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

Revision as of 14:28, 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 $x^3+ax^2+bx+c$ . 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 $x^3-3x^2+3x-1=(x-1)^3$ . Thus $x+y+z=1$ , and there are no other solutions.

@@ Line 10: / Line 10: @@
 ==See also==
 [[Newton's Sums]]
-{{USAMO newbox|year=1973|num-b=3|num-a=5}}
+{{USAMO oldbox|year=1973|num-b=3|num-a=5}}
 [[Category:Olympiad Algebra Problems]]

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

Revision as of 14:28, 4 October 2008

Problem

Solution

See also