We rolled a truncated conical cup around a circular table in such a way that it never reached the edge of the table during its movement. The diameter of the table is $1.6 \mathrm{~m}$. The bottom of the cup has a diameter of $5 \mathrm{~cm}$, and the top has a diameter of $6.5 \mathrm{~cm}$. How tall can the cup be?
Solution. When a right circular cone rolls, it moves along a circular path with a radius equal to the length of the cone's slant height (the vertex of the cone remains at the center of the circle). Thus, when a frustum of a cone rolls, it moves along a circular path with a radius equal to the slant height of the cone formed by extending the slant heights of the frustum. Since the cup never reaches the edge of the table, the slant height of the cone formed by extending the frustum is less than the radius of the table, which is $80 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_c0809b0c6e4520d219ebg-1.jpg?height=344&width=437&top_left_y=316&top_left_x=844)

Let the height of the cup be $h_{1}$, the height of the cone formed by extending the slant heights be $h_{2}$, and the slant height of the cone be $x$. We can draw a cross-sectional diagram. Let $E C = h_{1}$ be the height of the frustum, $A C = h_{2}$ be the height of the cone, and $A B = x$. On the diagram, $E D = 2.5 \mathrm{~cm}$ and $C B = 3.25 \mathrm{~cm}$.

The triangles $A E D$ and $A C B$ are similar, as their sides are pairwise parallel. Therefore,

$$
\frac{B C}{A C} = \frac{D E}{A E}, \quad \text{which means} \quad \frac{3.25}{h_{2}} = \frac{2.5}{h_{2} - h_{1}}
$$

From this, we get $h_{1} = \frac{0.75}{3.25} h_{2}$. We also know that $h_{2}^{2} = \sqrt{x^{2} - 3.25^{2}}$, and since $x < 80 \mathrm{~cm}$, it follows that $h_{2} < \sqrt{6389.4375} \mathrm{~cm}$. Thus, we can determine the possible values for the height of the cup: $h_{1} < \frac{0.75}{3.25} \cdot \sqrt{6389.4375} = h$, which means $0 < h_{1} < h$. Therefore, the height of the cup is less than $h$, where $h \approx 18.45 \mathrm{~cm}$.

Based on the work of Kristóf Bérczi (Szeged, Ságvári Endre Gyak. Gimn., 12th grade)