1967 IMO Problems/Problem 5
Let be reals, not all equal to zero. Let for . Given that among the numbers of the sequence , there are infinitely many equal to zero, determine all the values of for which
Solution
must be zero for all odd .
Proof: WLOG suppose that . If then for sufficiently high odd , will be dominated by alone i.e. it will always be positive. Similarly if ; hence . Now for odd these terms cancel, so we can repeat for the remaining values. Now all the terms cancel for all odd . Since some is nonzero for even .
The above solution was written by Fiachra and can be found here: [1]
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 |