Draw a secant in two concentric circles such that the chord falling in the larger circle is divided into three equal parts by the smaller one.
I. solution. Let us assume that the problem has a solution, and denote the points of intersection of the secant with the circles in succession by $A, B, C, D$. Thus, $A$ and $D$ are on the larger circle, $k$, and $B$ and $C$ are on the smaller circle, $k_{0}$, and $A B=B C=C D=A D / 3$. Let the common center of our circles be $O$ and $O A=r, O B=r_{0}(0$ because $B C=B A>0$. Finally, $D$ is the reflection of $A$ over the perpendicular bisector of the chord $B C$, which is the common axis of symmetry of $k$ and $k_{0}$, so $D C=A B=B C$.

The number of common points of the circles $k$ and $k_{1}$ is 2, 1, or 0, depending on whether the farthest point $T$ of $k_{1}$ from $O$ is outside $k$, or exactly on it - that is, $k_{1}$ touches $k$ internally - or if $T$ is inside $k$. Since it is obviously $O T=3 r_{0}$, the condition for solvability is $3 r_{0} \geq r$. In the case of equality, the secant passes through $O$. If $k$ and $k_{1}$ have 2 common points, the two secants derived from them have the same $d$ radius, so they do not lead to secants of different lengths.

László Tábori (Nagykanizsa, Landler J. Gymn., I. o. t.)

Remarks. 1. We can also state this solution as follows: $C$ - as the image of $A$ - is cut out from $k_{0}$ by the reflection of $k$ about $B$ (which is, of course, concentric with $k_{1}$).

2. We can solve the problem in several ways using similarity transformations based on the simple ratios of the distances between the points $A, B, C, D$. (In fact, the two reflections mentioned so far are also such, with a magnification ratio of 1.)

One such is written down starting from the choice of the point $A$ of the secant on $k$. Then, based on $A C: A B=2$, $C$ is cut out from $k_{0}$ by the image of $k_{0}$ under a dilation centered at $A$ with a magnification ratio of 2, and the secant is the line $A C$.

II. solution. The following is a neat, even quite simple solution using familiar knowledge. Let the point opposite to the chosen $A$ on $k$ be $E$ and the points dividing the diameter $A E$ into thirds be $L$ and $M$. Then $B L$ and $C M$ are parallel to $D E$, so they are perpendicular to the sought chord. Thus, $C$ is cut out from $k_{0}$ by the Thales circle $k^{*}$ above the diameter $A M$ (whose center is $L$).

$k^{*}$ only intersects $k_{0}$ if $M$ is not outside $k_{0}$, that is, if

$$
O M=\frac{O E}{3}=\frac{r}{3} \leq r_{0}
$$

the requirement is the same as above.

Tibor Vándor (Budapest, Kossuth L. Gymn., II. o. t.)

III. solution (outline). Let the common midpoint of the chords $B C, A D$ be $F$, and $O F=d, B F=h$, so $B C=2 h$, $A D=6 h$. From the right triangles $O F B$ and $O F A$

$$
\begin{gathered}
d^{2}=r_{0}^{2}-h^{2}=r^{2}-9 h^{2} \\
h^{2}=\frac{r^{2}-r_{0}^{2}}{8} \quad \text { and } \quad 2 h=\frac{\sqrt{2}}{2} \sqrt{r^{2}-r_{0}^{2}} \\
d^{2}=\frac{9 r_{0}^{2}-r^{2}}{8} \quad \text { and } \quad d=\frac{\sqrt{2}}{4} \sqrt{9 r_{0}^{2}-r^{2}}
\end{gathered}
$$

Based on these, either the length of $h$ or $d$ can be constructed using appropriate right triangles and squares, and the known conditions also follow.