Difference between revisions of "1967 IMO Problems/Problem 5"

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Take |a1| >= |a2| >= ... >= |a8|. Suppose that |a1|, ... , |ar| are all equal and greater than |ar+1|. Then for sufficiently large n, we can ensure that |as|n < 1/8 |a1|n for s > r, and hence the sum of |as|n for all s > r is less than |a1|n. Hence r must be even with half of a1, ... , ar positive and half negative.
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==Problem==
  
If that does not exhaust the ai, then in a similar way there must be an even number of ai with the next largest value of |ai|, with half positive and half negative, and so on. Thus we find that cn = 0 for all odd n.
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Let <math>a_1,\ldots,a_8</math> be reals, not all equal to zero. Let <cmath>c_n = \sum^8_{k=1} a^n_k</cmath> for <math>n=1,2,3,\ldots</math>. Given that among the numbers of the sequence <math>(c_n)</math>, there are infinitely many equal to zero, determine all the values of <math>n</math> for which <math>c_n = 0.</math>
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==Solution==
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<math>c_n</math> must be zero for all odd <math>n</math>.
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Proof:
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WLOG suppose that <math>a_1 \geq a_2 \geq ... \geq a_8</math>. If <math>a_1+a_8 > 0</math> then for sufficiently high odd <math>n</math>, <math>c_n</math> will be dominated by <math>a_1</math> alone i.e. it will always be positive. Similarly if <math>a_1+a_8 < 0</math>; hence <math>a_1=-a_8</math>. Now for odd <math>n</math> these terms cancel, so we can repeat for the remaining values. Now all the terms cancel for all odd <math>n</math>. Since some <math>a_i</math> is nonzero <math>c_n > 0</math> for even <math>n</math>.
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The above solution was written by Fiachra and can be found here: [https://aops.com/community/p138611]
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----
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<math>\textbf{Note:}\hspace{4000pt}</math> Problem 5 on this (https://artofproblemsolving.com/wiki/index.php/1967_IMO_Problems) page is equivalent to this since the only difference is that they are phrased differently.
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== See Also ==
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{{IMO box|year=1967|num-b=4|num-a=6}}

Latest revision as of 19:23, 10 November 2024

Problem

Let $a_1,\ldots,a_8$ be reals, not all equal to zero. Let \[c_n = \sum^8_{k=1} a^n_k\] for $n=1,2,3,\ldots$. Given that among the numbers of the sequence $(c_n)$, there are infinitely many equal to zero, determine all the values of $n$ for which $c_n = 0.$

Solution

$c_n$ must be zero for all odd $n$.

Proof: WLOG suppose that $a_1 \geq a_2 \geq ... \geq a_8$. If $a_1+a_8 > 0$ then for sufficiently high odd $n$, $c_n$ will be dominated by $a_1$ alone i.e. it will always be positive. Similarly if $a_1+a_8 < 0$; hence $a_1=-a_8$. Now for odd $n$ these terms cancel, so we can repeat for the remaining values. Now all the terms cancel for all odd $n$. Since some $a_i$ is nonzero $c_n > 0$ for even $n$.

The above solution was written by Fiachra and can be found here: [1]



$\textbf{Note:}\hspace{4000pt}$ Problem 5 on this (https://artofproblemsolving.com/wiki/index.php/1967_IMO_Problems) page is equivalent to this since the only difference is that they are phrased differently.

See Also

1967 IMO (Problems) • Resources
Preceded by
Problem 4
1 2 3 4 5 6 Followed by
Problem 6
All IMO Problems and Solutions