Olympiad Maths Prep

Library / /6 of 9

, 2010

Algebra Difficulty 6.0 AIME, harder Prove it Ukraine

n5n \ge 5 real numbers are written in a row. It turns out, that sum of any three consecutive numbers is positive and sum of any five consecutive numbers is negative. Find the largest nn for which it is possible?

Answer: n=6n = 6.

Solution

We first construct the example for n=6n = 6: 3,5,3,3,5,33, -5, 3, 3, -5, 3.

Suppose that there exist n7n \ge 7 real numbers, that satisfy the conditions of the problem and choose any 5 consecutive numbers a,b,c,d,ea, b, c, d, e. Using the conditions we get: a+b+c>0a+b+c > 0, c+d+e>0(a+b+c+d+e)+c>0c+d+e > 0 \Rightarrow (a+b+c+d+e) + c > 0 and a+b+c+d+e<0a+b+c+d+e < 0. This implies that c>0c > 0, therefore for any 5 consecutive numbers one, which is in the middle, is always positive. From the last observation, it follows that all numbers except last two from both ends are positive.

Let us now consider 6 consecutive numbers: a,b,c,d,e,fa, b, c, d, e, f. Using again our given conditions we get: a+b+c+d+e<0a+b+c+d+e < 0 and (a+b+c)+(d+e+f)>0(a+b+c)+(d+e+f) > 0. Thus, f>0f > 0. By analogy, we have a>0a > 0. This implies that all nn numbers must be positive and we get a contradiction.

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