A $10 \mathrm{~cm}$ edge cube has one of its edges on the plane of the tabletop, and one of its faces forms a $20^{\circ}$ angle with the tabletop. The cube is filled with water. Determine the height of the water level above the tabletop with an accuracy of one-tenth of a millimeter, if the cube contains a) 0.1 liter b) 0.45 liter, c) 0.9 liter of water.
The cube can be divided into parts with volumes $V_{1}, V_{2}, V_{3}$ by planes parallel to the table surface and passing through the edges parallel to the table. (See Figure 1, which shows a cross-section perpendicular to the edge on the table.)

![](https://cdn.mathpix.com/cropped/2024_05_02_6c2e50fa310823807a9bg-1.jpg?height=586&width=713&top_left_y=268&top_left_x=684)

Figure 1

Due to symmetry, it is clear that $V_{1}=V_{3}$. Let the volume of the water inside the cube be denoted by $T$. The height we are looking for is calculated as follows:

$$
\begin{aligned}
& TV_{1}+V_{2}
\end{aligned}
$$

The volume of the cube is $K=1000 \mathrm{~cm}^{3}$.

$V_{1}=V_{3}=\left(\frac{1}{2} 10 \cdot 10 \operatorname{tg} 20^{\circ}\right) \cdot 10=500 \operatorname{tg} 20^{\circ}=182 \mathrm{~cm}^{3}$

$V_{1}+V_{2}=K-V_{3}=818 \mathrm{~cm}^{3}$, and thus in cases a), b), and c), we proceed according to (1), (2), and (3) respectively.
a) $T=100 \mathrm{~cm}^{3}$, and thus the area of the triangular base $A P Q$ (Figure 2) of the 3-sided prism formed by the water is $\frac{100 \mathrm{~cm}^{3}}{10 \mathrm{~cm}}=10 \mathrm{~cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_6c2e50fa310823807a9bg-1.jpg?height=418&width=675&top_left_y=1569&top_left_x=714)

Figure 2

Let the hypotenuse of the base $P Q$ be $c$, and the height $A R$ be $m_{1}$, then

$$
c=\frac{P A}{\cos 20^{\circ}}=\frac{m_{1}}{\sin 20^{\circ} \cos 20^{\circ}}
$$

and thus the area of the base is

$$
\frac{c m_{1}}{2}=\frac{m_{1}^{2}}{2 \sin 20^{\circ} \cos 20^{\circ}}=\frac{m_{1}^{2}}{\sin 40^{\circ}}=10
$$

from which

$$
m_{1}=\sqrt{10 \sin 40^{\circ}}=2.53 \mathrm{~cm}
$$

b) $T=450 \mathrm{~cm}^{3}$, and thus the area of the base $A B P_{1} Q_{1}$ (Figure 3) of the 4-sided prism formed by the water is $\frac{450 \mathrm{~cm}^{3}}{10 \mathrm{~cm}}=45 \mathrm{~cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_6c2e50fa310823807a9bg-2.jpg?height=515&width=729&top_left_y=154&top_left_x=682)

Figure 3

Let the height $A R_{1}=A B_{0}+B_{0} R_{1}=m_{2}$.

$$
A B_{0}=10 \sin 20^{\circ}=3.42 \mathrm{~cm}
$$

The area of $\triangle A B B^{\prime}$ is, as we have seen, $\frac{182 \mathrm{~cm}^{3}}{10 \mathrm{~cm}}=18.2 \mathrm{~cm}^{2}$, and thus the area of the parallelogram $B B^{\prime} Q_{1} P_{1}$ is $45-18.2=26.8 \mathrm{~cm}^{2}$. Therefore,

$$
B B^{\prime} \cdot B_{0} R_{1}=\frac{10}{\cos 20^{\circ}} \cdot B_{0} R_{1}=26.8
$$

From this,

$$
B_{0} R_{1}=2.68 \cos 20^{\circ}=2.52 \mathrm{~cm}
$$

so the height we are looking for is

$$
m_{2}=3.42+2.52=5.94 \mathrm{~cm}
$$

c) In this case, the part of the cube not filled with water, due to central symmetry, is congruent to the prism formed by the water in case a), so the height we are looking for is $m_{3}=m-m_{1}$, where $m$ is the distance from the farthest edge of the cube to the table, i.e., (Figure 1)

$$
m=10 \sin 20^{\circ}+10 \cos 20^{\circ}=10\left(\sin 20^{\circ}+\cos 20^{\circ}\right)=12.82 \mathrm{~cm}
$$

and thus

$$
m_{3}=m-m_{1}=12.82-2.53=10.29 \mathrm{~cm}
$$

Éva Papp (Bp. VIII., Ságvári E. High School II. Class)