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"

(→‎Solution)
(→‎Solution 2)
 
(29 intermediate revisions by 7 users not shown)
Line 9: Line 9:
 
given (a) <math>A=\sqrt{2}</math>, (b) <math>A=1</math>, (c) <math>A=2</math>, where only non-negative real numbers are admitted for square roots?
 
given (a) <math>A=\sqrt{2}</math>, (b) <math>A=1</math>, (c) <math>A=2</math>, where only non-negative real numbers are admitted for square roots?
  
−
== Solution ==
+
== Solution 1 ==
  
−
Firstly, the square roots imply that a valid domain for x  is <math>x\ge \frac{1}{2}</math>.
+
The square roots imply that <math>x\ge \frac{1}{2}</math>.
  
−
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>
+
Square both sides of the given equation: <cmath>A^2 = \Big( x + \sqrt{2x - 1}\Big) + 2 \sqrt{x + \sqrt{2x - 1}}  \sqrt{x - \sqrt{2x - 1}} +  \Big( x - \sqrt{2x - 1}\Big) </cmath>
  
−
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>
+
Add the first and the last terms to get:
 +
<cmath>A^2 = 2x + 2 \sqrt{x + \sqrt{2x - 1}}  \sqrt{x - \sqrt{2x - 1}}</cmath>
  
−
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>
+
Multiply the middle terms, and use <math>(a + b)(a - b) = a^2 - b^2</math> to get:
 +
<cmath>A^2 = 2x + 2 \sqrt{x^2 - 2x + 1}</cmath>
  
−
Simplify to obtain <cmath>A^2 = 2(x+|x-1|)</cmath>
+
Since the term inside the square root is a perfect square, and by factoring 2 out, we get
 +
<cmath>A^2 = 2(x + \sqrt{(x-1)^2})</cmath>
 +
Use the property that <math>\sqrt{x^2}=|x|</math> to get
 +
<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
+
'''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>
  
−
<cmath>x = \frac{A^2 + 2}{4} > 1,</cmath>
+
'''Case II:''' If <math>x > 1</math>, then <math>|x-1| = x - 1</math> and we have
 +
<cmath>x = \frac{A^2 + 2}{4} > 1</cmath>
 +
which simplifies to
 
<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}</math>.  Q.E.D.
+
This tells there that there is no solution for (b), since we must have <math>A^2 \ge 2</math>
 +
 
 +
For (c), we have <math>A = 2</math>, which means that <math>A^2 = 4</math>, so the only solution is <math> x=\frac{3}{2}</math>.
  
 
~flamewavelight (Expanded)
 
~flamewavelight (Expanded)
 +
 +
 +
~phoenixfire (edited)
 +
 +
 +
 +
== Solution 2 ==
 +
Note that the equation can be rewritten to
 +
<cmath>\sqrt{(\sqrt{2x-1}+1)^2} + \sqrt{(\sqrt{2x-1}-1)^2}=A\sqrt{2}</cmath>
 +
i.e., <math>\sqrt{2x-1}+1 + |\sqrt{2x-1}-1|=A\sqrt{2}</math>.
 +
 +
'''Case I:''' when <math>2x-1\ge 1</math> (i.e., <math>x\ge 1</math>), the equation becomes <math>2\sqrt{2x-1}=\sqrt{2}A</math>. For (a), we have <math>x=1</math>; for (b) we have <math>x=\frac{3}{4}</math>; for (c) we have <math>x=\frac{3}{2}</math>. Since <math>x\ge 1</math>, (b) <math>x=\frac{3}{4}</math> is not what we want.
 +
 +
'''Case II:''' when <math>0\le 2x-1 <1</math> (i.e., <math>1/2\le x <1</math>), the equation becomes <math>2=\sqrt{2}A</math>, which only works for (a) <math>A=\sqrt{2}</math>.
 +
 +
In summary, any <math>x \in \left[\frac{1}{2}, 1\right]</math> is a solution for (a); there is no solution for (b); there is one solution for (c), which is <math>x=\frac{3}{2}</math>.
 +
  
 
{{Alternate solutions}}
 
{{Alternate solutions}}

Latest revision as of 00:29, 5 November 2024

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 1

The square roots imply that $x\ge \frac{1}{2}$.

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

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

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

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

Case I: If $x \le 1$, then $|x-1| = 1 - x$, and the equation reduces to $A^2 = 2$. This is precisely part (a) of the question, for which the valid interval is now $x \in \left[ \frac{1}{2}, 1 \right]$

Case II: If $x > 1$, then $|x-1| = x - 1$ and we have \[x = \frac{A^2 + 2}{4} > 1\] which simplifies to \[A^2 > 2\]

This tells there that there is no solution for (b), since we must have $A^2 \ge 2$

For (c), we have $A = 2$, which means that $A^2 = 4$, so the only solution is $x=\frac{3}{2}$.

~flamewavelight (Expanded)


~phoenixfire (edited)


Solution 2

Note that the equation can be rewritten to \[\sqrt{(\sqrt{2x-1}+1)^2} + \sqrt{(\sqrt{2x-1}-1)^2}=A\sqrt{2}\] i.e., $\sqrt{2x-1}+1 + |\sqrt{2x-1}-1|=A\sqrt{2}$.

Case I: when $2x-1\ge 1$ (i.e., $x\ge 1$), the equation becomes $2\sqrt{2x-1}=\sqrt{2}A$. For (a), we have $x=1$; for (b) we have $x=\frac{3}{4}$; for (c) we have $x=\frac{3}{2}$. Since $x\ge 1$, (b) $x=\frac{3}{4}$ is not what we want.

Case II: when $0\le 2x-1 <1$ (i.e., $1/2\le x <1$), the equation becomes $2=\sqrt{2}A$, which only works for (a) $A=\sqrt{2}$.

In summary, any $x \in \left[\frac{1}{2}, 1\right]$ is a solution for (a); there is no solution for (b); there is one solution for (c), which is $x=\frac{3}{2}$.


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=231647"