Difference between revisions of "1959 IMO Problems/Problem 2"
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<cmath>A^2 = 2(x+|x-1|)</cmath> | <cmath>A^2 = 2(x+|x-1|)</cmath> | ||
− | Case I: If <math>x \le 1</math>, then <math>|x-1| = 1 - x</math>, and the equation reduces to <math>A^2 =2</math>. | + | Case I: If <math>x \le 1</math>, then <math>|x-1| = 1 - x</math>, and the equation reduces to <math>A^2 = 2</math>. This is precisely part (a) of the question, for which the valid interval is now <math>x \in \left[ \frac{1}{2}, 1 \right]</math> |
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+ | Case II: If <math>x > 1</math>, we have | ||
<cmath>x = \frac{A^2 + 2}{4} > 1,</cmath> | <cmath>x = \frac{A^2 + 2}{4} > 1,</cmath> | ||
<cmath>A^2 > 2 </cmath> | <cmath>A^2 > 2 </cmath> | ||
− | Hence for | + | Hence for , 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. |
~flamewavelight (Expanded) | ~flamewavelight (Expanded) |
Revision as of 13:58, 15 December 2019
Problem
For what real values of is
given (a) , (b) , (c) , where only non-negative real numbers are admitted for square roots?
Solution
Firstly, the square roots imply that a valid domain for x is .
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 , we have
Hence for , for (b) there is no solution, since we must have , and for (c), the only solution is . Q.E.D.
~flamewavelight (Expanded)
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 |