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

Revision as of 15:04, 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: $A^2 = \Big( x + \sqrt{2x - 1}\Big) + 2 \sqrt{x + \sqrt{2x - 1}} \sqrt{x - \sqrt{2x - 1}} + \Big( x - \sqrt{2x - 1}\Big)$

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)

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

@@ Line 11: / Line 11: @@
 == 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
+Firstly, the square roots imply that a valid domain for x  is <math>x\ge \frac{1}{2}</math>.
-<center>
+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>
-<math>A^2 = 2(x+|x-1|)</math>
-</center>
-If <math>x \le 1</math>, then we must clearly have <math>A^2 =2</math>.  Otherwise, we have
+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>
-<center>
+Multiply the middle terms, and use <math>(a + b)(a - b) = a^2 - b^2</math> to get:
-<math>x = \frac{A^2 + 2}{4} > 1,</math>
+<cmath>A^2 = 2x + 2 \sqrt{x^2 - 2x + 1}</cmath>
-<math>A^2 > 2 </math>
+Since the term inside the square root is a perfect square, and by factoring 2 out, we get
-</center>
+<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>
-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.
+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
+<cmath>x = \frac{A^2 + 2}{4} > 1</cmath>
+which simplifies to
+<cmath>A^2 > 2 </cmath>
+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)
 {{Alternate solutions}}
 == 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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Difference between revisions of "1959 IMO Problems/Problem 2"

Revision as of 15:04, 15 December 2019

Problem

Solution

See Also