Difference between revisions of "2002 Indonesia MO Problems/Problem 3"

Revision as of 23:09, 3 August 2018

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.

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 ==See Also==
-{{Indonesia MO 7p box|year=2002|num-b=2|num-a=4}}
+{{Indonesia MO box|year=2002|num-b=2|num-a=4|eight=}}
 [[Category:Intermediate Algebra Problems]]

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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Difference between revisions of "2002 Indonesia MO Problems/Problem 3"

Revision as of 23:09, 3 August 2018

Problem

Solution

See Also