Cylindrical, flaring upwards, 1 liter household measuring vessel, the $1 / 2$ liter mark is found at $2 / 3$ of the vessel's height. What is the ratio of the diameter of the base circle to the diameter of the top circle?
Let the height of the frustum be $m$, the radius of the base circle be $r$, the radius of the top circle be $R$, and the radius of the circle intersecting at $\frac{1}{2}$ liter height be $r_{1}$.

The triangle $ABC$ is similar to the triangle $ADE$. From the similarity:

$$
(R-r): m=\left(r_{1}-r\right): \frac{2}{3} m
$$

from which

$$
r_{1}=\frac{2}{3}(R-r)+r=\frac{2}{3} R+\frac{1}{3} r
$$

The volume of the frustum: $V=\frac{m \pi}{3}\left(R^{2}+R r+r^{2}\right)$, so the total volume (1 liter $=1000 \mathrm{~cm}^{3}$):

$$
1000=\frac{m \pi}{3}\left(R^{2}+R r+r^{2}\right)
$$

The volume corresponding to $\frac{1}{2}$ liter: $500=\frac{\frac{2}{3} m \pi}{3}\left(r_{1}^{2}+r_{1} r+r^{2}\right)$.

From each of the two equations, solve for $m$, set them equal, and then substitute $r_{1}=\frac{2 R+r}{3}$:

$$
\begin{gathered}
\frac{3000}{\left(R^{2}+R r+r^{2}\right)}=\frac{4500}{2\left(r_{1}^{2}+r_{1} r+r^{2}\right)} \\
12\left[\left(\frac{2 R+r}{3}\right)^{2}+\frac{(2 R+r) r}{3}+r^{2}\right]=9 R^{2}+9 R r+9 r^{2}
\end{gathered}
$$

After rearranging, we get the following equation: $25 r^{2}+13 R r-11 R^{2}=0$. Divide by $R^{2}$ and introduce the variable $u$ for the desired ratio $\frac{r}{R}$: $25 u^{2}+13 u-11=0$, from which the positive solution is $u=\frac{\sqrt{1269}-13}{50}$.

The ratio of the diameters of the base circle and the top circle, $u=\frac{r}{R} \approx 0.452$.

![](https://cdn.mathpix.com/cropped/2024_05_02_bf4b61b673b167323a3cg-1.jpg?height=415&width=463&top_left_y=1424&top_left_x=820)