Difference between revisions of "1959 IMO Problems/Problem 2"
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If <math>x \le 1</math>, then we must clearly have <math>A^2 =2</math>. Otherwise, we have | If <math>x \le 1</math>, then we must clearly have <math>A^2 =2</math>. Otherwise, we have | ||
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<cmath>x = \frac{A^2 + 2}{4} > 1,</cmath> | <cmath>x = \frac{A^2 + 2}{4} > 1,</cmath> | ||
Revision as of 13:10, 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
The square roots imply that ![\[x\ge \frac{1}{2}\]](http://latex.artofproblemsolving.com/8/8/c/88c75d86b11405939a81621c08d1b64e0c1364ad.png)
Square both sides and simplify to obtain ![\[A^2 = 2(x+|x-1|)\]](http://latex.artofproblemsolving.com/8/0/a/80aa38fd1f8fd60f0b24ee1d2cf5cb4486b5ab49.png)
If 
, then we must clearly have 
. Otherwise, we have
![\[x = \frac{A^2 + 2}{4} > 1,\]](http://latex.artofproblemsolving.com/6/3/e/63e3fa17646a88fd17dacf722e355f4f04391608.png)
![\[A^2 > 2\]](http://latex.artofproblemsolving.com/2/2/a/22ab9dfef568f64933f1d1b717b53e0b597fb233.png)
Hence for (a) the solution is ![$x \in \left[ \frac{1}{2}, 1 \right]$](http://latex.artofproblemsolving.com/c/c/b/ccb86723dfdd4a38b0ae3a8ec3ec4815cf6c5ade.png)
, 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.
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 | ||
