IMO 1961 · Problem 1

Solve the system of equations: \[ \begin{aligned} x+y+z&=a\\ x^2+y^2+z^2&=b^2\\ xy&=z^2 \end{aligned} \] 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.

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

Answer

For real constants \(a,b\), all real solutions are as follows:

  • If \(a=b=0\), the unique solution is \((x,y,z)=(0,0,0)\).
  • If \(a\ne0\) and \[ \frac{a^2}{3}\le b^2\le3a^2, \] put \(R=(3a^2-b^2)(3b^2-a^2)\). The solutions are \[ \boxed{ (x,y,z)= \left( \frac{a^2+b^2\pm\sqrt R}{4a}, \frac{a^2+b^2\mp\sqrt R}{4a}, \frac{a^2-b^2}{2a} \right). } \] The two choices coincide when \(R=0\).
  • Otherwise there are no real solutions.

The necessary and sufficient conditions for \(x,y,z\) to be pairwise distinct and positive are \[ \boxed{a>0,\qquad \frac{a^2}{3}<b^2<a^2,} \] equivalently, \[ \boxed{|b|<a<\sqrt3\,|b|.} \]

Solution

Using \(xy=z^2\) and \(x+y=a-z\), we obtain \[ b^2=x^2+y^2+z^2 =(x+y)^2-2xy+z^2 =(a-z)^2-z^2 =a^2-2az. \]

If \(a=0\), this identity forces \(b=0\). Then the second equation gives \[ x^2+y^2+z^2=0, \] so \(x=y=z=0\). Thus the case \(a=0\) is completely settled.

Suppose now that \(a\ne0\). The identity above gives \[ z=\frac{a^2-b^2}{2a}, \qquad x+y=a-z=\frac{a^2+b^2}{2a}, \qquad xy=z^2=\frac{(a^2-b^2)^2}{4a^2}. \] Consequently, \(x,y\) must be the two roots, counted with multiplicity, of \[ t^2-\frac{a^2+b^2}{2a}\,t +\frac{(a^2-b^2)^2}{4a^2}=0. \] Its discriminant is \[ \begin{aligned} \Delta &=\frac{(a^2+b^2)^2-4(a^2-b^2)^2}{4a^2}\\ &=\frac{(3a^2-b^2)(3b^2-a^2)}{4a^2}. \end{aligned} \] Writing \(r=b^2/a^2\), the condition \(\Delta\ge0\) becomes \[ (3-r)(3r-1)\ge0, \] which holds exactly when \(1/3\le r\le3\). The quadratic formula therefore gives precisely the solutions stated in the Answer.

Conversely, every listed pair of roots satisfies \(x+y=a-z\) and \(xy=z^2\). Hence \[ x+y+z=a, \qquad x^2+y^2+z^2=(a-z)^2-z^2=a^2-2az=b^2. \] Thus every listed triple satisfies the original system, and no solutions have been omitted. At the discriminant endpoints, the solutions are respectively \[ b^2=\frac{a^2}{3}:\quad (x,y,z)=\left(\frac a3,\frac a3,\frac a3\right), \] and \[ b^2=3a^2:\quad (x,y,z)=(a,a,-a). \]

It remains to determine when the entries are distinct and positive.

Necessarily, positivity gives \(a=x+y+z>0\). Since \[ z=\frac{a^2-b^2}{2a}>0, \] we have \(b^2<a^2\). Also, distinctness gives \(x\ne y\), so \(\Delta>0\). Because \(3a^2-b^2>0\), this forces \(3b^2-a^2>0\). Therefore \[ a>0,\qquad \frac{a^2}{3}<b^2<a^2 \] is necessary.

Conversely, suppose these inequalities hold. Then \(\Delta>0\), so \(x,y\) are real and unequal, while \[ z>0,\qquad xy=z^2>0,\qquad x+y=\frac{a^2+b^2}{2a}>0. \] The positive product and positive sum imply \(x,y>0\). Finally, if \(x=z\), then \(xy=z^2\) and \(z>0\) imply \(y=z\), contradicting \(x\ne y\). The same argument excludes \(y=z\). Thus all three entries are pairwise distinct and positive, proving sufficiency.

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