: (√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. 45th Putnam 1984 © John Scholes jscholes@kalva.demon.co.uk 7 Jan 2001
45th Putnam 1984 Problem B6 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