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Cutting Corners (Posted on 2007-02-27) Difficulty: 5 of 5
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()=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    
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Summary and a bonus question | Comment 19 of 20 |
 All four questions have been answered for this problem.

The main credit goes to Brian Smith. Several people have also contributed to the solution. Charlie's numerical calculations were valuable for verifying various assumptions.

The function F(r) for any polygon is:

F(r)=1-(1-k)*(4r^2)/(4r^2-2r+1) for r=[0,1/2)
F(1/2)=0

where k is the areal ratio of the midpoint polygon to the initial polygon.

Answers:

(a) F(r) for k=1/4

(b) F(r) for k=1/2

(c) Not possible

(I have checked for regular triangle)

(d) The function is the same for all triangles, and for all quadrilaterals, but different for different n-gons.

An additional question:

What function y(r,x) describes the final figure for the square with vertices (0,0)(0,2)(2,2)(2,0)?

Edited on March 4, 2007, 4:03 pm
  Posted by Art M on 2007-03-04 15:58:00

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