1970 Canadian MO Problems/Problem 1: Difference between revisions
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From the first equation minus the second, we get : | From the first equation minus the second, we get : | ||
<cmath> \begin{matrix}x-y + yz - xz &=& 0 \\(1-z)(x | <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>. | ||
Latest revision as of 05:54, 30 December 2012
Problem
Find all number triples 
such that when any of these numbers is added to the product of the other two, the result is 
.
Solution
We have:
From the first equation minus the second, we get :
![\[\begin{matrix}x-y + yz - xz &=& 0 \\(1-z)(x-y) &=& 0\end{matrix}\]](http://latex.artofproblemsolving.com/6/2/0/620c8551cdc003c897cd60e620e0f6a453f94137.png)
So either 
or 
. For , since the equations were symmetric, we have one solution of 
.
From , substituting it into the original three derived equations, we have:
![\[\begin{matrix} x+xz &=& 2\\ z + x^2 &=& 2 \end{matrix}\]](http://latex.artofproblemsolving.com/f/6/c/f6c4322d96e4d83b378519dfe13c679f43644cc6.png)
We then get ![\[z = 2-x^2\]](http://latex.artofproblemsolving.com/f/d/6/fd6381d3040210d943035e8fd4b7f2388c689814.png)
Substituting this into 
, ![\[\begin{matrix} x + x(2-x^2) &=& 2\\ x^3 - 3x + 2 &=& 0\\ (x-1)^2(x+2) = 0 \end{matrix}\]](http://latex.artofproblemsolving.com/4/c/8/4c86c53830aec8163507f4db372088fda15fbaa2.png)
Thus, either 
, or 
. Since the equations were symmetric, we then get the full solution set of:
![\[(x,y,z) = {(1,1,1), (-2,-2,-2)}\]](http://latex.artofproblemsolving.com/3/2/4/324fe63375a7dbda6240c2f1ee0e19c97973e8f2.png)
See Also
| 1970 Canadian MO (Problems) | ||
| Preceded by First Question |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 2 |
