1959 IMO Problems/Problem 2

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 $A^2 > 2$

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 $x=\frac{3}{2}$ .

Solution 3

For real $x$ , $x \ge \frac{1}{2}$ . Also,

$\sqrt{x+\sqrt{2x-1}} = \frac{1}{\sqrt{2}} |\sqrt{2x-1}+1|$ $= \frac{1}{\sqrt{2}} (\sqrt{2x-1}+1),$ $\sqrt{x-\sqrt{2x-1}} = \frac{1}{\sqrt{2}} |\sqrt{2x-1}-1|$

It is easy to know:

(1) When $x \in \left[\frac{1}{2}, 1\right]$ ,

$|\sqrt{2x-1}-1| = 1-\sqrt{2x-1}.$

(2) When $x \ge 1,$

$|\sqrt{2x-1}-1| = \sqrt{2x-1}-1.$

Let's define

$A = \sqrt{x+\sqrt{2x-1}} + \sqrt{x-\sqrt{2x-1}} =$ $= \frac{1}{\sqrt{2}} (\sqrt{2x-1}+1+|\sqrt{2x-1}-1|).$

Hence, for $x \in \left[\frac{1}{2}, 1\right]$ , $A = \sqrt{2};$ $x \ge 1,$ $A \ge \sqrt{2}.$

To sum up, the solution for (a) $x \in \left[\frac{1}{2}, 1\right]$ ; (b) No solution; (c) $x=\frac{3}{2}$ .

~EUCLID_2003

Solution 4

We first Let $M = \sqrt{x+\sqrt{2x-1}}$ , and $N = \sqrt{x-\sqrt{2x-1}}$

As given by the question, we have $M + N = A$ .

Then, the square root expression in $M$ and $N$ motivates us to square $M$ and $N$ , we have: $M^2 = x+\sqrt{2x-1}$ $N^2 = x-\sqrt{2x-1}$

Manipulating with the algebraic properties, we have:

$M^2-N^2=2\sqrt{2x-1}$ $M^2+N^2=2x$ $M \times N = |x-1|$

Now, we have 3 variables, and 4 equations, and the question is almost solved:

$\begin{cases} M + N = A \\ M^2-N^2=2\sqrt{2x-1} \\ M^2+N^2=2x\\ M \times N = |x-1| \end{cases}$

Since $M^2-N^2 = (M+N) \times (M-N)$ , we have $M-N = \frac{M^2-N^2}{M+N} = \frac{2\sqrt{2x-1}}{A}$

We know that $(M+N)^2+(M-N)^2 = 2(M^2+N^2)$ , so $A^2 + (\frac{2\sqrt{2x-1}}{A})^2 = 2 \times 2x \Rightarrow A^2+\frac{8x-4}{A^2} = 4x$

Also, we have certain restrictions on the range of $x$ :

1. Since $\sqrt{2x-1}$ is defined: $2x-1 \geq 0 \Rightarrow x \geq \frac{1}{2}$

2. Since $\sqrt{x-\sqrt{2x-1}}$ is defined: $x-\sqrt{2x-1} \geq 0 \Rightarrow x \geq \sqrt{2x-1} \Rightarrow x^2 \geq 2x-1$ (This part does not give any restriction as $x^2-2x+1 = (x-1)^2$ is always greater than or equal to 0.

3. By using RMS-AM inequality on the original equation, we have: $\sqrt{x} \geq (\sqrt{x+\sqrt{2x-1}} + \sqrt{x-\sqrt{2x-1}})/2 = A/2 \Rightarrow x \geq A^2/4$

4. By using AM-GM inequality on the original equation, we have: $\sqrt(|x-1|) \leq A/2 \Rightarrow |x-1| \leq A^2/4$

Now, we can solve for x, given three values of A:

1. When $A = \sqrt{2}$ , $2+\frac{8x-4}{2} = 4x$ , it seems that $x$ can be any value, but we need to verify with $M \times N = |x-1|$ , since $M \times N = \frac{(M+N)^2-(M^2+N^2)}{2} = \frac{2-2x}{2} \Rightarrow \frac{2-2x}{2} = |x-1| \Rightarrow 1-x=|x-1|$ , meaning that $x \leq 1$ , from the restrictions, we also have $x \geq \frac{1}{2}$ , so $x \in \left[1/2, 3/2 \right]$ .

2. When $A = 1$ , $1+\frac{8x-4}{1} = 4x \Rightarrow x = \frac{3}{4}$ , still, testing with $M \times N = |x-1|$ , we have $\frac{1-2x}{2} = |x-1|$ , there is no solution.

3. When $A = 2$ , $4+\frac{8x-4}{4} = 4x \Rightarrow x = \frac{3}{2}$ , verifying with $M \times N = |x-1|$ , we have $\frac{4-2x}{2} = |x-1|$ , we can see that indeed $x = \frac{3}{2}$ is a solution.

Therefore, our answer is:

\begin{cases} A = \sqrt{2}, x \in \left[1/2, 1 \right]\\ A = 1, \text{No solution}\\ A = 2, x = \frac{3}{2} \end{cases}

Comments: the question is elegant in its algebraic properties, in fact, there are so many ways to utilize the properties of $M^2+N^2$ , $M+N$ , $M-N$ , $M^2-N^2$ , $M \times N$ . This is an especially weird problem, there are 3 variables and 4 equations; when $A = \sqrt{2}$ , x seems to take any value, surely the core equation is $M \times N = |x-1|$ , but why not the other 3? Can you solve the problem using the other 3 equations along? Those are some of the questions to consider. For the restrictions, surely some of the restrictions are inherent. My method seemed chaotic, but is probably what contestants with some experience will do in solving this. This solution is in fact middle school to high school level.

~IDKHowtoaddsolution

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

1959 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
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