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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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    $\begingroup$ What about a rectangle? $\endgroup$ – Peter Foreman Jun 21 '19 at 15:22
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    $\begingroup$ Wow, what an obvious example I missed! Of course, now I must ask are there any others, an infinite number? $\endgroup$ – Jamminermit Jun 21 '19 at 15:27
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    $\begingroup$ Any shape with an even number of sides can be made regular and then 'stretched' parallel to one of the sides to create an irregular shape with each angle the same. $\endgroup$ – Peter Foreman Jun 21 '19 at 15:29
  • $\begingroup$ Draw a regular polygon. Move one side so it still runs parallel to the original (then another side etc). The equilateral triangle is a special case. $\endgroup$ – Mark Bennet Jun 21 '19 at 15:30
  • $\begingroup$ This is, in my opinion, another reason why the triangle is special. Of course in hindsight it's not very surprising, but at least it's something that beyond triangles we need more criteria to assure regularity. In this sense it would seem that there are more $n$-gons (with $n>3$) than triangles, for each $n.$ $\endgroup$ – Allawonder Jun 21 '19 at 16:32
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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.

enter image description here

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.

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  • $\begingroup$ @NateEldredge Done. Note convexity not required. $\endgroup$ – Ethan Bolker Jun 21 '19 at 16:24
  • $\begingroup$ I hope you don't need this CMS-1500 claim form any more... $\endgroup$ – Marco13 Jun 22 '19 at 10:37
  • $\begingroup$ @Marco13 I think you posted your comment on the wrong site ... $\endgroup$ – Ethan Bolker Jun 22 '19 at 12:00
  • $\begingroup$ It referred to the reverse of the page that your image was scanned from, but true: It certainly wasn't "on topic", in this regard... $\endgroup$ – Marco13 Jun 22 '19 at 19:34
  • $\begingroup$ @Marco13 Now I understand. Recycled paper. $\endgroup$ – Ethan Bolker Jun 22 '19 at 19:40
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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.

enter image description here

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    $\begingroup$ +1. Only polygon where this not works is a regular triangle. $\endgroup$ – M. Winter Jun 22 '19 at 12:34
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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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