A unit-edged cube-shaped closed glass container has a base $A B C D$, side edges $A A_{1}, B B_{1}, C C_{1}$, and $D D_{1}$. The container is filled with water up to $2 / 3$ of its height, with air above it. We position the container so that the body diagonal $A C_{1}$ is vertical (with $\mathrm{A}$ at the bottom). At which points on which edges will the water-air interface be in this position? (The positions of the points in question are given by the approximate values of the segments they define, rounded to two decimal places.)
Since only air is above the water in the container, the interface between them will be horizontal after the desired setting, perpendicular to $A C_{1}$. Draw a horizontal line through the points $B_{1}, C$, and $D_{1}$ in the planes of the right-angled triangles $C_{1} A B_{1}$, $C_{1} A C$, and $C_{1} A D_{1}$, i.e., draw a perpendicular to the common hypotenuse $C_{1} A$. These lines intersect $C_{1} A$ at the same point $E$, because the equality of their sides implies that the three triangles are congruent, and their vertices correspond to each other in the order listed.
![](https://cdn.mathpix.com/cropped/2024_05_02_deb2a6d627737868a07cg-1.jpg?height=622&width=1196&top_left_y=354&top_left_x=451)

Thus, the plane $B_{1} C D_{1}=H$ is horizontal. Since the volume of the part of the cube above $H$ is $\left(C_{1} B_{1} \cdot C_{1} C \cdot C_{1} D_{1}\right) /(2 \cdot 3)=1 / 6$ unit, the part below it is $5 / 6$ unit, so the water surface will be below $H$. On the other hand, reflecting the above in the center $O$ of the cube, the surface will be above the horizontal plane $B D A_{1}=H_{1}$, so the boundary of the water surface will have its breakpoints, vertices, and the segments between them on the edges $B_{1} B, B C, C D, D D_{1}, D_{1} A_{1}, A_{1} B_{1}$ of the cube, and these segments will be parallel to the face diagonals $B_{1} C, B D, \ldots, A_{1} B$.

Let the breakpoint on the edge $B_{1} B$ be $T$, and on $B_{1} A_{1}$ be $U$. Extending the boundary segments in the planes $C_{1} B_{1} B C$ and $C_{1} B_{1} A_{1} D_{1}$, they intersect at a point $P$ on the extension of the edge $C_{1} B_{1}$, since $C_{1} B_{1}$ is the common line of the two planes, and from the above, $B_{1} P=B_{1} T=B_{1} U$. It is sufficient to determine the length $x$ of these segments, because the cube, by rotating $120^{\circ}$ around the axis $C_{1} A$, fits onto itself $\left(B_{1} C=C D_{1}=D_{1} B_{1}=A B \sqrt{2}\right)$, so the water surface divides each of the 6 edges into segments of length $x$ and $(1-x)$. Denoting the images of $P$ under these rotations by $Q$ and $R$, the volume of the air in the container is obtained by subtracting three times the volume of the pyramid $B_{1} P T U$ from the volume of the pyramid $C_{1} P Q R$, and this, according to our data, is:

$$
V=\frac{1}{6}(1+x)^{3}-3 \cdot \frac{1}{6} x^{3}=\frac{1}{3}
$$

Accordingly, we need to give one of the roots of the following equation, rounded to two decimal places:

$$
f(x)=(1+x)^{3}-3 x^{3}-2=0
$$

We can only reach the goal by systematic trial and error, for which we will use the cube table from the school function table.

Since $f(0)=-10$, let's try $x=0.25$ which lies between them. Using powers of 5 as well:

$$
f(0.25)=1.25^{3}-3 \cdot 0.25^{3}-2=\left(5^{3} \cdot 10^{-2}\right)^{3}-3 \cdot\left(5^{2} \cdot 10^{-2}\right)^{3}-2=-0.0938
$$

Since $f(0.25)$ is negative, let's try $x=0.3$:

$$
f(0.3)=1.3^{3}-3 \cdot 0.3^{3}-2=2.197-2.081=+0.116>0
$$

In absolute value, the latter has a slightly larger deviation from 0 than $f(0.25)$, so we expect $x$ to be in the first half of the interval $(0.25 ; 0.3)$. Let's try $x=0.27$:

$$
f(0.27)=1.27^{3}-3 \cdot 0.27^{3}-2=-0.0107
$$

According to all this, the value of $x$ is in the interval $(0.27 ; 0.275)$, so its value rounded to two decimal places (and indeed rounded down) is 0.27, and the value of $(1-x)$ rounded to two decimal places (rounded up) is 0.73. With this, we have answered the question of the problem.

Páles Zsolt (Sátoraljaújhely, Kossuth L. Gimn. IV. o. t. )