Can we have an equilateral polygon $n \geq 5$, which is not equiangular? Ot does every odd n-gon which is equilateral must be equiangular? Is a construction of an equilateral but not equiangular n-gon possible with ruler and compass?
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$\begingroup$For the first part, note that a regular pentagon is not rigid - so if a model is made with rigid sides and flexible joints it can be adjusted to any number of differently shaped pentagons.
For the second part, take a square (constructible) and construct an equilateral triangle on one side - that gives an equiliateral pentagon which is not equiangular. One can keep adding equilateral triangles to increase the number of sides - or base constructions on other constructible figures with $3\cdot 2^n$ sides. I think it may prove to be harder work to construct a convex figure of the kind you are looking for.
You can construct a rectangle with sides $1$ and $n-1$ which is a degenerate $2n$-gon - stick an equilateral triangle on the end and you have a degenerate $2n+1 $-gon.
If you place an equilateral triangle on one end and $n-2$ equilateral triangles on each of the long sides you get a sawtooth arrangement with $4n-3$ sides. If you add $n-1$ equilateral triangles on each side and one on the end you get $4n-1$ sides - that deals with all sufficiently large odd numbers. But the figures constructed are not convex - did you want convexity?
Why not see if you can create a general construction mainly of squares with just one triangle or pentagon (or $17$-gon or $257$-gon or $65537$-gon added). Can you see why an equilateral figure made wholly of squares will not do?
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