Proving lemma about regular pentagons?

[quackdesk]

Proving lemma about regular polygons?

Homework Statement

Can a n-1 sided regular polygon be inscribed in n sided regular polygon for
\forall n \in \mathbb {N} \gt 3

Homework Equations

N/A
The area of n-1 sided regular polygon may be the largest of any n-1 polygon which is to be inscribed.

The Attempt at a Solution

PS. It is polygon of course not pentagon MY BAD!

 

[mfb]

A pentagon has always 5 sides.

It should be possible to derive an equation for the "height" of those polygons. If they fit in each other, one side can be identical, and it will become interesting on the opposite side(s), I think.

 

[LCKurtz]

A pentagon has always 5 sides.

It should be possible to derive an equation for the "height" of those polygons. If they fit in each other, one side can be identical, and it will become interesting on the opposite side(s), I think.

I don't think having a common side will work even with trying to inscribe an equilateral triangle into a square.

 

[mfb]

Hmm... I am unsure how to interpret the question now.

- does the inscribed polygon have to touch all sides, with arbitrary side lengths?
- do we have the same side length, and the polygon just have to fit inside? <- my interpretation in post 2

 

[LCKurtz]

Hmm... I am unsure how to interpret the question now.

- does the inscribed polygon have to touch all sides, with arbitrary side lengths?
- do we have the same side length, and the polygon just have to fit inside? <- my interpretation in post 2

He said they had to be regular polygons so the sides must be equal length. They can't touch all sides; consider the equilateral triangle in a square. My interpretation would be the the inscribed polygon must have its vertices touching the outer polygon. That said, I have nothing to say about how to proceed.

 

[mfb]

The polygons are regular, so all n sides have the same length, of course. But what about the inscribed polygon? Does it have to have the same side length as well?

They can't touch all sides; consider the equilateral triangle in a square.

That is a part of the solution (n=4), if the inner polygon has to touch all sides of the outer polygon.

Independent of the interpretation, there is a symmetry we can use.

 

[LCKurtz]

. They can't touch all sides; consider the equilateral triangle in a square.

That is a part of the solution (n=4), if the inner polygon has to touch all sides of the outer polygon.

Independent of the interpretation, there is a symmetry we can use.

I think I overlooked the possibility of a vertex of the triangle touching a vertex of the square.