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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.


(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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Some Thoughts Links and stuff | Comment 6 of 20 |

A link to a picture from gsp showing a plot of (r,F(r))
[Well not exactly F(r) because I only did 3 iterations, but it's really close.]

A link to the geometers sketchpad file

Since F(r) is defined as the figure's area after infinite steps.  Let's define F1(r) as the figure after 1 step, F2(r) after 2 steps, etc.

For a square I've figured
F0(r) = 1
F1(r) = 1 - 2r^2
F2(r) = 1 - 2r^2 - 4r^3 + 8r^4

It appears the polynomial increases in degree, perhaps doubling in degree after every step.  I'm not sure F3 will continue this pattern.   It seems very hard to find.

The answer to part (a) may just turn out to be a polynomial of infinite degree that doesn't simplify to anything nice.

  Posted by Jer on 2007-02-28 12:40:52
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