From the right triangle $POK$, we find that $\cos \alpha = \frac{OK}{OP} = \frac{1}{x}$. Then $\sin \alpha = \sqrt{1 - \frac{1}{x^2}} = \frac{\sqrt{x^2 - 1}}{x}$. From the right triangle $KOM$, we find that

$$
\frac{a}{2} = OM = \frac{OK}{\sin \angle PMO} = \frac{1}{\sin \alpha} = \frac{x}{\sqrt{x^2 - 1}}
$$

Thus, $a = \frac{2x}{\sqrt{x^2 - 1}}$. Let $S(x)$ be the total surface area of the pyramid. Then

$$
\begin{gathered}
S(x) = a^2 + \frac{x^2}{\sin \alpha} = \left(\frac{2x}{\sqrt{x^2 - 1}}\right)^2 + \frac{2x}{\sqrt{x^2 - 1}} \cdot x = \\
= \frac{4x^2}{x^2 - 1} + \frac{4x^3}{x^2 - 1} = \frac{4x^2(1 + x)}{x^2 - 1} = \frac{4x^2}{x - 1}.
\end{gathered}
$$

Thus, the problem reduces to finding the minimum value of the function $S(x) = \frac{4x^2}{x - 1}$ on the interval $[h; +\infty)$, where $h > 1$. Let's find the derivative of this function and its critical points within the given interval:

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-33.jpg?height=111&width=532&top_left_y=2077&top_left_x=763)

If $h \geq 2$, then there are no critical points on this interval, and the derivative is positive on this interval, meaning the function is increasing. Therefore, $S_{\min} = S(h) = \frac{4h^2}{h - 1}$. If $1 < h < 2$, then there is one critical point $x = 2$ on the interval $[h; +\infty)$. As we pass through this point, the derivative changes sign from "-" to "+", so the point $x = 2$ is a point of minimum. In this case, $S_{\min} = S(2) = 16$. This is the minimum value of $S(x)$ on the interval $[h; +\infty)$ when $h \geq 2$.

## Answer

If $h < 2$, then $S_{\min} = S(2) = 16$; if $h \geq 2$, then $S_{\min} = S(h) = \frac{4h^2}{h - 1}$.