Difference between revisions of "1973 USAMO Problems/Problem 4"

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==Solution==
 
==Solution==
Let <math>x</math>, <math>y</math>, and <math>z</math> be the roots of the cubic <math>x^3+ax^2+bx+c</math>. Let <math>S_1=x+y+z=3</math>, <math>S_2=x^2+y^2+z^2=3</math>, and <math>S_3=x^3+y^3+z^3=3</math>. From this, <math>S_1+a=0</math>, <math>S_2+aS_1+b=0</math>, and <math>S_3+aS_2+bS_1+c=0</math>. Solving each of these, <math>a=-3</math>, <math>b=3</math>, and <math>c=-1</math>. Thus <math>x</math>, <math>y</math>, and <math>z</math> are the roots of the polynomial <math>x^3-3x^2+3x-1=(x-1)^3</math>. Thus <math>x+y+z=1</math>, and there are no other solutions.
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Let <math>x</math>, <math>y</math>, and <math>z</math> be the roots of the cubic <math>t^3+at^2+bt+c</math>. Let <math>S_1=x+y+z=3</math>, <math>S_2=x^2+y^2+z^2=3</math>, and <math>S_3=x^3+y^3+z^3=3</math>. From this, <math>S_1+a=0</math>, <math>S_2+aS_1+b=0</math>, and <math>S_3+aS_2+bS_1+c=0</math>. Solving each of these, <math>a=-3</math>, <math>b=3</math>, and <math>c=-1</math>. Thus <math>x</math>, <math>y</math>, and <math>z</math> are the roots of the polynomial <math>t^3-3t^2+3t-1=(t-1)^3</math>. Thus <math>x=y=z=1</math>, and there are no other solutions.
  
 
==See also==
 
==See also==

Revision as of 14:33, 4 October 2008

Problem

Determine all the roots, real or complex, of the system of simultaneous equations

$x+y+z=3$,

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

$x^3+y^3+z^3=3$.

Solution

Let $x$, $y$, and $z$ be the roots of the cubic $t^3+at^2+bt+c$. Let $S_1=x+y+z=3$, $S_2=x^2+y^2+z^2=3$, and $S_3=x^3+y^3+z^3=3$. From this, $S_1+a=0$, $S_2+aS_1+b=0$, and $S_3+aS_2+bS_1+c=0$. Solving each of these, $a=-3$, $b=3$, and $c=-1$. Thus $x$, $y$, and $z$ are the roots of the polynomial $t^3-3t^2+3t-1=(t-1)^3$. Thus $x=y=z=1$, and there are no other solutions.

See also

Newton's Sums

1973 USAMO (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5
All USAMO Problems and Solutions
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