Let's select 4 vertices from a regular 13-sided polygon in every possible way such that the quadrilateral defined by them has all sides and diagonals of different lengths. (Selections that can be transformed into each other by rotation or reflection are not considered different. Write down the systematic search method by which we arrived at the suitable quadruples.)





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Since the task does not distinguish between diagonals and sides, it is sufficient to always consider the $N$ convex quadrilateral determined by the 4 selected vertices of the regular 13-sided polygon $S$. In this, the total number of sides and diagonals is 6. The number of different lengths of diagonals and sides in $S$ is the same, because it is sufficient to examine the segments drawn from one vertex; the line connecting the vertex to the midpoint of the opposite side is a symmetry axis of $S$, and with respect to this, the segments drawn to the other 12 vertices are pairwise mirror images; however, any two of the 6 segments on one side of the axis are different, because the shorter arcs of the circle circumscribed around $S$ corresponding to them are different.

Let the lengths of the chords over the $1/13, 2/13, \ldots, 6/13$ arcs of the circle be denoted by $d_{1}, d_{2}, \ldots, d_{6}$. The segment of length $d_{1}$ can only be a side in $N$, because there is no other vertex of $S$ on either side of it. The chord of length $d_{2}$ is also a side, because if it were a diagonal in $N$, its endpoints would be connected by the only vertex of $S$ on one side, thus $N$ would have two equal sides, which the problem statement does not allow.

Let's form the possible $N$s starting from the fact that the sides of lengths $d_{1}$ and $d_{2}$ are adjacent or opposite. The two remaining sides together still have 10/13 of the circle left. This can only be composed of different arcs from the remaining 10/13, specifically in the combinations $4/13$ and $6/13$, or $3/13$ and $7/13$, over which chords of lengths $d_{4}$ and $d_{6}$, or $d_{3}$ and $d_{6}$ are found (since the arc greater than a semicircle, $7/13$, complements to $6/13$ of the circle), so the segment of length $d_{6}$ must always be a side.

If the sides of lengths $d_{1}$ and $d_{2}$ are chosen as adjacent, the diagonal of length $d_{3}$ connects their non-common endpoints, so only sides of lengths $d_{4}$ and $d_{6}$ can appear here. The side of length $d_{4}$ cannot be adjacent to the side of length $d_{2}$, because then the diagonal connecting their non-common endpoints would be equal to the side of length $d_{6}$. Connecting the side of length $d_{4}$ to the free endpoint of the side of length $d_{1}$, however, is successful, as the diagonal connecting their endpoints is exactly the missing length $d_{5}$ (Figure 1).
![](https://cdn.mathpix.com/cropped/2024_05_02_362f02473c34064ba7b6g-1.jpg?height=778&width=422&top_left_y=972&top_left_x=840)

Now let the sides of lengths $d_{1}$ and $d_{2}$ be opposite. In this case, $N$ cannot have a side of length $d_{4}$, because it would be adjacent to the side of length $d_{2}$, which, as we have seen, is not suitable. Trying the combination of lengths $d_{3}$ and $d_{6}$, we again get a suitable $N$, as the shorter arc of the circle over one of its diagonals is $1/13 + 3/13 = 4/13$ of the circle, and over the other, $2/13 + 3/13 = 5/13$ (Figure 2). The exchange of sides of lengths $d_{3}$ and $d_{6}$ is not considered a new solution according to the conditions. The number of selection possibilities is therefore 2.

János Körner (Budapest, Apáczai Csere J. Gyak. G. II. o. t.)

Remark. Most of the received solutions tried all the possibilities for the length of one side after selecting it. This results in more unsuccessful attempts. In this problem, it seems more effective to focus on the arrangement by answering the question "what to put where?" rather than "where to put what?".