1993 USAMO Problems/Problem 1
Problem
For each integer , determine, with proof, which of the two positive real numbers and satisfying is larger.
Solution
Square and rearrange the first equation and also rearrange the second. It is trivial that since clearly cannot equal (Otherwise ). Thus where we substituted in equations (1) and (2) to achieve (5). If , then since , , and are all positive. Adding the two would mean , a contradiction, so . However, when equals or , the first equation becomes meaningless, so we conclude that for each integer , we always have .
See also
1993 USAMO (Problems • Resources) | ||
Preceded by First question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |