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Hexagon: Diagonals and sides (If and only if?)


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1) No. Here is a counterexample. Start with an equilateral triangle with a side length of 2 (or even slightly less than 2). The vertices of the triangle will become 3 of the vertices of the hexagon. The other 3 vertices will be found by extending the midpoints of the triangle's sides outward just enough, so that each new side formed is longer than 1, but all the diagonals remain 2 or shorter. 

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2) Yes. The easiest way to see it is to reverse the necessary condition into a sufficient condition - "A convex hexagon with all sides equal to (or less than) 1 cannot have all main diagonals longer than 2". It's kind of obvious to me, so I'm not sure what exactly needs proving. All main diagonals are the same when the hexagon is regular and are equal 2. Just imagine a hexagon made out of rigid sticks with swivel connections - making one diagonal longer forces other diagonals to be shorter, etc. It's impossible to stretch all main diagonals without stretching at least one side of the hexagon.

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I want to say 1) yes and 2) no - proof forthcoming.

 

For (1), my first thought is that, assuming that all the sides are 1+, making one diagonal shorter makes another longer, so the hexagon with the shortest length that none of its diagonals exceed I imagine would be a regular hexagon. On a regular hexagon, if all the sides are 1+, then all the diagonals must be 2+. I will try to come up with more rigorous proof.

 

However, for (2), it's easy to come up with a counterexample. One way is to first draw a line shorter than 1, then draw draw crossing diagonals longer than 2 from each endpoint. The endpoints of these two diagonals are vertices number 1, 2, 4, and 5 of a counterexample hexagon.

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