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?

Before creating the function F(r), first a function f(r), which calculates the amount of area lost, will be created.

Let V be any corner to be cut. Let VA be one half of one side of the corner and let VB be one half of the other side. Let C be on VA according to the cutting algorithm and let D be on VB similarily. The ratio of VC:VA = VD:VB = r:(1/2) = (2r):1. Triangle VCD is the area which is cut off in the first round of cuts.

Let M be the midpoint of CD. CM and MD are each one half of CD and AC and DB are one half of the remainder of the sides they are on. Then triangles ACM and BDM are the vertex areas subject to the next round of cuts and triangle ABM will not be cut.

The ratio of the cut and uncut sections of ACM and BDM are the same as the ratio for VAB. That ratio can be found by the ratio of the areas of VCD and MAB.

If the altitude of VAB is h, then the altitude of VCD is 2rh and the altitude of MAB is (1-2r)h. If the corresponding base of VAB is b, then the corresponding base of VCD is 2rb and the corresponding base of MAB is b. The area of VCD is then hb*(2r^2) and the area of MAB is then hb*(1-2r)/2.

The function which calculates the portion of VAB which is ultimately cut off can be formulated as:

f(r) = (2r^2)/((2r^2)+((1-2r)/2))

f(r) = 4r^2/(4r^2-2r+1)

This function is valid for the interval [0,1/2). Note that r=1/2 is not included. This is because the behavior of the cutting algorithm changes to ultimately removing the entire polygon at r=1/2 instead of just removing a portion of each corner.

f(0) = 0 and f(1/2) = 1 as expected and f(1/4) = 1/3, which agrees with Charlie's numeric simulations.

In addition to areas S1 and S2 defined in the problem, let the area of the "midpoint polygon" of the original polygon be S3.

The area S1-S3 is the area which the cutting function f(r) applies. Then the area of S2 can be calculated as S1-f(r)*(S1-S3). Let the ratio S3/S1 = k. The value of k for all triangles is 1/4, for quadrilaterals k=1/2, but for pentagons k can vary from 1/2 to 3/4, and for hexagons and larger, k can approach 1.

The function F(r) can then be expressed as:

F(r) = 1-(1-k)*(4r^2)/(4r^2-2r+1)

F(r) = (4k*r^2-2r+1)/(4r^2-2r+1)

Midpoint polygons and how to calculate their areas is discussed at

"http://techhouse.brown.edu/~mdp/midpoint/"

Sides r Charlie's data k F(r)

3 .05 0.991758241758228 1/4 361/364 = 0.991758241758242

3 .15 0.914556962053026 1/4 289/316 = 0.914556962025316

3 .25 0.750000000931304 1/4 3/4 = 0.750000000000000

3 .35 0.534810126733177 1/4 169/316 = 0.534810126582278

3 .45 0.332417582417581 1/4 121/364 = 0.332417582417582

4 .05 0.994505494505502 1/2 181/182 = 0.994505494505494

4 .15 0.943037974702018 1/2 149/158 = 0.943037974683544

4 .25 0.833333333954223 1/2 5/6 = 0.833333333333333

4 .35 0.689873417822104 1/2 109/158 = 0.689873417721518

4 .45 0.554945054945063 1/2 101/182 = 0.554945054945054

Charlie's numeric data agrees with this formula for at least nine decimal places!