A concave polygon can be equilateral. For example, the concave octagon to the right is equilateral.
What is the least number of sides a concave equilateral polygon can have?
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As pointed out in the comments, another way to form an equilateral concave pentagon is to start with a regular pentagon and flip one of the points inward:
I didn't expect the equilateral shape to look so strange...
My preference for the pentagon was just a regular pentagon with one point flipped inward. :)
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I like that one. It looks like a stylized pac-man that way.
Also there is a square with an equilateral triangle cut out from one side
what about a "cross"?! the area of the polygon is simply zero? it is just a 4 sided polygon (left, up, right, down) with equal side length. you simply reduce the angle of the left up right and down in the red shape from the question. So looking at becoming infinity small angle the sides should be "melted" together to all in all 4. this should be also concave
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That would still have 8 sides - polygons don't have half-connected vertices. Furthermore, the area should be more than 0, otherwise 2 would be the answer - just two lines that are on top of each other
Doesn't a polygon have to have a measurable area? I believe creating a "shape with sides" with no measurable area causes it to stop being a shape, and it doesn't meet the criteria of the problem.
I did not understand anything, nor the question, nor the explanation.
You can take a hexagon, slice it in half from corner to corner, put a point in the center of the longest side and claim it an angle a=180° + lim x where x->0.
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If angle a is measurable, even immeasurably small, then its two legs will still result in a total of five legs. As angle a approaches zero, the definition of the polygon as "concave" becomes less convincing. At x = 0, the polygon is no longer concave.
Also you would need to adjust the other angles of the hexagon in order for the resulting shape to be equilateral.
Nice solution !
What about rhombus ?
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The word "rhombus" refers to any equilateral quadrilateral. The equality of the sides forces it to be a parallelogram. This fact is logically equivalent to the solution above.
A rhombus can’t be concave
So it appears the answer to this question is incorrect. I answered 4 and it said I was wrong.
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No. You were wrong.
You are replying to a proof of why there is no convex equilateral quadrilateral.
That's because the answer is 5. See the proof above.
The proof that that there is no equilateral concave quadrilateral is called a proof by contradiction . You start with the premise that the opposite of what you are trying to prove is true. Then you show that this leads to a contradiction.
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*Proof by contradiction I don’t think a proof can be made of just contractions... hehe.
the quadrilateral concave polygon cannot be equilateral
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Suppose that it were possible to have a concave equilateral quadrilateral.
Consider the triangles △ A B C and △ A D C . If A B C D is equilateral, then A B = B C = A D = D C . In addition, the triangles share the side A C . The triangles are congruent by SSS congruence, but this implies they are coincident. This is a contradiction. Thus, a concave equilateral quadrilateral is impossible.
A concave equilateral pentagon is possible, although they typically look kind of weird.
This particular concave equilateral pentagon was formed by taking a rhombus whose smallest angle is less than 60 ∘ , and then placing an equilateral triangle on top of it. However, there are other ways of forming a concave equilateral pentagon.