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

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~phoenixfire (edited)
 
~phoenixfire (edited)
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An alternate solution: Note that the equation can be rewritten to
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<cmath>\sqrt{(\sqrt{2x-1}+1)^2} + \sqrt{(\sqrt{2x-1}-1)^2}=A\sqrt{2}</cmath>
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i.e., <math>\sqrt{2x-1}+1 + |\sqrt{2x-1}-1|=A\sqrt{2}</math>.
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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=3/4</math> but <math>x\ge 1</math>, thus no solution; for (c) we have <math>x=\frac{3}{2}</math>.
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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>.
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In summary, any <math>x \in \left[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>.
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{{Alternate solutions}}
 
{{Alternate solutions}}

Revision as of 18:33, 6 September 2021

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 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)


An alternate solution: 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 $x=3/4$ but $x\ge 1$, thus no solution; for (c) we have $x=\frac{3}{2}$.

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 $x \in \left[1/2, 1\right]$ 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
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