Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 11"

Latest revision as of 20:55, 20 October 2019

Find the sum of the squares of the roots, real or complex, of the system of simultaneous equations

$x+y+z=3,~x^2+y^2+z^2=3,~x^3+y^3+z^3 =3.$

The roots are $x$ , $y$ , and $z$ , and we add the squares:

$x^2+y^2+z^2=\boxed{003}$

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 ==Solution==
-We take note of the obvious solution: (1,1,1). Now we manipulate the equations a bit:
+The roots are <math>x</math>, <math>y</math>, and <math>z</math>, and we add the squares:
+<cmath>x^2+y^2+z^2=\boxed{003}</cmath>
+==See Also==
-<math>(x+y+z)^2-(x^2+y^2+z^2)=2xy+2xz+2yz=6</math>
+http://www.artofproblemsolving.com/Wiki/index.php/1973_USAMO_Problems/Problem_4
+{{Mock AIME box|year=2006-2007|n=2|num-b=10|num-a=12}}
-<math>(x+y+z)(x^2+y^2+z^2)=x^3+y^3+z^3+x^2(y+z)+y^2(x+z)+z^2(y+x)=9=3+x^2(y+z)+y^2(x+z)+z^2(y+x)</math>
-<math>(x+y+z)^3=x^3+y^3+z^3+3(x^2(y+z)+y^2(x+z)+z^2(y+x))+6xyz=3+18+6xyz=27</math>
-<math>x+y+z=3</math>
-<math>xy+xz+zy=3</math>
-<math>xyz=1</math>
-Therefore, x, y, and z are the roots of
-<math>a^3-3a^2+3a-1=(a-1)^3</math>
-Therefore, the only solution to those three equations is (1,1,1). The sum of the squares of the roots is <math>1^2+1^2+1^2=3</math>
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-*[[Mock AIME 2 2006-2007/Problem 10 | Previous Problem]]
-*[[Mock AIME 2 2006-2007/Problem 12 | Next Problem]]
-*[[Mock AIME 2 2006-2007]]
-== Problem Source==
-This problem was given to 4everwise by a friend, Henry Tung. Upper classmen bullying freshmen. (Just kidding; it's a nice problem. [[Image:Razz.gif]])