MHB Is there a non-regular hexagon

[highmath]

How can I prove that If there the only hexagon that his sides equal one to each other then it angles can't be different one from each other?

 

[Greg]

I don't think that's true.

 

[Olinguito]

Indeed it is not true. For example, the hexagon on the left below is equiangular but not equilateral while the one on the right is equilateral but not equiangular.

In general, for any integer $n\geqslant3$, if $n$ is odd, then any $n$-gon is equilateral if and only if it is equilateral, but if $n$ is even, there will be $n$-gons that are equilateral but not equiangular and ones that are equiangular but not equilateral. For instance, in the case of quadrilaterals, a rectangle is equiangular but not equilateral, while a rhombus is equilateral but not equiangular.

I suspect you want to prove your result for a heptagon ($n=7$) rather than hexagon. Can you do that? Hint:

Spoiler

If $n$ is even, then it is possible to have an $n$-gon with all opposite sides parallel. This is why such a polygon can be equiangular but not equilateral (you simply make one pair of opposite sides longer than the others). Also it is possible to have a polygon with all vertices forming diagonally opposite pairs. This explains how such a polygon can be equilateral but not equiangular (you can alter its interior angles making sure that diagonally opposite ones are equal).

Can you see why neither of the above scenarios is possible for a polygon with an odd number of sides?

 

[highmath]

Indeed it is not true. For example, the hexagon on the left below is equiangular but not equilateral while the one on the right is equilateral but not equiangular.
[/SPOILER]

First of All - thanks!

Can you post a picture of equiangular polygon (for example, heagox) that is not equilateral?

 

[HOI]

How can I prove that If there the only hexagon that his sides equal one to each other then it angles can't be different one from each other?

You can't because that is not true! Imagine a regular hexagon- all sides the same length and all angle the same, made of sticks "pinned" together- that is, two sticks are held together by a simple pin through holes in the two sticks so that the sticks can rotate around that pin. Now grab two opposite sides and move one left and the other right. The regular hexagon will "warp" into a hexagon still with all sides the same length but non-equal angles.

Only triangles are "rigid". Any figure with more that three sides can be "warped" into figures with the same side length but different angles.

 

[Olinguito]

Can you post a picture of equiangular polygon (for example, heagox) that is not equilateral?

Equiangular non-equilateral octagon:

Equiangular non-equilateral decagon:

And an equilateral non-equiangular (and non-convex) dodecagon:

 

[Olinguito]

In general, for any integer $n\geqslant3$, if $n$ is odd, then any $n$-gon is equilateral if and only if it is equilateral …

I should point out that this is only true for convex polygons, not concave ones.

 

[Opalg]

In general, for any integer $n\geqslant3$, if $n$ is odd, then any $n$-gon is equilateral if and only if it is equilateral

As written, that statement is true (but vacuous)! From the rest of the thread, it seems that the second "equilateral" should have been "equiangular". The statement would then say "if $n$ is odd, then any $n$-gon is equilateral if and only if it is equiangular". But that statement is not true (as Country Boy pointed out in comment #5). In fact, a child's drawing of a house shows that you can have an equilateral convex pentagon that is not equiangular.

[TIKZ]\draw (0,0) -- (0,2) -- (1,3.732) -- (2,2) -- (2,0) -- cycle;[/TIKZ]

 

[I like Serena]

And an equilateral non-equiangular (and non-convex) dodecagon:

This is an equilateral equiangular polygon.
It's also a regular polygon and a concave polygon.
See for instance wiki on regular polygon.

 

[Olinguito]

This is an equilateral equiangular polygon.

It’s not. By equiangular I mean that all the interior angles are equal. (Wink)

Anyway, thanks to everyone for pointing out my mistakes. I should have thought more carefully before posting.