We have three equal volume containers and a pipette. The first container (I.) is half-filled with a solution that dissolves in water, while the second (II.) and third (III.) containers are empty. We transfer one full pipette of the solution from I. to II., then add enough water to make it half full. After mixing, we transfer one full pipette of the solution to III., fill it halfway, mix it, then use the new mixture to fill I. halfway, and pour the remainder into II. Now, the dilution in I. and II. is the same. - What fraction of the container volume is the pipette volume?
Let the volume of the containers be $2v$, and that of the pipette be $x$. Thus, in the I. container, there was a volume of $v$ of liquid, and then $v-x$ remained. In the II. container, we mixed $x$ of the liquid and $v-x$ of water, so in the transferred mixture to the III. container, for each volume unit, there was $x / v$ unit of the original liquid, thus we transferred $x^{2} / v$ unit of the original liquid, and after filling, in the III. container, for each volume unit, $(x / v)^{2}$ of the original liquid was present.

For the half-filling of the I. container, again $x$ volume of the mixture from the III. container was needed; accordingly, of the original liquid, $x^{3} / v^{2}$ volume units returned, so the dilution ratio of the I. mixture is:

$$
\frac{1}{v} \cdot\left(v-x+\frac{x^{3}}{v^{2}}\right)=1-\frac{x}{v}+\left(\frac{x}{v}\right)^{3}
$$

After the second filling and the transfers, the volume of the mixture in the II. container is the same as the volume of the used water, that is, $2(v-x)$. (It does not fill the container.) The amount of the original liquid here is as much as is missing from the I. container, so the dilution ratio is:

$$
\frac{x-\frac{x^{3}}{v^{2}}}{2(v-x)}=\frac{x\left(v^{2}-x^{2}\right)}{2 v^{2}(v-x)}=\frac{x(v+x)}{2 v^{2}}=\frac{1}{2} \cdot \frac{x}{v}+\frac{1}{2}\left(\frac{x}{v}\right)^{2}
$$

(We excluded the case $x=v$, which is uninteresting, as it would mean filling the original liquid without adding water.)

From the equality of (1) and (2), with $x / v=q$, and after rearranging,

$$
2 q^{3}-q^{2}-3 q+2=0
$$

The sum of the coefficients is 0, so $q=1$ is a root of the equation, and according to the previous consideration, it is a trivial solution to the problem. The appropriate $q-1$ factor can be factored out from the left side of (3) by transforming it:

$$
\begin{gathered}
\left(2 q^{3}-2 q\right)-\left(q^{2}+q-2\right)=2 q(q-1)(q+1)-(q-1)(q+2)= \\
=(q-1)\left(2 q^{2}+q-2\right)=0
\end{gathered}
$$

After the separation, the remaining quadratic equation $2 q^{2}+q-2=0$ has roots of opposite signs, since their product is -1, so we take only the positive root:

$$
q=\frac{x}{v}=\frac{-1+\sqrt{17}}{4} \approx 0.781
$$

thus the volume of the pipette is $x / 2 v=q / 2 \approx 0.390$ of the volume of the containers.

Németh Tamás (Pannonhalma, Bencés g. III. o. t.)