45th Putnam 1984

Problem B6

Define a sequence of convex polygons Pn as follows. P0 is an equilateral triangle side 1. Pn+1 is obtained from Pn by cutting off the corners one-third of the way along each side (for example P1 is a regular hexagon side 1/3). Find limn→∞area(Pn).



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 P0. 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.



45th Putnam 1984

© John Scholes
7 Jan 2001