[ Systems of Points and Segments. Examples and Counterexamples ]

Author: Saghafian $M$.

Find all configurations of six points in general position on a plane such that the triangle formed by any three of them is equal to the triangle formed by the other three.
Let , , , be given points. If not stated otherwise, by a triangle we mean a triangle with vertices at , by a segment we mean a segment with endpoints at , and by length we mean the length of such a segment. We will prove a series of lemmas.

1) For any length , at least one of the following statements is true:

(A) There exists an equilateral triangle with side ;

(B) There exist three segments of length with six distinct endpoints.

![](https://cdn.mathpix.com/cropped/2024_05_06_bca77ef9e87e9e47ba11g-51.jpg?height=54&width=1946&top_left_y=2614&top_left_x=16)

![](https://cdn.mathpix.com/cropped/2024_05_06_bca77ef9e87e9e47ba11g-51.jpg?height=51&width=2012&top_left_y=2667&top_left_x=0)
(A), otherwise we get a five-segment broken line, all segments of which are equal to . Its extreme and middle segments will give the construction (B).

2) Let be the maximum length. Then (A) is false, and therefore (B) is true.

Indeed, if _ is equilateral with side , then inside the Reuleaux triangle constructed on it lie the vertices of an equal , which is impossible. 3) Any two segments of maximum length , intersect.

Indeed, if $A$ we draw through and lines perpendicular to . In the strip between them lie the other points. Doing the same for $A$ we get that and - are non-parallel heights of some rhombus, connecting internal points of its sides. Such heights, obviously, intersect.

![](https://cdn.mathpix.com/cropped/2024_05_06_bca77ef9e87e9e47ba11g-52.jpg?height=828&width=1006&top_left_y=318&top_left_x=526)

4) Let segments , and intersect at three points (see figure). Then the perpendiculars and are equal as heights of equal

![](https://cdn.mathpix.com/cropped/2024_05_06_bca77ef9e87e9e47ba11g-52.jpg?height=83&width=1989&top_left_y=1199&top_left_x=0)

![](https://cdn.mathpix.com/cropped/2024_05_06_bca77ef9e87e9e47ba11g-52.jpg?height=51&width=2058&top_left_y=1275&top_left_x=0)
that this construction satisfies the conditions.

Exactly the same reasoning in the case of segments intersecting at one point shows that they intersect at a common midpoint, forming angles of

## Answer

Two equilateral triangles inscribed in the same circle.

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