## Sunday, March 12, 2006

### Fig Newtons. Topic: Algebra/Polynomials. Level: AIME/Olympiad.

Problem: (2003 MOP Lecture, Reid Barton) If $x < y < z$ are real numbers such that $x+y+z = 5$, $x^2+y^2+z^2 = 11$, and $x^3+y^3+z^3 = 26$, find $y$.

Solution: Consider the monic polynomial $P(a)$ of degree $3$ with roots $x,y,z$.

To find the coefficients, we use Newton Sums:

$x+y+z = 5$
$xy+yz+zx = \frac{(x+y+z)^2-(x^2+y^2+z^2)}{2} = \frac{25-11}{2} = 7$.
$xyz = \frac{(x+y+z)^3-3(x+y+z)(x^2+y^2+z^2)+2(x^3+y^3+z^3)}{6} = \frac{125-3 \cdot 5 \cdot 11+2 \cdot 26}{6} = 2$.

So $P(a) = a^3-5a^2+7a-2 = (a-2)(a^2-3a+1)$. Since the roots of the quadratic are

$a = \frac{3 \pm \sqrt{5}}{2}$,

we know that $y = 2$ because exactly one of the other roots is bigger than it and exactly one is smaller. QED.

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Comment: Newton Sums are a powerful idea and can help solve systems of equations very easily. Usually they are an intermediate step, but in the following USAMO problem Newton Sums give you the solution immediately.

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Practice Problem: (1973 USAMO - #4) Find all complex solutions to the system of equations

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