In a polygon, if all the sides are equal, it doesn’t necessarily mean that the polygon is regular (eg. a rhombus). Is this also true for angles? Meaning can you draw a polygon whose interior angles are equal, but the shape is still not regular? I couldn’t think of any examples, but I’m sure there is one.
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$\begingroup$Here are four pentagons all with interior angles of $108^\circ$. Only the largest is regular. The generalization to any regular polygon should be clear.
Start with any polygon that has more than three edges. Move one of the edges parallel to itself a little, extend or contract the adjacent edges appropriately and you will have a new polygon with the same edge directions but different relative side lengths. If you start with a regular polygon the angles will remain all the same.
The idea behind this construction is generic. If you start with any sequence of $n > 3$ vectors that span the plane there will be an $n-2$ dimensional space of linear combinations that vanish. Each such linear combination defines a polygon with the same edge directions: form the partial sums in order to find the vertices.
$\endgroup$ 5 $\begingroup$The side lengths need not be equal. It is amenable to a vector representation.
If segments of length of an isotropic force set $F_i$ acts on a particle ( components ${F_{xi},F_{yi}}$ of consecutive positive integers $i$ ) then the vectors sum up to zero:
$$ \Sigma_i^n F_{x}=0 $$$$ \Sigma_i^n F_{y}=0 $$
Each vector joins tail to head of vector and the force polygon comes back to point of start after all $n$ rotations of $\dfrac{2 \pi}{n}$.
So.. irregular polygons of arbitrary parallel translation of each vector displaced from regular polygons because of force equilibrium.
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