Difference between revisions of "1959 IMO Problems/Problem 2"
(→Solution) |
(→Solution) |
||
| Line 11: | Line 11: | ||
== Solution == | == Solution == | ||
| − | The square roots imply that <cmath>x\ge \frac{1}{2}</cmath> Square both sides and simplify to obtain | + | The square roots imply that <cmath>x\ge \frac{1}{2}</cmath> Square both sides and simplify to obtain <cmath>A^2 = 2(x+|x-1|)</cmath> |
| − | |||
| − | < | ||
| − | |||
| − | |||
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 | ||
Revision as of 13:09, 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
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 | ||
