## Task 10/83

Given is a straight circular frustum of a cone, whose generatrices are inclined at $60^{\circ}$ to the base, and for whose base and top areas the relationship $A_{G}=4 A_{D}$ holds. A rubber band is stretched taut between a point $P$ on the edge of the base and a point $Q$ on the edge of the top.

How long can the rubber band be at most if it is not to leave the lateral surface at any point?
Since the rubber thread should be taut, it will mark the shortest connection on the mantle of the frustum. As the mantle of the frustum can be unwrapped into the plane without distortion, this shortest connection in the unwrapped form is the line segment $P O$. The unwrapped mantle of the frustum is a sector of an annulus. If $r_{G}$ and $r_{D}$ are the radii of the base and top, respectively, then

$$
A_{G}=r_{G}^{2} \pi=4 A_{D}=4 r_{D}^{2} \pi \rightarrow r_{G}=2 r_{D}
$$

Furthermore, if $s_{G}$ is the slant height of the frustum and $s_{D}$ is the slant height of the complementary cone, then

$$
\frac{r_{G}}{s_{G}+s_{D}}=\frac{r_{D}}{s_{D}}=\cos 60^{\circ}=\frac{1}{2}
$$

thus $s_{D}=2 r_{D}=r_{G}=s_{G}$. For the central angle $\alpha$ of the unwrapped form, we have

$$
\frac{\alpha}{360^{\circ}}=\frac{r_{G}}{s_{G}+s_{D}}=\frac{1}{2}
$$

so $\alpha=180^{\circ}$. The unwrapped mantle of the frustum thus consists of two concentric semicircles with radii $s_{G}=r_{G}$ and $2 s_{G}=2 r_{G}$. The connecting line segment between the outer and inner semicircle, which lies entirely within the semicircular ring, is the tangent from any point $P$ on the outer semicircle to the inner one (the point $P$ is to be chosen such that this tangent exists). It has (by the Pythagorean theorem) the length

$$
l=\sqrt{\left(2 s_{G}\right)^{2}-s_{G}^{2}}=s_{G} \sqrt{3}=2 r_{G} \sqrt{3}
$$