IMO 1959 Problem 2

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Verdicts: PASS + FAIL
Solve time: 15m36s

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?

Problem Understanding

We are asked to determine all real numbers $x$ such that

$$\sqrt{x+\sqrt{2x-1}} + \sqrt{x-\sqrt{2x-1}} = A$$

for three specific values of $A$: $\sqrt{2}$, $1$, and $2$, under the restriction that all square roots represent non-negative real numbers. The inner radical $\sqrt{2x-1}$ imposes the domain constraint $2x-1 \ge 0$, which gives $x \ge \frac12$. The outer radicals require $x \pm \sqrt{2x-1} \ge 0$; each of these must be non-negative individually, not just their product, to avoid taking the square root of a negative number.

This problem requires both finding candidate solutions algebraically and rigorously checking that these candidates lie in the domain of the original equation.

Key Observations

First, the domain is determined by the inner radical: $2x-1 \ge 0 \implies x \ge \frac12$. Then, for the outer radicals, define

$$u = \sqrt{x+\sqrt{2x-1}}, \quad v = \sqrt{x-\sqrt{2x-1}}.$$

By construction, $u \ge 0$ and $v \ge 0$ provided that $x-\sqrt{2x-1} \ge 0$. This second inequality is equivalent to

$$x \ge \sqrt{2x-1} \implies x^2 \ge 2x-1 \implies (x-1)^2 \ge 0,$$

which is always true for all real $x$. Therefore, $x-\sqrt{2x-1} \ge 0$ holds for $x \ge \frac12$. Consequently, the full domain of the equation is $x \ge \frac12$. The substitution $u$ and $v$ allows us to write

$$u^2 + v^2 = (x+\sqrt{2x-1}) + (x-\sqrt{2x-1}) = 2x$$

and

$$uv = \sqrt{(x+\sqrt{2x-1})(x-\sqrt{2x-1})} = \sqrt{x^2-(2x-1)} = \sqrt{(x-1)^2} = |x-1|.$$

Squaring both sides of the original equation gives

$$(u+v)^2 = A^2 \implies u^2 + 2uv + v^2 = A^2 \implies 2x + 2|x-1| = A^2.$$

This reduces the problem to solving the absolute value equation

$$|x-1| = \frac{A^2 - 2x}{2}$$

while respecting the domain $x \ge \frac12$.

Solution

The absolute value equation requires a case-by-case analysis.

Case 1: $x \ge 1$. Then $|x-1| = x-1$, giving

$$x-1 = \frac{A^2-2x}{2} \implies 2x-2 = A^2 - 2x \implies 4x = A^2 + 2 \implies x = \frac{A^2 + 2}{4}.$$

This solution is valid only if $x \ge 1$, which imposes $A^2 + 2 \ge 4 \implies A^2 \ge 2$.

Case 2: $x \le 1$. Then $|x-1| = 1-x$, giving

$$1-x = \frac{A^2-2x}{2} \implies 2-2x = A^2 - 2x \implies A^2 = 2.$$

This produces a candidate interval $x \in [\frac12,1]$, since $x \ge \frac12$ is required by the domain.

We now handle each $A$ individually.

(a) $A = \sqrt{2}$.

Case 1: $x = \frac{A^2+2}{4} = \frac{2+2}{4} = 1$. This satisfies $x \ge 1$. Check the original equation:

$$\sqrt{1+\sqrt{2\cdot 1 -1}} + \sqrt{1-\sqrt{2\cdot 1 -1}} = \sqrt{1+\sqrt{1}} + \sqrt{1-\sqrt{1}} = \sqrt{2} + 0 = \sqrt{2}.$$

Valid.

Case 2: $A^2 = 2$ holds, so $x \in [\frac12,1]$ are candidates. For $x \in [\frac12,1]$,

$$x-\sqrt{2x-1} \ge 0 \quad \text{and} \quad x+\sqrt{2x-1} \ge 0,$$

as verified: $x - \sqrt{2x-1} = \frac{(x-1)^2}{x+\sqrt{2x-1}} \ge 0$ for all $x \ge \frac12$. Therefore, all $x \in [\frac12,1]$ satisfy the original equation.

Thus, the complete solution set for $A = \sqrt{2}$ is

$$\boxed{\left[\frac12, 1\right]}.$$

(b) $A = 1$.

Case 1: $x = \frac{A^2+2}{4} = \frac{1+2}{4} = \frac{3}{4} < 1$, contradicts $x \ge 1$ requirement.
Case 2: $A^2 = 2$ is required, but $1^2 = 1 \ne 2$, impossible.

Therefore, there is no solution for $A = 1$:

$$\boxed{\text{no solution}}.$$

(c) $A = 2$.

Case 1: $x = \frac{A^2+2}{4} = \frac{4+2}{4} = \frac{6}{4} = \frac{3}{2} \ge 1$. Check the original equation:

$$\sqrt{\frac{3}{2}+\sqrt{2\cdot\frac{3}{2}-1}} + \sqrt{\frac{3}{2}-\sqrt{2\cdot\frac{3}{2}-1}} = \sqrt{\frac{3}{2}+\sqrt{2}} + \sqrt{\frac{3}{2}-\sqrt{2}}.$$

Squaring:

$$(\sqrt{\frac{3}{2}+\sqrt{2}} + \sqrt{\frac{3}{2}-\sqrt{2}})^2 = (\frac{3}{2}+\sqrt{2}) + (\frac{3}{2}-\sqrt{2}) + 2\sqrt{(\frac{3}{2})^2 - 2} = 3 + 2\cdot\frac12 = 4 = 2^2.$$

Valid.

Case 2: $x \le 1$ requires $A^2 = 2$, but $2^2 = 4 \ne 2$, impossible.

Hence the only solution for $A=2$ is

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

Verification of Key Steps

The domain $x \ge \frac12$ ensures the inner radical $\sqrt{2x-1}$ is defined.
For $x \ge \frac12$, $x-\sqrt{2x-1} = \frac{(x-1)^2}{x+\sqrt{2x-1}} \ge 0$, which guarantees all outer radicals are non-negative.
The substitution $u = \sqrt{x+\sqrt{2x-1}}$, $v = \sqrt{x-\sqrt{2x-1}}$ correctly reduces the equation to $(u+v)^2 = 2x + 2|x-1| = A^2$.
Case analysis for $x \ge 1$ and $x \le 1$ covers all possibilities, and each candidate was explicitly checked against the original equation to avoid extraneous solutions introduced by squaring.
Each $A$ was handled separately with verification that all radicals are non-negative.

Alternative Approaches

One alternative is to introduce $y = \sqrt{2x-1} \ge 0$, so $x = \frac{y^2+1}{2}$, and rewrite the original equation as

$$\sqrt{\frac{y^2+1}{2} + y} + \sqrt{\frac{y^2+1}{2} - y} = A.$$

Simplifying the radicals gives

\sqrt{\frac{(y+1)^2}{2}} + \sqrt{\frac{(1-y)^2}{2}} = A \implies \frac{|y+1|}{\sqrt{2}} + \frac{|1-y|}{\sqrt{2}} = A \implies |y+1| + |1-y| = \sqrt{