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?

    If we join the midpoints of the sides of a simple n-gon, we will get another n-gon, let's call it internal. If the areas are S1 and S3, respectively, S3/S1=1/4 for triangles, S3/S1=1/2 for quadrilaterals. It will be interesting to see if S3/S1 can be generalized for n-gons.

Thus the n-gon consists of the internal n-gon and n triangles with areas T1, T2...Tn. For the purpose of our problem each triangle is being cut independently from others, but only at one corner. So the problem is simplified if we can find a new function F0(r), which is similar by definition to F(r) with only one corner being cut. If F0(r) is same for all triangles, which is probably, the problem can be generalized for all polygons. Also F0(r) is a continuous function.


S1=S3+T1+T2+...+Tn
S2=S3+F0(r)(T1+T2+...+Tn)=S3+F0(r)(S1-S3)

F(r)=S2/S1=S3/S1+F0(r)(1-S3/S1)

For triangles:

F(r)=1/4+3/4*F0(r)

For quadrilaterals:

F(r)=1/2+1/2*F0(r)

Posted by Art M on 2007-02-28 16:10:06