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Difference between revisions of "1970 Canadian MO Problems/Problem 1"

(Created page with "== Problem == Find all number triples <math> (x,y,z)</math> such that when any of these numbers is added to the product of the other two, the result is <math>2</math>. == Soluti...")
 
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From the first equation minus the second, we get :
 
From the first equation minus the second, we get :
  
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<cmath> \begin{matrix}x-y + yz - xz &=& 0 \\(1-z)(x+y) &=& 0\end{matrix} </cmath>
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<cmath> \begin{matrix}x-y + yz - xz &=& 0 \\(1-z)(x-y) &=& 0\end{matrix} </cmath>
  
 
So either <math>z=1</math> or <math>x=y</math>. For <math>z=1</math>, since the equations were symmetric, we have one solution of <math> (x,y,z)=(1,1,1)</math>.
 
So either <math>z=1</math> or <math>x=y</math>. For <math>z=1</math>, since the equations were symmetric, we have one solution of <math> (x,y,z)=(1,1,1)</math>.
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<cmath> (x,y,z) = {(1,1,1), (-2,-2,-2)} </cmath>
 
<cmath> (x,y,z) = {(1,1,1), (-2,-2,-2)} </cmath>
  
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{{Old CanadaMO box|num-b=1|num-a=2|year=1970}}
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== See Also ==
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{{Old CanadaMO box|before=First Question|num-a=2|year=1970}}
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[[Category:Olympiad Algebra Problems]]

Latest revision as of 06:54, 30 December 2012

Problem

Find all number triples $(x,y,z)$ such that when any of these numbers is added to the product of the other two, the result is $2$.

Solution

We have:

$\begin{matrix} x + yz &=& 2\\ y + xz &=& 2\\ z + yx &=& 2 \end{matrix}$

From the first equation minus the second, we get :

\[\begin{matrix}x-y + yz - xz &=& 0 \\(1-z)(x-y) &=& 0\end{matrix}\]

So either $z=1$ or $x=y$. For $z=1$, since the equations were symmetric, we have one solution of $(x,y,z)=(1,1,1)$. From $x=y$, substituting it into the original three derived equations, we have: \[\begin{matrix} x+xz &=& 2\\ z + x^2 &=& 2 \end{matrix}\] We then get \[z = 2-x^2\] Substituting this into $x+xz = 2$, \[\begin{matrix} x + x(2-x^2) &=& 2\\ x^3 - 3x + 2 &=& 0\\ (x-1)^2(x+2) = 0 \end{matrix}\] Thus, either $x = 1$, or $x = -2$. Since the equations were symmetric, we then get the full solution set of:

\[(x,y,z) = {(1,1,1), (-2,-2,-2)}\]

See Also

1970 Canadian MO (Problems)
Preceded by
First Question 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • Followed by
Problem 2
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