IMO 1961 Problem 1
The problem asks for conditions on the parameters $a$ and $b$ under which the system admits three distinct positive numbers $x,y,z$.
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 4m48s
Problem
(Hungary) Solve the system of equations:
$\begin{matrix} \quad x + y + z !!! &= a ; , \ x^2 +y^2+z^2 !!! &=b^2 \ \qquad \qquad xy !!! &= z^2 \end{matrix}$
where $a$ and $b$ are constants. Give the conditions that $a$ and $b$ must satisfy so that $x, y, z$ (the solutions of the system) are distinct positive numbers.
Exploration
The problem asks for conditions on the parameters $a$ and $b$ under which the system admits three distinct positive numbers $x,y,z$. Since the equations are symmetric in $x$ and $y$ but not in $z$, the relation $xy=z^2$ suggests reducing the number of variables.
The first five numerical checks reveal useful structure. Take $z=1$. If $x=4$ and $y=\tfrac14$, then $xy=1=z^2$, and
$$a=4+\frac14+1=\frac{21}{4}, \qquad b^2=16+\frac1{16}+1=\frac{273}{16}.$$
The numbers are positive and distinct. Compute
$$a^2-b^2=\frac{441}{16}-\frac{273}{16}=\frac{168}{16}=\frac{21}{2}.$$
Also,
$$xy+xz+yz = 1+4+\frac14 = \frac{21}{4} = az.$$
This hints that eliminating $x+y$ may produce a quadratic condition on $z$.
From
$$x+y=a-z$$
and
$$x^2+y^2=b^2-z^2,$$
the identity
$$(x+y)^2=x^2+y^2+2xy$$
gives
$$(a-z)^2=b^2-z^2+2z^2=b^2+z^2.$$
Hence
$$a^2-2az=z^2+b^2-z^2=b^2,$$
which simplifies to
$$z=\frac{a^2-b^2}{2a}.$$
This reduction appears decisive provided $a>0$.
After determining $z$, the numbers $x,y$ are roots of
$$t^2-(a-z)t+z^2=0.$$
Positivity and distinctness require the quadratic to have two distinct positive roots. Since the sum is $a-z$ and the product is $z^2>0$, positivity follows once the discriminant is positive and $a-z>0$. The discriminant equals
$$(a-z)^2-4z^2=a^2-2az-3z^2.$$
Using $a^2-2az=b^2$, this becomes
$$b^2-3z^2>0.$$
Substituting
$$z=\frac{a^2-b^2}{2a}$$
should yield a condition purely in $a,b$.
A likely danger is forgetting that $z$ itself must be positive and that $x\neq y$ requires strict inequality in the discriminant. Another subtlety is that $x,y,z$ are all distinct, so one must exclude $x=z$ or $y=z$. Since $xy=z^2$, equality $x=z$ forces $y=z$, contradicting distinctness once the discriminant is positive.
Problem Understanding
We are given a system in three real variables,
$$x+y+z=a,\qquad x^2+y^2+z^2=b^2,\qquad xy=z^2,$$
where $a$ and $b$ are constants. The task is to determine exactly which values of $a$ and $b$ allow the system to possess three distinct positive solutions $x,y,z$.
This is a Type A problem because we must characterize all admissible parameter pairs $(a,b)$. A complete solution requires two directions: first, proving that the derived conditions are sufficient to produce distinct positive numbers satisfying the system, and second, proving that any system with distinct positive solutions necessarily satisfies those conditions.
The central difficulty lies in translating positivity and distinctness of three unknown numbers into inequalities involving only $a$ and $b$. A naive elimination of variables produces complicated algebra, but the relation $xy=z^2$ and the identity
$$(x+y)^2=x^2+y^2+2xy$$
allow the system to collapse to a single quadratic equation for $x$ and $y$ after expressing $z$ in terms of $a$ and $b$.
The answer will be that $a$ and $b$ must satisfy
$$a>0,\qquad a^2>b^2,\qquad 3(a^2-b^2)^2<4a^2b^2.$$
Intuitively, the first two inequalities guarantee a positive value of $z$, while the last inequality guarantees that the quadratic defining $x$ and $y$ has two distinct positive roots.
Proof Architecture
We proceed through three claims.
Lemma 1. Any solution of the system satisfies
$$z=\frac{a^2-b^2}{2a}.$$
The proof combines the first and third equations with the identity for $(x+y)^2$.
Lemma 2. Once $z$ is fixed, the numbers $x$ and $y$ are the roots of
$$t^2-(a-z)t+z^2=0,$$
and they are distinct positive numbers if and only if
$$z>0$$
and
$$(a-z)^2>4z^2.$$
The proof uses the quadratic formula and the sign conditions on roots.
Lemma 3. The discriminant condition in Lemma 2 is equivalent to
$$3(a^2-b^2)^2<4a^2b^2.$$
The proof substitutes the expression for $z$ from Lemma 1.
The hardest direction is sufficiency, because every positivity and distinctness requirement must be verified after constructing the roots. The most delicate point is proving that the discriminant condition exactly translates into the inequality in $a$ and $b$.
Solution
We seek all pairs $(a,b)$ for which the system
$$x+y+z=a,$$
$$x^2+y^2+z^2=b^2,$$
$$xy=z^2$$
has distinct positive solutions.
Lemma 1
Every solution satisfies
$$z=\frac{a^2-b^2}{2a}.$$
Proof
From the first equation,
$$x+y=a-z.$$
Squaring both sides gives
$$(x+y)^2=(a-z)^2.$$
Using the identity
$$(x+y)^2=x^2+y^2+2xy,$$
together with
$$x^2+y^2=b^2-z^2$$
and
$$xy=z^2,$$
we obtain
$$(a-z)^2=(b^2-z^2)+2z^2=b^2+z^2.$$
Expanding the left-hand side,
$$a^2-2az+z^2=b^2+z^2.$$
Hence
$$a^2-2az=b^2.$$
Since
$$x,y,z>0,$$
the first equation yields $a>0$, so division by $2a$ is legitimate:
$$z=\frac{a^2-b^2}{2a}.$$
∎
This establishes that every admissible solution determines a unique value of $z$; attempting to treat $z$ as free would ignore a hidden algebraic constraint.
Lemma 2
For fixed $z$, the numbers $x$ and $y$ are distinct positive solutions if and only if
$$z>0$$
and
$$(a-z)^2>4z^2.$$
Proof
The equations
$$x+y=a-z, \qquad xy=z^2$$
show that $x$ and $y$ are roots of the polynomial
$$t^2-(a-z)t+z^2.$$
The discriminant equals
$$\Delta=(a-z)^2-4z^2.$$
The roots are real and distinct precisely when
$$\Delta>0.$$
Their product equals
$$z^2>0,$$
and their sum equals
$$a-z.$$
If $\Delta>0$ and $a-z>0$, both roots are positive.
Since
$$x+y=a-z>z>0,$$
positivity of $z$ already implies $a-z>0$. Thus distinct positive roots occur exactly when
$$z>0$$
and
$$(a-z)^2>4z^2.$$
∎
This establishes the exact positivity and distinctness criterion; replacing the strict inequality by a weak one would allow the forbidden case $x=y$.
Lemma 3
The condition
$$(a-z)^2>4z^2$$
is equivalent to
$$3(a^2-b^2)^2<4a^2b^2.$$
Proof
By Lemma 1,
$$z=\frac{a^2-b^2}{2a}.$$
Also,
$$a-z = a-\frac{a^2-b^2}{2a} = \frac{a^2+b^2}{2a}.$$
Substituting into the discriminant condition,
$$(a-z)^2>4z^2,$$
gives
$$\left(\frac{a^2+b^2}{2a}\right)^2 > 4\left(\frac{a^2-b^2}{2a}\right)^2.$$
Multiplying by $4a^2>0$,
$$(a^2+b^2)^2 > 4(a^2-b^2)^2.$$
Expanding,
$$a^4+2a^2b^2+b^4 > 4a^4-8a^2b^2+4b^4.$$
Rearranging,
$$0 > 3a^4-10a^2b^2+3b^4.$$
This factors as
$$0 > (3(a^2-b^2)^2-4a^2b^2).$$
Hence
$$3(a^2-b^2)^2<4a^2b^2.$$
∎
This establishes the required inequality in $a$ and $b$ alone; expanding without factoring risks obscuring the decisive structure.
We now prove necessity and sufficiency.
Assume first that the system has distinct positive solutions. By positivity,
$$a=x+y+z>0.$$
Lemma 1 gives
$$z=\frac{a^2-b^2}{2a}>0,$$
hence
$$a^2>b^2.$$
Since $x$ and $y$ are distinct positive numbers, Lemma 2 applies, and Lemma 3 yields
$$3(a^2-b^2)^2<4a^2b^2.$$
Thus every admissible pair $(a,b)$ satisfies
$$a>0,\qquad a^2>b^2,\qquad 3(a^2-b^2)^2<4a^2b^2.$$
Conversely, suppose
$$a>0,\qquad a^2>b^2,\qquad 3(a^2-b^2)^2<4a^2b^2.$$
Define
$$z=\frac{a^2-b^2}{2a}.$$
Since $a>0$ and $a^2>b^2$,
$$z>0.$$
Consider the quadratic
$$t^2-(a-z)t+z^2=0.$$
Its discriminant satisfies
$$(a-z)^2-4z^2>0$$
by Lemma 3, so it has two distinct real roots $x,y$. Their product is
$$xy=z^2>0,$$
and their sum is
$$x+y=a-z>0,$$
hence both roots are positive.
By construction,
$$x+y+z=a, \qquad xy=z^2.$$
Finally,
$$x^2+y^2+z^2 = (x+y)^2-2xy+z^2 = (a-z)^2-z^2.$$
Lemma 1 implies
$$(a-z)^2=b^2+z^2,$$
hence
$$x^2+y^2+z^2=b^2.$$
Thus the system possesses distinct positive solutions.
Hence the necessary and sufficient conditions are
$$\boxed{ a>0,\qquad a^2>b^2,\qquad 3(a^2-b^2)^2<4a^2b^2 }.$$
Verification of Key Steps
The first delicate step is the derivation of
$$z=\frac{a^2-b^2}{2a}.$$
Re-derive it independently:
$$x+y=a-z, \qquad xy=z^2,$$
so
$$(a-z)^2=(x+y)^2=x^2+y^2+2xy.$$
Substitute
$$x^2+y^2=b^2-z^2$$
and
$$xy=z^2,$$
obtaining
$$(a-z)^2=b^2+z^2.$$
After cancellation,
$$a^2-2az=b^2.$$
A careless argument may forget that division by $a$ requires $a\neq0$. Positivity of the variables supplies $a>0$.
The second delicate step is positivity of the roots. For a quadratic
$$t^2-St+P,$$
real distinct roots with
$$P>0$$
need not both be positive; they could both be negative. The extra condition
$$S>0$$
is essential. Here,
$$S=a-z=x+y>0,$$
which follows from positivity of $z$ and
$$a=x+y+z.$$
The third delicate step is the conversion of the discriminant inequality into a condition on $a,b$. Starting from
$$(a-z)^2>4z^2$$
and substituting
$$z=\frac{a^2-b^2}{2a},$$
a sign mistake in clearing denominators can reverse the inequality. Since
$$a>0,$$
multiplication by
$$4a^2$$
preserves the inequality direction.
Alternative Approaches
A different route introduces
$$s=x+y, \qquad p=xy.$$
Since
$$p=z^2$$
and
$$s=a-z,$$
the second equation becomes
$$s^2-2p+z^2=b^2,$$
which simplifies to
$$(a-z)^2-z^2=b^2.$$
This again produces the same formula for $z$, after which positivity and distinctness reduce to the quadratic condition
$$s^2>4p.$$
Another possible strategy uses symmetric polynomials. Since
$$x+y+z=a$$
and
$$x^2+y^2+z^2=a^2-2(xy+yz+zx),$$
one may derive
$$xy+xz+yz=\frac{a^2-b^2}{2}.$$
Using
$$xy=z^2,$$
this becomes
$$z(x+y+z)=\frac{a^2-b^2}{2},$$
hence
$$az=\frac{a^2-b^2}{2}.$$
The remainder proceeds through the quadratic for $x,y$. The main approach is preferable because it reaches the key formula for $z$ more directly and keeps every condition tied to a concrete quadratic discriminant.