A torus has an inner diameter of $\delta$ and an outer diameter of $d$. It is placed over the lateral surface of a right circular cone with height $m$, passing through the apex of the cone. If the radius of the base circle of the cone is $r(\delta<2 r)$, in what radius circle does the torus touch the lateral surface of the cone?
The diameter of the meridian circle of the ring test (1. the figure) is $\frac{d}{2}-\frac{\delta}{2}$, so its radius is $O D=\frac{d-\delta}{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_25e56c781067c3829f8bg-1.jpg?height=369&width=531&top_left_y=233&top_left_x=794)

If we denote the radius of the contact circle by $\varrho$ and draw a perpendicular from it to the $O^{\prime} O$ line, we see that

$$
\varrho=\frac{\delta}{2}+\frac{d-\delta}{4}-E O
$$

Since $E O D_{\triangle} \sim B A C_{\triangle}$ (because the angles marked by the arcs are right angles), we have

$$
E O: \frac{d-\delta}{4}=m: \sqrt{m^{2}+r^{2}}
$$

Substituting the value of $E O$ into (1), we get

$$
\varrho=\frac{\delta}{2}+\frac{d-\delta}{4}\left(1-\frac{m}{\sqrt{m^{2}+r^{2}}}\right)
$$

result.

Jenő Keszei (Celldömölk, Gábor Áron g. IV. o. t.)[^0]


[^0]:    ${ }^{1}$ The meridian curve of a surface of revolution is the plane section of the surface whose plane passes through the axis of rotation of the surface.