Define a sequence of convex polygons P_{n} as follows. P_{0} is an equilateral triangle side 1. P_{n+1} is obtained from P_{n} by cutting off the corners one-third of the way along each side (for example P_{1} is a regular hexagon side 1/3). Find lim_{n→∞}area(P_{n}).

**Solution**

Answer: (√3)/7.

n = 0 and n = 1 are untypical cases. Later polygons have unequal sides and sides that do not touch the circle inscribed in P_{0}. The key is to look at a slightly more general case. Suppose that just before the nth series of cuts we have two adjacent sides, represented by vectors **a**, **b**. Then the new side formed when we cut off the corner is (**a**+**b**)/3, so the three new sides are **a**/3, (**a**+**b**)/3, **b**/3, and the area of corner removed is (**a x b**)/2. At step n+1 we remove two corners, with areas ( **a**/3 **x** (**a**+**b**)/3 + (**a**+**b**)/3 **x** **b**/3 )/2 = (**a x b**)/9 = 2/9 x area removed at step n.

The area removed in going from the triangle to the hexagon is 1/3 area triangle. So the total area removed is 1/3 (1 + 2/9 + (2/9)^{2} + ... ) = 3/7 area triangle. Hence the area remaining is 4/7 (√3)/4 = (√3)/7.

© John Scholes

jscholes@kalva.demon.co.uk

7 Jan 2001