All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Cutting Corners (Posted on 2007-02-27)
Take a polygon with area S1 and pick a number r in [0,1/2]. Take vertex A that connects sides AB and AC and add points M and N on these sides so that AM/AB=AN/AC=r. Cut corner A along MN. Cut all other corners the same way.

After repeating these steps infinite times we will get a figure with an area S2. Let's F(r)=S2/S1. It's clear that F(0)=1 and F(½)=0.

Questions:

(a) What is this function for square?

(b) What is this function for equilateral triangle?

(c) Is it possible to get a circle from a square or from an equilateral triangle this way?

(d) Is it possible that this function is universal for all triangles, or for all rectangles, or for all polygons?

 No Solution Yet Submitted by Art M No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Pick a side | Comment 2 of 20 |

If r<1/2, then one interesting observation is part of each original side remains, although that part goes to 0 as the number of cuts goes to infinity. So r=1/2 and r=0 are really just special cases -- the other cuts double the number of corners.

As far as going to a circle, it seems after each cycle, the angles would stay the same if r=1/4 for squares and r=1/3 for triangles.

It seems like if the curvature stays constant in the limit (ie the angles stay the same) then the limit could be a circle, even though the sides don't converge at the same rate. (ie 1/3k = 1/k = 0 as x->infinity)

 Posted by Gamer on 2007-02-27 22:24:32

 Search: Search body:
Forums (0)