Two circles touch each other externally at point $D$. We draw one of the external tangents $AB$ to these circles. We rotate the entire figure around the central line of the two circles. What is the ratio of the volumes of the solids formed by the rotation of the region defined by the lines $AB$, $AD$, and $BD$ (forming triangle $ABD$) and the region defined by the arcs $AD$ and $BD$ and the line $AB$?
Let $K(A B D)$ denote the volume of the solid formed by the rotation of triangle $A B D$ around the axis $O_{1} O_{2}$, and similarly, $K\left(A E_{1} E_{2} B\right)$ denote the volume of the solid formed by the rotation of $A E_{1} E_{2} B$, etc.

$$
K(A B D)=K\left(A E_{1} E_{2} B\right)-K\left(E_{1} A D\right)-K\left(E_{2} B D\right)
$$

and

$$
\begin{gathered}
K(\widehat{A D B})=K\left(A E_{1} E_{2} B\right)-K\left(E_{1} \widehat{A D}\right)-K\left(E_{2} \widehat{B D}\right) \\
O_{1} E_{1}=R \cos \alpha, O_{2} R_{2}=r \cos \alpha, A E_{1}=R \sin \alpha \text {, and } B E_{2}=r \sin \alpha
\end{gathered}
$$

But

$$
\cos \alpha=\frac{R-r}{R+r} \text { and } \sin \alpha=\frac{2 \sqrt{R r}}{R+r}
$$

so

$$
\begin{gathered}
E_{1} E_{2}=R+r-O_{1} E_{1}+O_{2} E_{2}=R+r-R \frac{R-r}{R+r}+r \frac{R-r}{R+r}=\frac{4 R r}{R+r}, \\
A E_{1}=\frac{2 R}{R+r} \sqrt{R r} \text { and } B E_{2}=\frac{2 R}{R+r} \sqrt{R r} \\
K\left(A E_{1} E_{2} B\right)=\frac{E_{1} E_{2} \pi}{3}\left[A E_{1}^{2}+A E_{1} \cdot B E_{2}+B E_{2}^{2}\right]= \\
=\frac{16 \pi}{3} \frac{R^{2} r^{2}}{(R+r)^{3}}\left(R^{2}+R r+r^{2}\right) \\
K\left(E_{1} A D\right)=\frac{A E_{1}^{2} \cdot E_{1} D}{3} \pi=\frac{A E_{1}^{2}\left(R-O_{1} E_{1}\right)}{3} \pi=\frac{8}{3} \pi \frac{R^{4} r^{2}}{(R+r)^{3}} \\
K\left(E_{2} B D\right)=\frac{B E_{2}^{2} \cdot E_{2} D}{3} \pi=\frac{B E_{2}^{2}\left(r+O_{2} E_{2}\right)}{3} \pi=\frac{8}{3} \pi \frac{R^{2} r^{4}}{(R+r)^{3}} \\
K\left(E_{1}\right)=\frac{E_{1} D^{2} \cdot \pi}{3}\left(3 R-E_{1} D\right)=\frac{\pi}{3}\left(R-O_{1} E_{1}\right)^{2}\left(2 R+O_{1} E_{1}\right)= \\
=\frac{4}{3} \pi \frac{R^{3} r^{2}}{(R+r)^{3}}(3 R+r),
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_60ef8711b429f1a4d95bg-1.jpg?height=391&width=724&top_left_y=1081&top_left_x=684)

and

$$
\begin{aligned}
K\left(E_{2} \widehat{B D}\right)=\frac{E_{2} D^{2} \cdot \pi}{3} & =\left(3 r-E_{2} D\right)=\frac{\pi}{3}\left(r+O_{2} E_{2}\right)^{2}\left(2 r-O_{2} E_{2}\right)= \\
& =\frac{4}{3} \pi \frac{R^{2} r^{3}}{(R+r)^{3}}(R+3 r)
\end{aligned}
$$

Therefore,

$$
\begin{gathered}
K(A B D)=\frac{16 \pi}{3} \frac{R^{2} r^{2}}{(R+r)^{3}}\left(R^{2}+R r+r^{2}\right)-\frac{8}{3} \pi \frac{R^{4} r^{2}}{(R+r)^{3}}- \\
-\frac{8}{3} \pi \frac{R^{2} r^{4}}{(R+r)^{3}}=\frac{8}{3} \pi \frac{R^{2} r^{2}}{R+r}
\end{gathered}
$$

and

$$
\begin{gathered}
K(\widehat{A D B})=\frac{16 \pi}{3} \frac{R^{2} r^{2}}{(R+r)^{3}}\left(R^{2}+R r+r^{2}\right)-\frac{4 \pi}{3} \frac{R^{3} r^{2}}{(R+r)^{3}}(3 R+r)- \\
-\frac{4 \pi}{3} \frac{R^{2} r^{3}}{(R+r)^{2}}(R+3 r)=\frac{4}{3} \pi \frac{R^{2} r^{2}}{(R+r)}
\end{gathered}
$$

thus the ratio of the volumes of the two solids of revolution is

$$
K(A B D): K(\widehat{A D B})=\frac{8}{3} \pi \frac{R^{2} r^{2}}{(R+r)}: \frac{4}{3} \pi \frac{R^{2} r^{2}}{(R+r)}=2: 1
$$

(Imre Dömény, Budapest.)

The problem was also solved by: Fodor H., Földes R., Fuchs I., Heimlich P., Jánosy Gy., Krampera Gy., Rosenberg J., Ruwald S., Székely J., Tandlich E.