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

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.

@@ 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?
-== 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>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>
@@ Line 23: / Line 23: @@
 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
+Use the property that <math>\sqrt{x^2}=|x|</math> to get
 <cmath>A^2 = 2(x+|x-1|)</cmath>
-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>
+'''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>
-Case II: If <math>x > 1</math>, then <math>|x-1| = x - 1</math> and we have
+'''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>
-Hence for , 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)
+~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}}

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

Latest revision as of 00:29, 5 November 2024

Contents

Problem

Solution 1

Solution 2

See Also