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Difference between revisions of "1961 IMO Problems/Problem 1"

 
(reverted to official wording (i.e., from <http://imo.math.ca/>); added solution; added formatting + category tag (although it's more like a long, hairy intermediate problem than an olympiad problem))
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Solve the following system of equations:
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== Problem ==
  
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<math>x + y + z = a</math>
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(''Hungary'')
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Solve the system of equations:
  
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<math>x^2</math><math>+y^2+z^2=b^2</math>
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<center>
 +
<math>
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\begin{matrix}
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\quad x + y + z \!\!\! &= a \; \, \\
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x^2 +y^2+z^2 \!\!\! &=b^2 \\
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\qquad \qquad xy \!\!\!  &= z^2
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\end{matrix}
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</math>
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</center>
  
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<math>xy = z^2</math>
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where <math> \displaystyle a </math> and <math> \displaystyle b </math> are constants.  Give the conditions that <math> \displaystyle a </math> and <math> \displaystyle b </math> must satisfy so that <math> \displaystyle x, y, z </math> (the solutions of the system) are distinct positive numbers.
  
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where ''a'' and ''b'' are given real numbers.  What conditions must hold on ''a'' and ''b'' for the solutions to be positive and distinct?
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== Solution ==
 +
 
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Note that <math> \displaystyle x^2 + y^2 = (x+y)^2 - 2xy = (x+y)^2 - 2z^2 </math>, so the first two equations become
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<center>
 +
<math>
 +
\begin{matrix}
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\quad (x + y) + z \!\!\! &= a \; \; (*) \\
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(x+y)^2 - z^2 \!\!\! &=b^2 (**)
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\end{matrix}
 +
</math>.
 +
</center>
 +
 
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We note that <math> \displaystyle (x+y)^2 - z^2 = \Big[ (x+y)+z \Big]\Big[ (x+y)-z\Big] </math>, so if <math> \displaystyle a </math> equals 0, then <math> \displaystyle b </math> must also equal 0.  We then have <math> \displaystyle  x+y = -z </math>; <math> \displaystyle xy = (x+y)^2 </math>.  This gives us <math> \displaystyle x^2 + xy + y^2 = 0 </math>.  Mutiplying both sides by <math> \displaystyle (x-y) </math>, we have <math> \displaystyle x^3 - y^3 = 0 </math>.  Since we want <math> \displaystyle x,y </math> to be real, this implies <math> \displaystyle x = y </math>.  But <math> \displaystyle x^2 + x^2 + x^2 </math> can only equal 0 when <math> \displaystyle x=0 </math> (which, in this case, implies <math> \displaystyle y,z = 0 </math>).  Hence there are no positive solutions when <math> \displaystyle a = 0 </math>.
 +
 
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When <math> \displaystyle a \neq 0 </math>, we divide <math> \displaystyle (**) </math> by <math> \displaystyle (*) </math> to obtain the system of equations
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<center>
 +
<math>
 +
\begin{matrix}
 +
(x+y)+z &= a \; \quad \\
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(x+y)-z &= b^2/a
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\end{matrix}
 +
</math>,
 +
</center>
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which clearly has solution <math> x+y = \frac{a^2 + b^2}{2a} </math>, <math> z = \frac{a^2 - b^2}{2a} </math>.  In order for these both to be positive, we must have positive <math> \displaystyle a </math> and <math> \displaystyle a^2 > b^2 </math>.  Now, we have <math> x+y = \frac{a^2 + b^2}{2a} </math>; <math> xy = \left(\frac{a^2 - b^2}{2a}\right)^2 </math>, so <math> \displaystyle x,y </math> are the roots of the quadratic <math> m^2 - \frac{a^2 + b^2}{2a}m + \left(\frac{a^2 - b^2}{2a}\right)^2 </math>.  The [[discriminant]] for this equation is
 +
<center>
 +
<math>
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\left(\frac{a^2 + b^2}{2a}\right)^2 - \left(2\frac{a^2 -b^2}{2a}\right)^2 = \frac{ (3a^2 - b^2)(3b^2 - a^2) }{4a^2}
 +
</math>.
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</center>
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If the expressions <math> \displaystyle (3a^2 - b^2), (3b^2 - a^2) </math> were simultaneously negative, then their sum, <math> \displaystyle 2(a^2 + b^2) </math>, would also be negative, which cannot be.  Therefore our quadratic's discriminant is positive when <math> \displaystyle 3a^2 > b^2 </math> and <math> \displaystyle 3b^2 > a^2 </math>.  But we have already replaced the first inequality with the sharper bound <math> \displaystyle a^2 > b^2 </math>.  It is clear that both roots of the quadratic must be positive if the discriminant is positive (we can see this either from <math> \left(\frac{a^2 + b^2}{2a}\right)^2 > \left(\frac{a^2 + b^2}{2a}\right)^2 - \left(2\frac{a^2 -b^2}{2a}\right)^2 </math> or from [[Polynomial#Descartes.27_Law_of_Signs | Descartes' Rule of Signs]]).  We have now found the solutions to the system, and determined that it has positive solutions if and only if <math> \displaystyle a </math> is positive and <math> \displaystyle 3b^2 > a^2 > b^2 </math>.  Q.E.D.
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{{alternate solutions}}
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== Resources ==
 +
 
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* [[1961 IMO Problems]]
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* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=343297#p343297 Discussion on AoPS/MathLinks]
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 +
 
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[[Category:Olympiad Algebra Problems]]

Revision as of 18:44, 6 April 2007

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 $\displaystyle a$ and $\displaystyle b$ are constants. Give the conditions that $\displaystyle a$ and $\displaystyle b$ must satisfy so that $\displaystyle x, y, z$ (the solutions of the system) are distinct positive numbers.

Solution

Note that $\displaystyle x^2 + y^2 = (x+y)^2 - 2xy = (x+y)^2 - 2z^2$, so the first two equations become

$\begin{matrix} \quad (x + y) + z \!\!\! &= a \; \; (*) \\ (x+y)^2 - z^2 \!\!\! &=b^2 (**) \end{matrix}$.

We note that $\displaystyle (x+y)^2 - z^2 = \Big[ (x+y)+z \Big]\Big[ (x+y)-z\Big]$, so if $\displaystyle a$ equals 0, then $\displaystyle b$ must also equal 0. We then have $\displaystyle x+y = -z$; $\displaystyle xy = (x+y)^2$. This gives us $\displaystyle x^2 + xy + y^2 = 0$. Mutiplying both sides by $\displaystyle (x-y)$, we have $\displaystyle x^3 - y^3 = 0$. Since we want $\displaystyle x,y$ to be real, this implies $\displaystyle x = y$. But $\displaystyle x^2 + x^2 + x^2$ can only equal 0 when $\displaystyle x=0$ (which, in this case, implies $\displaystyle y,z = 0$). Hence there are no positive solutions when $\displaystyle a = 0$.

When $\displaystyle a \neq 0$, we divide $\displaystyle (**)$ by $\displaystyle (*)$ to obtain the system of equations

$\begin{matrix} (x+y)+z &= a \; \quad \\ (x+y)-z &= b^2/a \end{matrix}$,

which clearly has solution $x+y = \frac{a^2 + b^2}{2a}$, $z = \frac{a^2 - b^2}{2a}$. In order for these both to be positive, we must have positive $\displaystyle a$ and $\displaystyle a^2 > b^2$. Now, we have $x+y = \frac{a^2 + b^2}{2a}$; $xy = \left(\frac{a^2 - b^2}{2a}\right)^2$, so $\displaystyle x,y$ are the roots of the quadratic $m^2 - \frac{a^2 + b^2}{2a}m + \left(\frac{a^2 - b^2}{2a}\right)^2$. The discriminant for this equation is

$\left(\frac{a^2 + b^2}{2a}\right)^2 - \left(2\frac{a^2 -b^2}{2a}\right)^2 = \frac{ (3a^2 - b^2)(3b^2 - a^2) }{4a^2}$.

If the expressions $\displaystyle (3a^2 - b^2), (3b^2 - a^2)$ were simultaneously negative, then their sum, $\displaystyle 2(a^2 + b^2)$, would also be negative, which cannot be. Therefore our quadratic's discriminant is positive when $\displaystyle 3a^2 > b^2$ and $\displaystyle 3b^2 > a^2$. But we have already replaced the first inequality with the sharper bound $\displaystyle a^2 > b^2$. It is clear that both roots of the quadratic must be positive if the discriminant is positive (we can see this either from $\left(\frac{a^2 + b^2}{2a}\right)^2 > \left(\frac{a^2 + b^2}{2a}\right)^2 - \left(2\frac{a^2 -b^2}{2a}\right)^2$ or from Descartes' Rule of Signs). We have now found the solutions to the system, and determined that it has positive solutions if and only if $\displaystyle a$ is positive and $\displaystyle 3b^2 > a^2 > b^2$. Q.E.D.


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Resources

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