The conical frustum-shaped glass we are considering can hold three times as much beverage when filled to its full height compared to when it is filled to only half its height. Which is larger: the diameter of the base of the glass, or the radius of the top? (The height, diameter, and radius refer to the internal dimensions.)
To solve the task, it is sufficient to know the formula for the volume of a frustum of a cone, and also that if we cut the frustum with a plane perpendicular to the base and passing through the center, it cuts out a symmetrical trapezoid from the frustum. The plane that is perpendicular to the height of the trapezoid and passes through its midpoint intersects the frustum in a circle with diameter $2 \varrho$, and since this is exactly the midline of the trapezoid, $2 \varrho = \frac{2 R + 2 r}{2}$.

Based on the above, we can write the following equation:

$$
\frac{m \pi}{3}\left(R^{2} + r R + r^{2}\right) = 3 \frac{\pi}{3} \cdot \frac{m}{2}\left(\varrho^{2} + \varrho r + r^{2}\right)
$$

( $R$ is the radius of the top circle, $r$ is the radius of the base circle, $m$ is the height of the frustum). Rearrange the equation and substitute $\varrho = \frac{R + r}{2}$, we get:

$$
5 R^{2} - 4 R r - 13 r^{2} = 0
$$

Divide the equation by $r^{2} \neq 0$, and introduce the new variable $a = \frac{R}{r}$. Then,

$$
5 a^{2} - 4 a - 13 = 0, \quad a = \frac{4 + \sqrt{276}}{10} \approx 2.06
$$

(considering only the positive root).

Thus, $R \approx 2.06 r > 2 r = d$, meaning the diameter of the base circle is smaller than the radius of the top circle.

Remark. Most of the submissions did not justify why the radius of the top circle of the half-filled glass is equal to $\frac{R + r}{2}$, and they received 4 points.

![](https://cdn.mathpix.com/cropped/2024_05_02_df7ed36d2df999fa7c7fg-1.jpg?height=385&width=417&top_left_y=1176&top_left_x=843)