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?

Since I don't know the answers I can submit my observations too. So far we know that:

- F(r) is discontinuous at r=1/2: F(1/2)=0, but for triangles F(r->1/2)->1/4; for rectangles F(r->1/2)->1/2, so F(r) is different for triangles and rectangles. As r changes from 0 to 1/2, the original polygon is smoothly transformed into another polygon, which is obtained by connecting the side midpoints of the original polygon.

- Charlie's observation about affine transformation is interesting and seems to be right. If so, then the function is the same for all triangles and for all rectangles.

- My observation is that all midpoints of all line segments formed during the cutting process will be on the final figure. In fact, the final figure IS the collection of these midpoints.
So for square, in order to get a circle, the midpoint of MN should be at distance 1/2 from the center (assuming unit square). From this condition r= 1-1/sqrt(2). I didn't have time to check the next step (whether the midpoint of the next segment is 1/2 unit from the center). If it is not then it is impossible to get a circle from square. If it is, then there is a hope. The same can be done with triangles.

Edited on February 28, 2007, 2:00 am

Posted by Art M on 2007-02-28 01:50:49