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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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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 <math>a </math> and <math>b </math> are constants.  Give the conditions that <math>a </math> and <math>b </math> must satisfy so that <math>x, y, z </math> (the solutions of the system) are distinct positive numbers.
  
 
== Solution ==
 
== 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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Note that <math>x^2 + y^2 = (x+y)^2 - 2xy = (x+y)^2 - 2z^2 </math>, so the first two equations become
 
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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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We note that <math>(x+y)^2 - z^2 = \Big[ (x+y)+z \Big]\Big[ (x+y)-z\Big] </math>, so if <math>a </math> equals 0, then <math>b </math> must also equal 0.  We then have <math> x+y = -z </math>; <math>xy = (x+y)^2 </math>.  This gives us <math>x^2 + xy + y^2 = 0 </math>.  Mutiplying both sides by <math>(x-y) </math>, we have <math>x^3 - y^3 = 0 </math>.  Since we want <math>x,y </math> to be real, this implies <math>x = y </math>.  But <math>x^2 + x^2 + x^2 </math> can only equal 0 when <math>x=0 </math> (which, in this case, implies <math>y,z = 0 </math>).  Hence there are no positive solutions when <math>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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When <math>a \neq 0 </math>, we divide <math>(**) </math> by <math>(*) </math> to obtain the system of equations
 
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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
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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>a </math> and <math>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>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
 
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</math>.
 
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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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If the expressions <math>(3a^2 - b^2), (3b^2 - a^2) </math> were simultaneously negative, then their sum, <math>2(a^2 + b^2) </math>, would also be negative, which cannot be.  Therefore our quadratic's discriminant is positive when <math>3a^2 > b^2 </math> and <math>3b^2 > a^2 </math>.  But we have already replaced the first inequality with the sharper bound <math>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>a </math> is positive and <math>3b^2 > a^2 > b^2 </math>.  Q.E.D.
  
  
 
{{alternate solutions}}
 
{{alternate solutions}}
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{{IMO box|year=1961|num-b=First question|num-a=2}}
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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]
 
  
  
 
[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Algebra Problems]]

Revision as of 19:13, 25 October 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 $a$ 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.

Solution

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

When $a \neq 0$, we divide $(**)$ by $(*)$ 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 $a$ and $a^2 > b^2$. Now, we have $x+y = \frac{a^2 + b^2}{2a}$; $xy = \left(\frac{a^2 - b^2}{2a}\right)^2$, so $x,y$ 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 $(3a^2 - b^2), (3b^2 - a^2)$ were simultaneously negative, then their sum, $2(a^2 + b^2)$, would also be negative, which cannot be. Therefore our quadratic's discriminant is positive when $3a^2 > b^2$ and $3b^2 > a^2$. But we have already replaced the first inequality with the sharper bound $a^2 > b^2$. 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 $a$ is positive and $3b^2 > a^2 > b^2$. Q.E.D.


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

1961 IMO (Problems) • Resources
Preceded by
Problem First question 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
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