1959 IMO Problems/Problem 2
Contents
Problem
For what real values of is
given (a) , (b) , (c) , where only non-negative real numbers are admitted for square roots?
Solution 1
The square roots imply that .
Square both sides of the given equation:
Add the first and the last terms to get:
Multiply the middle terms, and use to get:
Since the term inside the square root is a perfect square, and by factoring 2 out, we get Use the property that to get
Case I: If , then , and the equation reduces to . This is precisely part (a) of the question, for which the valid interval is now
Case II: If , then and we have which simplifies to
This tells there that there is no solution for (b), since we must have
For (c), we have , which means that , so the only solution is .
~flamewavelight (Expanded)
~phoenixfire (edited)
Solution 2
Note that the equation can be rewritten to i.e., .
Case I: when (i.e., ), the equation becomes . For (a), we have ; for (b) we have but , thus no solution; for (c) we have .
Case II: when (i.e., ), the equation becomes , which only works for (a) .
In summary, any is a solution for (a); there is no solution for (b); there is one solution for (c), which is .
~zhaoxiayu
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1959 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |