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?

(In reply to Corner Cutting Function by Brian Smith)

In your post:

"The areas of the four triangles created by the third generation of cuts (at corners R', P, etc) vs the area of the two in the second generation of cuts (corners R, T) share the same ratio as the second and first generations of cuts, which means the area of a single cut decreases geometrically with the ratio (r/2 - r^2).  In the nth generation, there are 2^(n-1) triangles, each with area ((r^2)/2) * (r/2 - r^2)^(n-1)."

But this is not true.  I assume you are starting with a square, or at least a right angled corner.

The triangle from the first generation of cuts is isosceles and right with legs of length r so area (r^2)/2

The two in the second generation are congruent with shorter sides r^2*sqrt(2) and r-2r^2 and a 135 degree angle.  The area of each is (r^3 - 2r^4)/2.

The ratio here is as you say.

The third generation of triangles (with 157.5 degree angles)

are not all congruent because the remaining sides have many different lengths.  The areas of 2 of them are
(r^3-2r^4)sqrt((4r^2-4r+2)(2-sqrt(2)))/4
and the other two
(4r^5-4r^4+r^3)sqrt((2r^2-2r+1)(2-sqrt(2)))/4

This is way too complicated to continue with a general value of r.  The next generation would have four different sizes of triangles with 168.75 degree angles.

 

Posted by Jer on 2007-03-01 13:34:11