Vojta Vodník enjoyed pouring water between three containers. First, he poured one third of the water from the second container into the first and third. Then he poured one quarter of the water from the first container into the second and third, and finally, he poured one fifth of the water from the third container into the first and second containers. After that, there was one liter of water in each container. How much water did Vojta originally have in each of the containers?

(M. Petrová)
(Solution "from the beginning") Let the amount of water in each container be $x, y$, and $z$. The entire process of pouring is briefly illustrated in the following table; all values are in liters.

|  | 1st container | 2nd container | 3rd container |
| :---: | :---: | :---: | :---: |
| at the beginning | $x$ | $y$ | $z$ |
| 1st pouring | $+\frac{y}{3}$ | $-2 \cdot \frac{y}{3}$ | $+\frac{y}{3}$ |
| after pouring | $x+\frac{y}{3}$ | $\frac{y}{3}$ | $z+\frac{y}{3}$ |
| 2nd pouring | $-2 \cdot \frac{x+\frac{y}{3}}{4}$ | $+\frac{x+\frac{y}{3}}{4}$ | $+\frac{x+\frac{y}{3}}{4}$ |
| after pouring | $\frac{x}{2}+\frac{y}{6}$ | $\frac{x}{4}+\frac{5 y}{12}$ | $\frac{x}{4}+\frac{5 y}{12}+z$ |
| 3rd pouring | $+\frac{\frac{x}{4}+\frac{5 y}{12}+z}{5}$ | $+\frac{\frac{x}{4}+\frac{5 y}{12}+z}{5}$ | $-2 \cdot \frac{\frac{x}{4}+\frac{5 y}{12}+z}{5}$ |
| at the end | $\frac{11 x}{20}+\frac{y}{4}+\frac{z}{5}$ | $\frac{3 x}{10}+\frac{y}{2}+\frac{z}{5}$ | $\frac{3 x}{20}+\frac{y}{4}+\frac{3 z}{5}$ |

We can now easily set up a system of equations to solve:

$$
\begin{aligned}
& \frac{11 x}{20}+\frac{y}{4}+\frac{z}{5}=1 \\
& \frac{3 x}{10}+\frac{y}{2}+\frac{z}{5}=1 \\
& \frac{3 x}{20}+\frac{y}{4}+\frac{3 z}{5}=1
\end{aligned}
$$

or

$$
\begin{aligned}
& 11 x+5 y+4 z=20, \\
& 6 x+10 y+4 z=20, \\
& 3 x+5 y+12 z=20 .
\end{aligned}
$$

Subtracting the first two equations gives us

$$
5 x-5 y=0, \text { i.e., } y=x .
$$

Substituting into the modified second and third equations:

$$
\begin{aligned}
& 16 x+4 z=20, \\
& 8 x+12 z=20 .
\end{aligned}
$$

Subtracting these equations from each other:

$$
8 x-8 z=0, \text { i.e., } z=x .
$$

This means that Vojta Vodník originally had the same amount of water in all three containers, i.e., one liter in each, as at the end of the pouring.

Evaluation. 3 points for deriving the amount of water in the containers after the third pouring and 3 points for solving the system of equations.

Alternative Solution. (Solution "from the end") We will proceed in reverse. At the end, after the third pouring, we have one liter of water in each container. One liter in the third container represents $\frac{3}{5}$ of the amount of water that was in this container before the third pouring. We poured $\frac{1}{3}$ liter each. We will "return" this water:

|  | 1st container | 2nd container | 3rd container |
| :---: | :---: | :---: | :---: |
| after 3rd pouring | 1 | 1 | 1 |
| pouring back | $-\frac{1}{3}$ | $-\frac{1}{3}$ | $+\frac{2}{3}$ |
| before 3rd pouring | $\frac{2}{3}$ | $\frac{2}{3}$ | $\frac{5}{3}$ |

In the first container, there are now $\frac{2}{3}$ liters, which represent $\frac{2}{4}$ of the amount of water in this container before the second pouring. Again, we poured $\frac{1}{3}$ liter each. We will return this water:

|  | 1st container | 2nd container | 3rd container |
| :---: | :---: | :---: | :---: |
| after 2nd pouring | $\frac{2}{3}$ | $\frac{2}{3}$ | $\frac{5}{3}$ |
| pouring back | $+\frac{2}{3}$ | $-\frac{1}{3}$ | $-\frac{1}{3}$ |
| before 2nd pouring | $\frac{4}{3}$ | $\frac{1}{3}$ | $\frac{4}{3}$ |

We will repeat the entire reasoning one more time. The amount of water in the second container represents $\frac{1}{3}$ of the original amount of water in this container. This means that we will return $\frac{1}{3}$ liter of water each:

|  | 1st container | 2nd container | 3rd container |
| :---: | :---: | :---: | :---: |
| after 1st pouring | $\frac{4}{3}$ | $\frac{1}{3}$ | $\frac{4}{3}$ |
| pouring back | $-\frac{1}{3}$ | $+\frac{2}{3}$ | $-\frac{1}{3}$ |
| at the beginning | 1 | 1 | 1 |

Evaluation. For each pouring, 1 point for determining what fraction of the water remained in the container and 1 point for determining the amount of water in the containers before the pouring.