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Minimum area of an odd number-sided, equilateral polygon with side lengths of 1 unit.

checkittwice

Member
Apr 3, 2012
37
Suppose you look at all of the equilateral (non-self-intersecting)
polygons** with an odd number of sides, and each side length is
equal to 1 unit.

For examples, the polygon with the fewest number of sides in this group
is the equilateral triangle, and then the next one is an equilateral pentagon.

Has anyone thought about this?

The area of the equilateral triangle is [tex]\dfrac{\sqrt{3}}{4}[/tex] square units.


Is the area of any of these certain polygons (beyond the equilateral triangle)

less than [tex]\dfrac{\sqrt{3}}{4}[/tex] square units?



Please, do not attempt any sort of a proof. The one I saw
(and did not fully digest), is about two and a half pages long.
And, to me, I couldn't see the motivations for using the
strategies in the proof.





** These are not limited to regular polygons in general.
The equilateral triangle happens to also be regular.
 

biffboy

New member
Jun 25, 2012
5
Since you dont want a proof I'll just say no!
 

checkittwice

Member
Apr 3, 2012
37
Since you dont want a proof I'll just say no!
From the problem statement, I would have (some type of past tense) gone with "yes."

I would have thought about the snake-like effect of the sides twisting around and what
I believed to be a decreasing net area.