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"

(Undo revision 112909 by Flamewavelight (talk))
(Tag: Undo)
(→‎Solution)
Line 13: Line 13:
 
Firstly, the square roots imply that a valid domain for x  is <math>x\ge \frac{1}{2}</math>.
 
Firstly, the square roots imply that a valid domain for x  is <math>x\ge \frac{1}{2}</math>.
  
−
Square both sides of the given equation:
+
Square both sides of the given equation: <cmath> \Big( x + \sqrt{2x - 1}\Big) + 2 \sqrt{x + \sqrt{2x - 1}}  \sqrt{x - \sqrt{2x - 1}} +  \Big( x - \sqrt{2x - 1}\Big) = A^2</cmath>
−
<cmath> \Big( x + \sqrt{2x - 1}\Big) + 2 \sqrt{x + \sqrt{2x - 1}}  \sqrt{x - \sqrt{2x - 1}} +  \Big( x - \sqrt{2x - 1}\Big) = A^2</cmath>
 
  
−
Add the first and the last terms to get
+
Add the first and the last terms to get <cmath>2x + 2 \sqrt{x + \sqrt{2x - 1}}  \sqrt{x - \sqrt{2x - 1}} = A^2</cmath>
−
<cmath>2x + 2 \sqrt{x + \sqrt{2x - 1}}  \sqrt{x - \sqrt{2x - 1}} = A^2</cmath>
 
  
−
Multiply the middle terms, and use <math>(a + b)(a - b) = a^2 - b^2</math> to get:
+
Multiply the middle terms, and use <math>(a + b)(a - b) = a^2 - b^2</math> to get: <cmath>2x + 2 \sqrt{x^2 - 2x + 1} = A^2</cmath>
−
<math>2x + 2 \sqrt{x^2 - 2x + 1} = A^2<cmath>
 
  
 
Since the term inside the square root is a perfect square, and by factoring 2 out, we get
 
Since the term inside the square root is a perfect square, and by factoring 2 out, we get
−
</cmath>2(x + \sqrt{(x-1)^2}) = A^2<cmath>
+
<cmath>2(x + \sqrt{(x-1)^2}) = A^2</cmath>
−
Use the property that </cmath>\sqrt{x^2}=x<cmath> to get
+
Use the property that <cmath>\sqrt{x^2}=x</cmath> to get
−
</cmath>A^2 = 2(x+|x-1|)</math><math>
+
<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
  
 
<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 (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}$.  Q.E.D.
+
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.
  
 
~flamewavelight (Expanded)
 
~flamewavelight (Expanded)

Revision as of 14:47, 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

Firstly, the square roots imply that a valid domain for x is $x\ge \frac{1}{2}$.

Square both sides of the given equation: \[\Big( x + \sqrt{2x - 1}\Big) + 2 \sqrt{x + \sqrt{2x - 1}} \sqrt{x - \sqrt{2x - 1}} + \Big( x - \sqrt{2x - 1}\Big) = A^2\]

Add the first and the last terms to get \[2x + 2 \sqrt{x + \sqrt{2x - 1}} \sqrt{x - \sqrt{2x - 1}} = A^2\]

Multiply the middle terms, and use $(a + b)(a - b) = a^2 - b^2$ to get: \[2x + 2 \sqrt{x^2 - 2x + 1} = A^2\]

Since the term inside the square root is a perfect square, and by factoring 2 out, we get \[2(x + \sqrt{(x-1)^2}) = A^2\] Use the property that \[\sqrt{x^2}=x\] to get \[A^2 = 2(x+|x-1|)\]

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.

~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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1959_IMO_Problems/Problem_2&oldid=112911"