Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1959 IMO Problems/Problem 2"

m
(→‎Solution)
Line 11: Line 11:
 
== Solution ==
 
== Solution ==
  
−
We note that the square roots imply that <math>x\ge \frac{1}{2}</math>. We now 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
  
 
<center>
 
<center>
Line 28: Line 28:
  
 
{{Alternate solutions}}
 
{{Alternate solutions}}
−
 
  
 
== See Also ==
 
== See Also ==

Revision as of 13:58, 15 December 2019

Problem

For what real values of $x$ is

$\sqrt{x+\sqrt{2x-1}} + \sqrt{x-\sqrt{2x-1}} = A,$

given (a) $A=\sqrt{2}$, (b) $A=1$, (c) $A=2$, where only non-negative real numbers are admitted for square roots?

Solution

The square roots imply that \[x\ge \frac{1}{2}\] Square both sides and simplify to obtain

$A^2 = 2(x+|x-1|)$

If $x \le 1$, then we must clearly have $A^2 =2$. Otherwise, we have

$x = \frac{A^2 + 2}{4} > 1,$

$A^2 > 2$

Hence for (a) the solution is $x \in \left[ \frac{1}{2}, 1 \right]$, for (b) there is no solution, since we must have $A^2 \ge 2$, and for (c), the only solution is $x=\frac{3}{2}$. 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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1959_IMO_Problems/Problem_2&oldid=112884"