1984 USAMO Problems/Problem 2
Problem
The geometric mean of any set of non-negative numbers is the -th root of their product.
For which positive integers is there a finite set of distinct positive integers such that the geometric mean of any subset of is an integer?
Is there an infinite set of distinct positive integers such that the geometric mean of any finite subset of is an integer?
Solution
a) We claim that for any numbers , , ... , will satisfy the condition, which holds for any number .
Since , we can separate each geometric mean into the product of parts, where each part is the th root of each member of the subset and the subset has members.
Assume our subset has members. Then, we know that the th root of each of these members is an integer (namely ), because and thus . Since each root is an integer, the geometric mean will also be an integer.
b) If we define as an arbitrarily large number, and and as numbers in set , we know that is irrational for large enough , meaning that it cannot be expressed as the fraction of two integers. However, both the geometric mean of the set of and other arbitrary numbers in and the set of and the same other numbers are integers, so since the other numbers cancel out, the geometric means divided, or , must be rational. This is a contradiction, so no such infinite is possible.
-aops111 (first solution dont bully me)
See Also
1984 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
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