[ $\quad$ Tangent Spheres $\quad]$

Inside the rectangular parallelepiped $A B C D A 1 B 1 C 1 D 1$, there are two spheres $\sigma 1$ and $\sigma 2$, touching each other externally; moreover, sphere $\sigma 1$ touches the faces $A B C D, A B B 1 A 1, A D D 1 A 1$, and sphere $\sigma 2$ touches the faces $A 1 B 1 C 1 D 1, B C C 1 B 1, C D D 1 C 1$. It is known that $A B=6-\sqrt{2}, A 1 D 1=6+\sqrt{2}, C C 1=6$. Find the distance between the centers of the spheres $\sigma 1$ and $\sigma 2$. Find the smallest and largest total volume of the spheres.
Let's introduce a rectangular coordinate system, directing the axis $Ox$ along the ray $AB$, the axis $Oy$ - along the ray $AD$, and the axis $Oz$ - along the ray $AA1$. Let $O1$ be the center of the sphere $\sigma1$ with radius $r1$, and $O2$ be the center of the sphere $\sigma2$ with radius $r2$. We will assume that $r1 \leqslant r2$. Since the spheres are inscribed in trihedral angles with vertices $A$ and $C1$ of the parallelepiped, their centers have coordinates $O1(r1; r1; r1)$ and $O2(AB - r2; BC - r2; AA1 - r2)$. The line of centers of the touching spheres passes through their point of tangency $P$, so $O1O2 = O1P + O2P = r1 + r2$, or

$$
\sqrt{\left(AB - r2 - r1\right)^2 + \left(BC - r2 - r1\right)^2 + \left(AA1 - r2 - r1\right)^2} = r1 + r2
$$

Substituting $AB = 6 - \sqrt{2}$, $BC = 6 + \sqrt{2}$, and $AA1 = 6$ into this equation, we get

$$
\sqrt{\left(6 - \sqrt{2} - r2 - r1\right)^2 + \left(6 + \sqrt{2} - r2 - r1\right)^2 + \left(6 - r2 - r1\right)^2} = r1 + r2
$$

Let $r1 + r2 = t$ and rewrite this equation as

$$
\begin{gathered}
\sqrt{(6 - \sqrt{2} - t)^2 + (6 + \sqrt{2} - t)^2 + (6 - t)^2} = t \Leftrightarrow (6 - \sqrt{2} - t)^2 + (6 + \sqrt{2} - t)^2 + (6 - t)^2 = t^2 \Leftrightarrow \\
\Leftrightarrow t^2 - 18t + 56 = 0
\end{gathered} \Leftrightarrow \left[\begin{array}{l}
t = 4 \\
t = 14
\end{array}\right.
$$

Both spheres are located inside the parallelepiped, so

$$
r1 \leqslant r2 < \frac{1}{2} \min(AB, BC, AA1) = \frac{1}{2}(6 - \sqrt{2})
$$

Under the conditions (1) and (2), spheres with radii $r1$ and $r2$ can be inscribed in the trihedral angles at vertices $A$ and $C1$, and all conditions of the problem will be satisfied. Since $t = r1 + r2 \leqslant 2r2 \leqslant 6 - \sqrt{2}$, the condition of the problem is satisfied only by $t = 4$. Let $V$ be the sum of the volumes of the spheres. Then

$$
\begin{aligned}
V & = \frac{4}{3} \pi r1^3 + \frac{4}{3} \pi r2^3 = \frac{4}{3} \pi (r1^3 + r2^3) = \frac{\pi}{3} (r1 + r2)(4r1^2 - 4r1r2 + 4r2^2) = \\
& = \frac{\pi}{3} (r1 + r2)((r1 + r2)^2 + 3(r2 - r1)^2) = \frac{\pi t}{3} (t^2 + 3(2r2 - t)^2) \geqslant \frac{\pi t^3}{3} = \frac{64 \pi}{3}
\end{aligned}
$$

with equality achieved when $r1 = r2 = 2$. At the same time,

$$
V = \frac{\pi t}{3} (t^2 + 3(2r2 - t)^2) \leqslant \frac{\pi t}{3} (t^2 + 3(AB - t)^2) = \frac{4 \pi}{3} (42 + 3(2 - \sqrt{2})^2) =
$$

with equality achieved when $r2 = \frac{1}{2} AB = 3 - \frac{1}{\sqrt{2}}$, $r1 = t - r2 = 1 + \frac{1}{\sqrt{2}}$.

## Answer

$d = 4, V_{\max} = \left(\frac{136}{3} - 16 \sqrt{2}\right) \pi, V_{\min} = \frac{64 \pi}{3}$.

## [Area and Volume (Problems on Extremum).]

Difficulty: 4

![](https://cdn.mathpix.com/cropped/2024_05_06_a98533106cc029c284d7g-20.jpg?height=143&width=1241&top_left_y=443&top_left_x=14)

Inside the rectangular parallelepiped $ABCD A1B1C1D1$, there are two spheres $\sigma1$ and $\sigma2$, touching each other externally; moreover, the sphere $\sigma1$ touches the faces $ABCD$, $ABB1A1$, and $BCC1B1$, while the sphere $\sigma2$ touches the faces $A1B1C1D1$, $ADD1A1$, and $CDD1C1$. It is known that $A1B1 = 14 - \sqrt{3}$, $BC = 14$, and $CC1 = 14 + \sqrt{3}$. Find the distance between the centers of the spheres $\sigma1$ and $\sigma2$. Find the smallest and largest sum of the volumes of the spheres.

## Answer

$d = 9, V_{\max} = (495 - 90 \sqrt{3}) \pi, V_{\min} = 243 \pi$.