IMO 1959 · Problem 2

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?

Official text of the problem. Official problem paper (PDF)

Answer

  • (a) \(x\in\left[\frac12,1\right]\).
  • (b) No real solutions.
  • (c) \(x=\frac32\).

Solution

The inner square root requires \(x\ge\frac12\). Conversely, for every such \(x\), we have \(x\ge0\) and \[ x^2-(2x-1)=(x-1)^2\ge0, \] so \(x\ge\sqrt{2x-1}\). Thus both outer radicands are non-negative, and the domain is exactly \(x\ge\frac12\).

Let \[ F(x)=\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}. \] On this domain, \[ \begin{aligned} F(x)^2 &=2x+2\sqrt{\bigl(x+\sqrt{2x-1}\bigr) \bigl(x-\sqrt{2x-1}\bigr)}\\ &=2x+2\sqrt{x^2-2x+1}\\ &=2x+2|x-1|. \end{aligned} \] Since \(F(x)\ge0\), it follows that \[ F(x)= \begin{cases} \sqrt2,&\frac12\le x\le1,\\[2mm] \sqrt{4x-2},&x>1. \end{cases} \]

For (a), the first branch equals \(\sqrt2\) throughout, while the second is strictly greater than \(\sqrt2\). Hence precisely \(\frac12\le x\le1\) satisfies the equation.

For (b), \(F(x)\ge\sqrt2>1\) everywhere on its domain, so there are no solutions.

For (c), the first branch cannot equal \(2\). On the second branch, \[ \sqrt{4x-2}=2 \quad\Longleftrightarrow\quad 4x-2=4 \quad\Longleftrightarrow\quad x=\frac32. \] This value lies in \(x>1\) and therefore is a solution.

The solution was checked against John Scholes' IMO solution archive (Kalva).