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

re: Corner Cutting Function by Jer)

Yeah, I see where I went wrong and have a new solution which yielded a different result.

Take any corner BAC, with A as the vertex to be cut. Let D be the midpoint of AB and let E be the midpoint of AC. All of the cuts in the process will be inside triangle ADE. The goal will be to construct a function f(r), defined on [0,1/2), which calculates the fraction of triangle ADE which will eventually be cut off.

Choose F on AB such that AF/AB = r and choose G on AC such that AG/AC = r. Triangle AFG is similar to triangle ADE with scale factor of 2r:1. Also, DE is parallel to FG. Triangle ADE is part of the cut off area and trapezoid DEGF is still indeterminate. Calculating the ratio of cut areas to uncut areas in DEGF will be key in creating f(r).

Let Z be the midpoint of FG, let H be the midpoint of DF, and let I be the midpoint of EG. Any future cuts occurs inside triangles GIZ and FHZ, and region DEIZH is an uncut area.

Choose J on FD such that FJ/FD = r, choose L on FG such that FL/FG = r, choose K on GE such that GK/GE = r, and choose M on GF such that GM/GF = r. Triangles GKM and FJL are cut areas and are similar to GIZ and FHZ respectively with scale factor of 2r:1. KM is parallel to IZ and JL is parallel to HZ.

Trapezoids IKMZ and HJLZ are indeterminate areas and will be split into cut and uncut areas in the same ratio as DEGF. DEGF is ultimately cut into two regions with a fractal cut, so the ratio of the areas will be the same as the ratio of (area GKM + area FJL) : (area DEIZH).

The bases of trapezoid DEGF are DE and FG and DE:FG = 2r:1 from similarity of ADE and AFG. Let DE = 2r*b and FG = b. Let the height be h. The area of DEGF is then h*b*(2r+1)/2.

H and I are midpoints of DF and GE. which means HI is parallel to DE and FG and is located exactly halfway inbetween. The length of HI is then b*(2r+1)/2 and the height of trapezoid DEIH is h/2, and the height of triangle HIZ (which is the same as the height of trapezoid FGIH) is also h/2. The area of DEIH is h*b*(3+2r)/8 and the area of HIZ is h*b*(1+2r)/8.

The area of DEIZH is h*b*(1+r)/2, and the sum of areas FHZ and GIZ is h*b*(r/2). The ratio of (area GKM + area FJL):(area FHZ + area GIZ) is (2r)^2:1 = 4r^2:1, which makes the sum of areas GKM and FJL equal to h*b*2r^3. This makes the ratio (area GKM + area FJL) : (area DEIZH) equal to (h*b*2r^3)/(h*b*(1+r)/2) = (4r^3)/(r+1).

Let x be the area of ADE. That would make the area of AFG = 4r^2*x and the area of DEGF = (1-4r^2)*x. Using the ratio mentioned earlier, the area removed from DEGF equals (1-4r^2)*x*(4r^3)/(4r^3+(r+1)) = x*(4r^3)*(1-4r^2)/(4r^3+r+1).

f(r) can then be constructed as:

f(r) = [x*4r^2+x*(4r^3)*(1-4r^2)/(4r^3+r+1))]/x

f(r) = (8r^3+4r^2)/(4r^3+r+1).

The limit of f(r) as r->0 is 0, as expected, but the limit as r->1/2 is 1, which is equivalent to simply clipping all the corners once. This illustrates a change in behavoir of r<1/2 and r>=1/2. Even when r=.499999999, the midpoints of the sides stay in the final figure, but when r>=1/2, those midpoints are removed after the first or second step of cutting.