$\mathrm{Az} r_{1}$ and $r_{2}$ radius circles touch each other externally. The segment of their common external tangent that lies between the points of tangency is rotated around the line connecting the centers of the circles. Express the area of the frustum of the cone generated by this rotation in terms of $r_{1}$ and $r_{2}$.
Let the centers of circles $k_{1}$ and $k_{2}$ be denoted by $O_{1}$ and $O_{2}$, their radii by $r_{1}$ and $r_{2}$, and let $r_{1} \geq r_{2}$. The common point of the two circles is $E$, the points of tangency on $k_{1}$ and $k_{2}$ are $E_{1}$ and $E_{2}$. The perpendicular projections of $E_{1}$ and $E_{2}$ onto the line $O_{1} O_{2}$ are $T_{1}$ and $T_{2}$. The radii of the bases of the frustum formed by rotation are $E_{1} T_{1}=R$ and $E_{2} T_{2}=r$, and the slant height of the frustum is $E_{1} E_{2}=a$.

It is known that the lateral surface area of the frustum is:

$$
P=\pi(R+r) a
$$

Since $O_{1} E_{1}$ and $O_{2} E_{2}$ are perpendicular to $E_{1} E_{2}$, if we draw a line parallel to $E_{1} E_{2}$ from $O_{2}$, intersecting $O_{2} E_{2}$ at $G$, then $O_{1} O_{2}=r_{1}+r_{2}$ and $O_{1} G=r_{1}-r_{2}$. From the right triangle $O_{2} G O_{1}$,

$$
a=E_{1} E_{2}=O_{2} G=\sqrt{\left(r_{1}+r_{2}\right)^{2}-\left(r_{1}-r_{2}\right)^{2}}=\sqrt{4 r_{1} r_{2}}
$$

In the right trapezoid $T_{1} T_{2} E_{2} E_{1}$, the legs are $E_{1} T_{1}=R$ and $E_{2} T_{2}=r$. Let the common tangent drawn from $E$ intersect the line $E_{1} E_{2}$ at $F$. We know that the lengths of the tangents drawn from an external point to a circle are equal, so $F E_{1}=F E=F E_{2}$. $F E$ is perpendicular to $O_{1} O_{2}$, so $F E$ is the midline of the trapezoid:

$$
\frac{R+r}{2}=F E=F E_{2}=F E_{1}
$$

We have found that $R+r$ is equal to the slant height $E_{1} E_{2}$. $E_{1} E_{2}=R+r=\sqrt{4 r_{1} r_{2}}$, and substituting into (1):

$$
P=\pi(R+r) a=\pi \sqrt{4 r_{1} r_{2}} \cdot \sqrt{4 r_{1} r_{2}}=4 \pi r_{1} r_{2}
$$

Béla Gajdos (Ukraine, Beregszász, Bethlen G. Gymnasium, 11th grade)

Note. Essentially, during the solution, we saw that both the common tangent $E_{1} E_{2}$ of the circles and the sum $R+r$ are equal to the geometric mean of the diameters of the two circles.

![](https://cdn.mathpix.com/cropped/2024_05_02_05fa255fd1e614cb32d4g-1.jpg?height=525&width=813&top_left_y=1193&top_left_x=645)