**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 $.

x,y,z are the roots of f(a)=(a-1)^3 -> x=y=z=1.

ReplyDeleteSome olympiad problems are too easy.

these olympiad problems can be really neat

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