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

Revision as of 09:22, 8 October 2007

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

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

We take note of the obvious solution: (1,1,1). Now we manipulate the equations a bit:

$(x+y+z)^2-(x^2+y^2+z^2)=2xy+2xz+2yz=6$

$(x+y+z)(x^2+y^2+z^2)=x^3+y^3+z^3+x^2(y+z)+y^2(x+z)+z^2(y+x)=9=3+x^2(y+z)+y^2(x+z)+z^2(y+x)$

$(x+y+z)^3=x^3+y^3+z^3+3(x^2(y+z)+y^2(x+z)+z^2(y+x))+6xyz=3+18+6xyz=27$

$x+y+z=3$

$xy+xz+zy=3$

$xyz=1$

Therefore, x, y, and z are the roots of

$a^3-3a^2+3a-1=(a-1)^3$

Therefore, the only solution to those three equations is (1,1,1). The sum of the squares of the roots is $1^2+1^2+1^2=3$

This problem was given to 4everwise by a friend, Henry Tung. Upper classmen bullying freshmen. (Just kidding; it's a nice problem. )

@@ Line 2: / Line 2: @@
 Find the sum of the squares of the roots, real or complex, of the system of simultaneous equations
-<math>\displaystyle x+y+z=3,~x^2+y^2+z^2=3,~x^3+y^3+z^3 =3.</math>
+<math>x+y+z=3,~x^2+y^2+z^2=3,~x^3+y^3+z^3 =3.</math>
 ==Solution==
-{{solution}}
+We take note of the obvious solution: (1,1,1). Now we manipulate the equations a bit:
+<math>(x+y+z)^2-(x^2+y^2+z^2)=2xy+2xz+2yz=6</math>
+<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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