Solutions to the 73rd William Lowell Putnam Mathematical Competition Saturday, December 1, 2012
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A–1 Without loss of generality, assume d1 ≤ d2 ≤ ·· · ≤ d12. If d2 i+2 < d 2 i + d 2 i+1 for some i ≤ 10, then di,di+1,di+2 are the side lengths of an acute triangle, since in this case d2 i < d 2 i+1+d 2 i+2 and d 2 i+1 < d 2 i +d 2 i+2 as well. Thus we may assume d2 i+2 ≥ d2 i + d2 i+1 for all i. But then by induction, d2 i ≥ Fid 1 for all i, where Fi is the i-th Fibonacci number (with F1 = F2 = 1): i = 1 is clear, i = 2 follows from d2 ≥ d1, and the induction step follows from the assumed inequality. Setting i = 12 now gives d2 12 ≥ 144d2 1 , contradicting d1 > 1 and d12 < 12. Remark. A materially equivalent problem appeared on the 2012 USA Mathematical Olympiad and USA Junior Mathematical Olympiad.
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