Difference between revisions of "1989 OIM Problems/Problem 1"
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== Solution == | == Solution == | ||
| − | {{ | + | Square the first equation: |
| + | <cmath>x^2+y^2+z^2+2xy-2xz-2yz=1</cmath> | ||
| + | Subtract the second equation: | ||
| + | <cmath>2y^2+2xy-2xz-2yz=0</cmath> | ||
| + | This factors as: | ||
| + | <cmath>2(y+x)(y-z)=0</cmath> | ||
| + | This implies that <math>x=-y</math> or <math>y=z</math>. | ||
| + | |||
| + | If <math>x=-y</math>, then substituting into the first equation yields <math>z=1</math>. Substituting all of this into the third equation gives <math>2y^3=-2</math>, so <math>y=-1</math> and <math>x=1</math>. Thus a valid triple is <math>\boxed{(1,-1,1)}</math>. | ||
| + | |||
| + | If <math>y=z</math>, then substituting into the first equation yields <math>x=-1</math>. Substituting all of this into the third equation gives <math>2y^3=-2</math>, so <math>y=-1</math> and <math>z=-1</math>. Thus the other valid triple is <math>\boxed{(-1,-1,-1)}</math>. | ||
| + | |||
| + | Since both triples obviously work, this finishes the proof. | ||
| + | |||
| + | ~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406] | ||
== See also == | == See also == | ||
https://www.oma.org.ar/enunciados/ibe4.htm | https://www.oma.org.ar/enunciados/ibe4.htm | ||
Latest revision as of 18:12, 22 March 2025
Problem
Find all triples of real numbers that satisfy the system of equations:
![\[x+y-z=-1\]](http://latex.artofproblemsolving.com/4/a/a/4aa734f24b8e5bd18fb4acfcda17cc192edc8b8b.png)
![\[x^2-y^2+z^2=1\]](http://latex.artofproblemsolving.com/7/c/e/7ce715737673fdea26da5fbbcab0f9eba639a911.png)
![\[-x^3+y^3+z^3=-1\]](http://latex.artofproblemsolving.com/3/e/b/3ebeca29bdbdc7c3f91a976dd61e14705b211349.png)
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
Square the first equation:
![\[x^2+y^2+z^2+2xy-2xz-2yz=1\]](http://latex.artofproblemsolving.com/b/d/3/bd3d65c7420ff2716f2796410598ae31d56880f7.png)
Subtract the second equation:
![\[2y^2+2xy-2xz-2yz=0\]](http://latex.artofproblemsolving.com/d/4/d/d4d58963c62a67b92cae6a0d4ecf2d93ffa560ad.png)
This factors as:
![\[2(y+x)(y-z)=0\]](http://latex.artofproblemsolving.com/e/4/9/e4940dec9ddb62a6378cb69c23c53f4992df71ef.png)
This implies that 
or 
.
If , then substituting into the first equation yields 
. Substituting all of this into the third equation gives 
, so 
and 
. Thus a valid triple is 
.
If , then substituting into the first equation yields 
. Substituting all of this into the third equation gives , so and 
. Thus the other valid triple is 
.
Since both triples obviously work, this finishes the proof.
