8,9

A right triangular pyramid has two spheres inscribed in it. The first sphere with radius $r$ touches the base of the pyramid and its lateral faces. The second sphere touches the first sphere externally and also the lateral faces of the pyramid. Find the sum of the volumes of the spheres if the volume of the pyramid is minimal.
Let the side of the base $ABC$ of the given regular triangular pyramid be $a$, and the height $PH$ of the pyramid be $h$. It is clear that $h > 2r$, and the first sphere touches the lateral face $APB$ at a point $K$ lying on the apothem $PM$ contained in the face $APB$. Consider the section of the pyramid by a plane passing through the height $PH$ and the apothem $PM$ lying in the lateral face $ABP$. Denote $\angle PMH = \alpha$. Then

$$
\begin{gathered}
MH = \frac{a}{2 \sqrt{3}}, \operatorname{tg} \alpha = \frac{PH \cdot MH}{MH} = \frac{\frac{h}{\Delta}}{2 \sqrt{3}} = \frac{2 h \sqrt{3}}{\Delta} \\
\cos \alpha = \frac{1}{\sqrt{1 + \operatorname{tg}^2 \alpha}} = \sqrt{\frac{1}{1 + \frac{12 h^2}{a^2}}} = \frac{a}{\sqrt{a^2 + 12 h^2}}
\end{gathered}
$$

Since the center of the sphere lies on the height of the pyramid and $\angle POK = \angle PMH = \alpha$, then $\frac{OK}{PO} = \cos \alpha$, or $\frac{\varphi}{h - \varphi} = \frac{a}{\sqrt{a^2 + 12 h^2}}$. From this equation, we find that $a^2 = \frac{12 \varphi^2 h}{h - 2r}$. Let $V(h)$ be the volume of the given pyramid with height $h$. Then

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-37.jpg?height=125&width=1003&top_left_y=1321&top_left_x=533)

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-37.jpg?height=95&width=2012&top_left_y=1479&top_left_x=20)
$+\infty)$. Let's find the derivative of this function and its critical points belonging to the given interval:

$$
V'(x) = r \sqrt{3} \cdot \frac{2h(h - 2r) - h^2}{(h - 2r)^2} = r \sqrt{3} \cdot \frac{h(h - 4r)}{(h - 2r)^2}
$$

The given interval contains one critical point $h = 4r$. As we pass through this point, the derivative changes sign from "-" to "+", meaning that the function decreases on the interval $(2r; 4r)$ and increases on the interval $(4r; +\infty)$. Therefore, $h = 4r$ is a point of minimum, i.e., when $h = 4r$, the pyramid has the smallest possible volume. Let $Q$ be the center of the second sphere, and $x$ be the radius of this sphere. The second sphere touches the lateral face $APB$ at some point $L$,

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-37.jpg?height=101&width=2021&top_left_y=1919&top_left_x=13)
$\frac{2\gamma - x}{3r}$, from which we find that $x = \frac{r}{2}$. Therefore, the sum of the volumes of the spheres is $\frac{4}{3} \pi r^3 + \frac{4}{3} \pi \left(\frac{r}{2}\right)^3 = \frac{3}{2} \pi r^3$.

## Answer

## [ Area and Volume (extremum problems) ] [ Difficulty: 4 [ inscribed sphere in a pyramid.  Classes: 8,9

In a regular triangular pyramid, two spheres are placed such that the first sphere touches the base of the pyramid and its lateral edges, and the second sphere touches the first sphere externally and the lateral faces of the pyramid. The radius of the first sphere is $R$. Find the radius of the second sphere if the volume of the pyramid under these conditions is

## [ Area and Volume (extremum problems) ] [ Problem 110991 Topics: Regular Pyramid  Difficulty: 4 [ Inscribed Sphere in a Pyramid ]  Classes: 8,9

In a regular triangular pyramid, two spheres are placed such that the first sphere touches the base of the pyramid and its lateral faces, and the second sphere touches the first sphere externally and the lateral edges of the pyramid. The radius of the first sphere is $r$. Find the radius of the second sphere if the volume of the pyramid under these conditions is

minimal

## Answer

$(\sqrt{3} - 1) r$.

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