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???

 Sep 6, 2020

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




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


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 laughlaugh )

 Sep 7, 2020

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)

 Sep 7, 2020
edited by hectictar  Sep 7, 2020



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!

 Sep 11, 2020

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