Difference between revisions of "2002 Indonesia MO Problems/Problem 3"
Rockmanex3 (talk | contribs) (Solution to Problem 3) |
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Multiply the second equation by the first equation to get | Multiply the second equation by the first equation to get | ||
<cmath>x^3 + xy^2 + xz^2 + x^2y + y^3 + yz^2 + x^2z + y^2z + z^3 = 72</cmath> | <cmath>x^3 + xy^2 + xz^2 + x^2y + y^3 + yz^2 + x^2z + y^2z + z^3 = 72</cmath> | ||
| − | Subtract the third equation | + | Subtract the third equation to get |
<cmath>xy^2 + xz^2 + x^2y + yz^2 + x^2z + y^2z = 48</cmath> | <cmath>xy^2 + xz^2 + x^2y + yz^2 + x^2z + y^2z = 48</cmath> | ||
Cube the first equation to get | Cube the first equation to get | ||
| Line 26: | Line 26: | ||
Factor the polynomial to get | Factor the polynomial to get | ||
<cmath>(a-2)^3 = 0</cmath> | <cmath>(a-2)^3 = 0</cmath> | ||
| − | Since <math>a = 2</math> is a triple root to the polynomial, the only solution to the system of equations is <math>\boxed{(2,2,2)}</math>. | + | Since <math>a = 2</math> is a triple root to the polynomial, the only solution to the system of equations is <math>\boxed{(2,2,2)}</math>, and plugging the values back in satisfies the system. |
==See Also== | ==See Also== | ||
| − | {{Indonesia MO | + | {{Indonesia MO box|year=2002|num-b=2|num-a=4|eight=}} |
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 00:09, 4 August 2018
Problem
Find all real solutions from the following system of equations:
Solution
Square the first equation to get
![\[x^2 + y^2 + z^2 + 2(xy + yz + xz) = 36\]](http://latex.artofproblemsolving.com/f/f/8/ff8416704f237151ace1a0a0c17a9db3d5798e35.png)
Subtract the second equation from the result to get
![\[2(xy+yz+xz) = 24\]](http://latex.artofproblemsolving.com/f/9/a/f9a9e4a45a07653c9b29d51d56a152ebc6faa6a8.png)
![\[xy+yz+xz = 12\]](http://latex.artofproblemsolving.com/0/f/4/0f4d1daf49b68bd5ff40021eff6213f10b505a01.png)
Multiply the second equation by the first equation to get
![\[x^3 + xy^2 + xz^2 + x^2y + y^3 + yz^2 + x^2z + y^2z + z^3 = 72\]](http://latex.artofproblemsolving.com/6/c/9/6c9a4dfb4c4dbd60f19eb5b7eba3c65972d30719.png)
Subtract the third equation to get
![\[xy^2 + xz^2 + x^2y + yz^2 + x^2z + y^2z = 48\]](http://latex.artofproblemsolving.com/3/e/d/3eddb39758be62b7ace2e77f3d12d5bc50a8fa3f.png)
Cube the first equation to get
![\[(x^3 + y^3 + z^3) + 3(x^2y + x^2z + xy^2 + y^2z + xz^2 + yz^2) + 6xyz = 216\]](http://latex.artofproblemsolving.com/c/a/1/ca1a8d81dfd953d20b00cc7282cd9eefa7c10400.png)
![\[24 + 3(48) + 6xyz = 216\]](http://latex.artofproblemsolving.com/f/8/3/f83c4978fb993dccfd62489abd556d819657019a.png)
![\[168 + 6xyz = 216\]](http://latex.artofproblemsolving.com/d/3/3/d33047b239f348c9a5d2a51931d5b2135def140e.png)
![\[6xyz = 48\]](http://latex.artofproblemsolving.com/a/0/3/a03c646b0b3845c290bd63f151a8580dcada1978.png)
![\[xyz = 8\]](http://latex.artofproblemsolving.com/9/4/f/94f7d6eaab85dca74ec09823d71f9d01485cc623.png)
If 
, 
, and 
, the solution triplet is the roots of the polynomial
![\[a^3 - 6a^2 + 12a - 8 = 0\]](http://latex.artofproblemsolving.com/a/2/e/a2e83c63f425781a1e4cb7ffbd94518551b61094.png)
Factor the polynomial to get
![\[(a-2)^3 = 0\]](http://latex.artofproblemsolving.com/f/b/4/fb436cac3d75da27c38d3a208f12c1e38302859c.png)
Since 
is a triple root to the polynomial, the only solution to the system of equations is 
, and plugging the values back in satisfies the system.
See Also
| 2002 Indonesia MO (Problems) | ||
| Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 • 7 | Followed by Problem 4 |
| All Indonesia MO Problems and Solutions | ||
