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There exist six points $A_1, A_2, ..., A_6$ in the plane, no three of which are collinear.

Suppose that any triangle formed by three points will be similar to the triangle formed by the remaining three points. If $A_1, A_2, ..., A_6$ are the vertices of a regular hexagon, we know that this property is verified.

The question is "is there any other possibility that verifies this propiety"

Thanks in advance

amWhy
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ZENG
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  • The property holds for any equiangular hexagon whise sides alternate between two lengths, i.e., ababab. – Steve B Oct 10 '18 at 02:58
  • Yes, this hexagon verify also the property. May be we can try to determine all hexagons that verify the property? – ZENG Oct 10 '18 at 05:48
  • I have an idea for a proof, but the number of cases I need to consider has tripled. By the way, where does this question come from? – Steve B Oct 21 '18 at 23:25
  • @Steve B It's comes from my mind, but I don't have solution. I hope you can share your idea with me. – ZENG Oct 24 '18 at 16:16
  • My first idea fell through. My current plan of attack is to prove that the two triangles in a pair of congruent triangles have the same circumcircle. – Steve B Oct 24 '18 at 16:27

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