Difference between revisions of "1959 IMO Problems/Problem 2"
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== Problem == | == Problem == | ||
− | For what real values of <math> | + | For what real values of <math>x</math> is |
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− | given (a) <math>A=\sqrt{2}</math>, (b) <math> | + | given (a) <math>A=\sqrt{2}</math>, (b) <math>A=1</math>, (c) <math>A=2</math>, we only non-negative real numbers are admitted for square roots? |
== Solution == | == Solution == | ||
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− | <math> | + | <math>A^2 = 2(x+|x-1|)</math> |
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− | If <math> | + | If <math>x \le 1</math>, then we must clearly have <math>A^2 =2</math>. Otherwise, we have |
<center> | <center> | ||
<math>x = \frac{A^2 + 2}{4} > 1,</math> | <math>x = \frac{A^2 + 2}{4} > 1,</math> | ||
− | <math> | + | <math>A^2 > 2 </math> |
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− | Hence for (a) the solution is <math> x \in \left[ \frac{1}{2}, 1 \right]</math>, for (b) there is no solution, since we must have <math> | + | Hence for (a) the solution is <math> x \in \left[ \frac{1}{2}, 1 \right]</math>, for (b) there is no solution, since we must have <math>A^2 \ge 2</math>, and for (c), the only solution is <math> x=\frac{3}{2}</math>. Q.E.D. |
{{Alternate solutions}} | {{Alternate solutions}} | ||
− | == | + | {{IMO box|year=1959|num-b=1|num-a=3}} |
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[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] |
Revision as of 20:21, 25 October 2007
Problem
For what real values of is
given (a) , (b)
, (c)
, we only non-negative real numbers are admitted for square roots?
Solution
We note that the square roots imply that . We now square both sides and simplify to obtain
If , then we must clearly have
. Otherwise, we have
Hence for (a) the solution is , for (b) there is no solution, since we must have
, and for (c), the only solution is
. Q.E.D.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
1959 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |