IMO 1959 Problem 2
Let
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 4m19s
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?
Exploration
Let
$$S=\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}.$$
The domain requires $2x-1\ge 0$, hence $x\ge \tfrac12$. Both radicals are real when additionally $x\pm \sqrt{2x-1}\ge 0$. Since $x\ge \tfrac12$, the quantity $x-\sqrt{2x-1}$ is nonnegative because $x^2-(2x-1)=(x-1)^2\ge 0$, implying $x\ge \sqrt{2x-1}$.
A natural strategy is to square once to remove the outer radicals. Symmetry suggests introducing
$$u=\sqrt{x+\sqrt{2x-1}}, \quad v=\sqrt{x-\sqrt{2x-1}},$$
so that $S=u+v$ and
$$u^2+v^2=2x, \quad u^2-v^2=2\sqrt{2x-1}.$$
This leads to a system in $u+v$ and $uv$. A second squaring yields an equation in $x$ alone.
A key simplification is that $S^2$ contains $2x$ plus a cross term involving $\sqrt{x^2-(2x-1)}=\sqrt{(x-1)^2}$, which reduces to $|x-1|$. This absolute value forces a case split at $x=1$, which is likely where different answers arise.
The structure suggests that each fixed $A$ produces a quadratic condition in $x$, but only some roots satisfy the original sign constraints.
Problem Understanding
This is a Type A problem, requiring determination of all real values $x$ satisfying a nested radical equation depending on a parameter $A\in{\sqrt2,1,2}$.
We must find all $x\ge \tfrac12$ such that
$$\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=A.$$
The core difficulty is that squaring introduces absolute values and potential extraneous solutions. The inner structure simplifies to expressions involving $|x-1|$, so the equation splits into different algebraic forms depending on whether $x\ge 1$ or $x\le 1$. A naive squaring approach risks losing or adding solutions unless both cases are handled rigorously.
The expected outcome is that each $A$ yields a finite explicit set of $x$-values determined by quadratic equations after case analysis.
Proof Architecture
Lemma 1 states that for all $x\ge \tfrac12$,
$$\sqrt{(x+\sqrt{2x-1})(x-\sqrt{2x-1})}=|x-1|.$$
It follows from direct expansion $x^2-(2x-1)=(x-1)^2$, identifying the square root.
Lemma 2 states that
$$S^2=2x+2|x-1|.$$
This follows by expanding $(u+v)^2$ and applying Lemma 1.
Lemma 3 states that for $x\ge 1$, $S^2=4x-2$, and for $\tfrac12\le x\le 1$, $S^2=2$. This follows by resolving the absolute value.
Lemma 4 states that solving $S=A$ reduces to solving a quadratic in each region obtained from Lemma 3, followed by verification against the original equation.
The hardest part is controlling extraneous solutions introduced when squaring, especially at the boundary $x=1$, where the expression changes regime.
Solution
Lemma 1
For $x\ge \tfrac12$,
$$(x+\sqrt{2x-1})(x-\sqrt{2x-1})=x^2-(2x-1)=x^2-2x+1=(x-1)^2.$$
Taking square roots yields
$$\sqrt{(x+\sqrt{2x-1})(x-\sqrt{2x-1})}=|x-1|.$$
This establishes that the product of the two inner radicals collapses to an absolute value, since square roots are nonnegative and must equal the nonnegative magnitude of $x-1$.
Lemma 2
Let
$$u=\sqrt{x+\sqrt{2x-1}}, \quad v=\sqrt{x-\sqrt{2x-1}}.$$
Then
$$S^2=u^2+v^2+2uv.$$
Since $u^2+v^2=2x$ and $uv=\sqrt{(x+\sqrt{2x-1})(x-\sqrt{2x-1})}=|x-1|$, it follows that
$$S^2=2x+2|x-1|.$$
This step is valid because both $u$ and $v$ are nonnegative, ensuring no sign ambiguity in squaring.
Lemma 3
If $x\ge 1$, then $|x-1|=x-1$, so
$$S^2=2x+2(x-1)=4x-2.$$
If $\tfrac12\le x\le 1$, then $|x-1|=1-x$, so
$$S^2=2x+2(1-x)=2.$$
This partitions the domain into two algebraically distinct regimes determined by the sign of $x-1$.
Case (a): $A=\sqrt2$
From Lemma 3, in the region $\tfrac12\le x\le 1$,
$$S^2=2 \implies S=\sqrt2.$$
Thus every $x\in[\tfrac12,1]$ satisfies the equation in this region.
For $x\ge 1$,
$$S^2=4x-2.$$
Setting $S=\sqrt2$ yields $2=4x-2$, hence $x=1$, which lies in this region and is consistent.
Therefore the full solution set is
$$x\in\left[\tfrac12,1\right].$$
Every value in this interval satisfies the original equation because substitution into Lemma 3 preserves equality without further restriction.
Case (b): $A=1$
In the region $\tfrac12\le x\le 1$, Lemma 3 gives $S^2=2$, hence $S=\sqrt2$, which is incompatible with $A=1$. No solutions occur in this region.
In the region $x\ge 1$, solving
$$1^2=4x-2$$
gives $x=\tfrac34$, which does not satisfy $x\ge 1$. No valid solutions arise.
Thus no real $x$ satisfies the equation.
Case (c): $A=2$
For $\tfrac12\le x\le 1$, Lemma 3 gives $S=\sqrt2$, so no solutions occur.
For $x\ge 1$, solving
$$4=4x-2$$
gives $x=\tfrac32$, which satisfies the region condition.
Verification: for $x=\tfrac32$,
$$\sqrt{2x-1}=\sqrt2,$$
so the expression becomes
$$\sqrt{\tfrac32+\sqrt2}+\sqrt{\tfrac32-\sqrt2}.$$
Lemma 2 gives
$$S^2=2x+2(x-1)=4\cdot\tfrac32-2=4,$$
so $S=2$, consistent.
Thus the only solution is $x=\tfrac32$.
Verification of Key Steps
The identity in Lemma 1 depends on expanding the product inside the square root; the critical point is recognizing that $(x-1)^2$ arises exactly, and any sign error would incorrectly remove the absolute value, leading to wrong domain splitting.
In Case (b), the potential root $x=\tfrac34$ emerges from algebra but fails the domain condition $x\ge 1$; overlooking the piecewise structure would incorrectly accept this extraneous solution.
In Case (c), verification ensures that squaring does not introduce spurious solutions; failure to re-evaluate $S$ directly would allow incorrect acceptance of non-satisfying candidates.
Alternative Approaches
A different approach introduces the substitution $x=\cosh^2 t$, transforming $\sqrt{2x-1}$ into $\sinh t$, which linearizes the nested radicals via hyperbolic identities. This converts the expression into a sum of square roots of exponentials, yielding a direct simplification to $\cosh t \pm \sinh t$. The algebraic method used here is preferable because it avoids transcendental substitutions and isolates the absolute value structure directly, producing a clean case split in elementary form.