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?
Any triangle is obtainable from any other by an affine transformation. That transformation will preserve lines and the ratio of distances along a given line. They will also preserve relative areas. So the function should be the same for all triangles.
The same would be true for rectangles; that is, the function should be the same for all rectangles. But this argument does not apply to all quadrilaterals, as to map a general quadrilateral to a rectangle could require distorting different portions of the plane by different amounts, changing relative areas. The more so, this rules out "for all polygons".
Also, during this puzzle's stay in the queue, Jer noticed and brought up the fact that, while F(1/2) = 0, as x approaches 1/2, F(x) does not approach zero. Thought experiments will show that as x approaches 1/2, F(x) approaches the relative area of one (rather than infinitely many) iteration of using the midpoints of the original sides as vertices for a new area the same shape (triangle, square, etc.) as the original. In the case of a triangle, as x approaches 1/2, F(x) approaches 1/4. In the case of a square or other rectangle, as x approaches 1/2, F(x) approaches 1/2 (a smaller square at 45degree angle from the original square). And it looks as if, as the number of sides increase, this limiting value gets larger and larger, approaching 1 as the number of sides increases.
The foregoing paragraph is intended to show that, as the function is different for squares from what it is for triangles, it is not the same for all polygons.
So we've determined it's the same for all triangles; it's the same for all rectangles (but different from triangles); it's different for other polygons. We haven't proved one way or the other for generalized quadrilaterals, or generalized ngons for a particular n.

Posted by Charlie
on 20070227 22:25:30 