I just noticed that the intersection points of any regular polygon, when connected with lines, form a smaller regular polygon of the same side count. How to Prove this???

Pangolin14 Sep 6, 2020

#1**0 **

I don't know how to prove it, but your question inspired me to create this regular polygon grapher on desmos:

https://www.desmos.com/calculator/br9gidymwe

I spent all afternoon working on it, and I thought it was really cool, so I wanted to share

You can select the number of sides with the slider for n , and go into the drawings folder to turn off/on certain parts of the graph

( Also another cool use for this graph is to create diagrams for questions )

hectictar Sep 7, 2020

#2**0 **

A slightly modified version: https://www.desmos.com/calculator/ot7gry5kld

(It looks really cool if you don't force the step of n to be 1, then play it)

hectictar Sep 7, 2020

#3**+1 **

Thanks!

Also, I just came up with a proof due to your VERY USEFUL desmos grapher I don't know how you did it but it is amazing ANYWAYS

Because the diagonals of a regular polygon radiating from the same side intersect, if drawn a line through, will be the median line of the side they share or one of the vertices' sides independent from the other vertex's side. For example, if we have a hexagon ABCEDF, diagonal EA and FB will intersect at a point with a median line of side AF going through that point. Notice AF doesn't include E, but does include the other endpoint of diagonal EA.

Thus we can make a generalization that there will be--for a polygon with n sides-- n intersections of this sort. This is because there are n possibilities to order each of the letters ABCDEF in ascending order in the alphabet. AB, BC, CD, DE, EF, and FA=6 ways, in a hexagon. These intersections of diagonals form n triangles. Congruent triangles. Mirror image each of these, and they will form your shape with n bases as sides! And because these are congruent, this is a regular shape, similar to your mother shape!

Pangolin14 Sep 11, 2020