In one container, there is one liter of wine, in another, one liter of water. From the first container, we pour one deciliter into the second, mix it, then pour one deciliter of the mixture back into the first container. Calculate the limit of the amount of wine in the first container if the above procedure is repeated infinitely. (We assume perfect mixing during each transfer, and there is no fluid loss during the process.)
Suppose we have already performed the procedure described in the problem $n$ times. Let $a_{n}$ be the amount of wine in the vessel that originally contained only wine. Let's examine how the next pouring and then the pouring back will change this amount.

According to the conditions, during the pouring, $\frac{a_{n}}{10}$ liters of wine will be transferred to the other vessel, where there was already $1-a_{n}$ liters of wine, so now it has $1-0.9 a_{n}$ liters. When pouring back, $\frac{1}{11}$ of this amount will return to the $0.9 a_{n}$ liters of wine that remained, so $0.9 a_{n}+\frac{1-0.9 a_{n}}{11}$ liters of wine will be in the first vessel. This is nothing but $a_{n+1}$, so

$$
a_{n+1}=\frac{9}{11} a_{n}+\frac{1}{11}
$$

Our task is to calculate the limit of the infinite sequence defined by this recursive formula and the initial term $a_{0}=1$.

Let's try to write the $n$-th term in closed form. Repeated application of formula (1) gives us

$$
\begin{aligned}
a_{1} & =\frac{9}{11}+\frac{1}{11} \\
a_{2} & =\left(\frac{9}{11}\right)^{2}+\frac{9}{11} \cdot \frac{1}{11}+\frac{1}{11} \\
a_{3} & =\left(\frac{9}{11}\right)^{3}+\left(\frac{9}{11}\right)^{2} \cdot \frac{1}{11}+\frac{9}{11} \cdot \frac{1}{11}+\frac{1}{11} \\
& \vdots \\
a_{n} & =\left(\frac{9}{11}\right)^{n}+\left[\left(\frac{9}{11}\right)^{n-1}+\left(\frac{9}{11}\right)^{n-2}+\ldots+\frac{9}{11}+1\right] \cdot \frac{1}{11}
\end{aligned}
$$

Inside the square brackets, a geometric series stands, whose sum is $\frac{1-\left(\frac{9}{11}\right)^{n}}{1-\frac{9}{11}}$. Based on this, $a_{n}=\frac{1}{2}\left(\frac{9}{11}\right)^{n}+\frac{1}{2}$. As $n$ grows beyond all bounds, $\left(\frac{9}{11}\right)^{n}$ tends to zero, so the sequence $a_{n}$ also has a limit, which is $1 / 2$. Therefore, the limit of the amount of wine in the first vessel is half a liter.

Péter Szalay (Zalaegerszeg, Zrínyi M. Gymnasium, 4th grade)

Remarks. 1. To solve the problem, it was necessary to show that the sought limit exists. If we already knew this, then the limit $a$ of the sequence $a_{n}$ satisfies the equation $a=\frac{9}{11} a+\frac{1}{11}$, i.e., $a=1 / 2$. Thus, if the limit exists, it is $1 / 2$. However, it must be separately proven that the limit exists.

2. The evaluation of the solutions was based on the following criteria. Correct: flawless solutions. Incomplete: They write down correct equations but fail to draw the final conclusion. a) They consider it sufficient to state that the amount given as the limit is a lower bound. b) They use the existence of the limit in calculating the limit but do not prove it. Incorrect: a) Due to calculation errors, they obtain an incorrect result. b) It is clear from their solution that they are not familiar with the necessary mathematical concepts. (L. L.)