Task B-3.4. (20 points) The mutual distance between the centers of spheres with radii $R$ and $r(r < R)$ is $a(R-r < a \leq R+r)$.

a) Express the volume $V$ of the regular circular cone circumscribed around these spheres in terms of the parameters $a, R$, and $r$.

b) Calculate the volume of the cone if $R=10, r=6$, and $a=8$.
Solution. a) Let the radius of the cone's base be $|B D|=x$, and the height $|C D|=y$. Points $E$ and $F$ are on the segment $\overline{B C}$, $T$ is on $\overline{P F}$, and quadrilateral $Q T F E$ is a rectangle.

From the similarity of triangles $Q E C$ and $P T Q$, we have $\frac{|Q C|}{|Q E|}=\frac{|P Q|}{|P T|}$, and thus

$$
\frac{y-R-a}{r}=\frac{a}{R-r}
$$

After rearranging, we get

$$
y=\frac{(a+R-r) R}{R-r}
$$

![](https://cdn.mathpix.com/cropped/2024_05_30_978576e245d98ac92b08g-16.jpg?height=607&width=360&top_left_y=735&top_left_x=733)

From the similarity of triangles $B D C$ and $P T Q$, we have $\frac{|B D|}{|C D|}=\frac{|P T|}{|Q T|}$, that is

$$
\frac{x}{y}=\frac{R-r}{\sqrt{a^{2}-(R-r)^{2}}}
$$

From (1) and (2), we get $x=y \cdot \frac{R-r}{\sqrt{a^{2}-(R-r)^{2}}}$, or

$$
x=R \sqrt{\frac{a+R-r}{a-R+r}}
$$

From (1), (2), and (3), we have

$$
V=\frac{x^{2} \pi \cdot y}{3}=\frac{R^{3}(a+R-r)^{2} \pi}{3(R-r)(a-R+r)}
$$

b) In the special case for $R=10, r=6, a=8$, we get $V=3000 \pi$.