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Difference between revisions of "2002 Indonesia MO Problems/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>
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Subtract the third equation and factor to get
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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
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<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>, and plugging the values back in satisfies the system.
 
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.
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==Solution 2==
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We can use Newton's Sums (https://artofproblemsolving.com/wiki/index.php/Newton%27s_Sums) to solve this problem -- we can say the three variables are roots to a cubic monic polynomial (so <math>a_n = a_3 = 1</math>). From the problem we have <math>S_1 = 6, S_2 = 12, S_3 = 24</math> and using Newton's Sums we have <cmath>6 + a_2 = 0\\ 12 + 6a_2 + 2a_1 = 0\\ 24 + 12a_2 + 6a_1 + 3a_0 = 0</cmath>We can find <math>a_2</math>, then <math>a_1, a_0</math> respectively to get the polynomial <cmath>x^3 - 6x^2 + 12x - 8 = 0</cmath> Using the Rational Root Theorem (or trial and error) we can easily find one of the roots -- <math>2</math>, and see that the other two roots are <math>2</math> as well (eg by factoring out <math>x-2</math>) yielding the only solution <math>\boxed{(2, 2, 2)}</math>.
  
 
==See Also==
 
==See Also==
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{{Indonesia MO 7p box|year=2002|num-b=2|num-a=4}}
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{{Indonesia MO box|year=2002|num-b=2|num-a=4|eight=}}
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Latest revision as of 22:38, 13 July 2022

Problem

Find all real solutions from the following system of equations:

$\left\{\begin{array}{l}x+y+z = 6\\x^2 + y^2 + z^2 = 12\\x^3 + y^3 + z^3 = 24\end{array}\right.$

Solution

Square the first equation to get \[x^2 + y^2 + z^2 + 2(xy + yz + xz) = 36\] Subtract the second equation from the result to get \[2(xy+yz+xz) = 24\] \[xy+yz+xz = 12\] Multiply the second equation by the first equation to get \[x^3 + xy^2 + xz^2 + x^2y + y^3 + yz^2 + x^2z + y^2z + z^3 = 72\] Subtract the third equation to get \[xy^2 + xz^2 + x^2y + yz^2 + x^2z + y^2z = 48\] Cube the first equation to get \[(x^3 + y^3 + z^3) + 3(x^2y + x^2z + xy^2 + y^2z + xz^2 + yz^2) + 6xyz = 216\] \[24 + 3(48) + 6xyz = 216\] \[168 + 6xyz = 216\] \[6xyz = 48\] \[xyz = 8\] If $x+y+z=6$, $xy+yz+xz = 12$, and $xyz = 8$, the solution triplet is the roots of the polynomial \[a^3 - 6a^2 + 12a - 8 = 0\] Factor the polynomial to get \[(a-2)^3 = 0\] Since $a = 2$ is a triple root to the polynomial, the only solution to the system of equations is $\boxed{(2,2,2)}$, and plugging the values back in satisfies the system.

Solution 2

We can use Newton's Sums (https://artofproblemsolving.com/wiki/index.php/Newton%27s_Sums) to solve this problem -- we can say the three variables are roots to a cubic monic polynomial (so $a_n = a_3 = 1$). From the problem we have $S_1 = 6, S_2 = 12, S_3 = 24$ and using Newton's Sums we have \[6 + a_2 = 0\\ 12 + 6a_2 + 2a_1 = 0\\ 24 + 12a_2 + 6a_1 + 3a_0 = 0\]We can find $a_2$, then $a_1, a_0$ respectively to get the polynomial \[x^3 - 6x^2 + 12x - 8 = 0\] Using the Rational Root Theorem (or trial and error) we can easily find one of the roots -- $2$, and see that the other two roots are $2$ as well (eg by factoring out $x-2$) yielding the only solution $\boxed{(2, 2, 2)}$.

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
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