## Saturday, June 10, 2006

### Five Whole Numbers. Topic: Algebra/Polynomials. Level: AIME.

Problem: (1999 JBMO - #1) Let $a, b, c, x, y$ be five real numbers such that $a^3+ax+y = 0$, $b^3+bx+y = 0$, and $c^3+cx+y = 0$. If $a, b, c$ are all distinct numbers, prove that their sum is zero.

Solution: Consider the polynomial $f(t) = t^3+tx+y$. Since it is a third degree polynomial, it can have at most three real roots. But we are given that

$f(a) = 0$, $f(b) = 0$, $f(c) = 0$

so we know that it in fact has exactly three distinct real roots and no other ones. But by Vieta's Formulas, we know that the sum of the roots is the coefficient of the $t^2$ term, which is zero. Thus

$a+b+c = 0$,

as desired. QED.

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Comment: A pretty simple problem to be on an olympiad, even if it is the Junior Balkan Math Olympiad (JBMO). The solution is almost immediate if you have worked with polynomials a lot.

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Practice Problem: (2000 JBMO - #1) Let $x$ and $y$ be positive reals such that

$x^3 + y^3 + (x + y)^3 + 30xy = 2000$.

Show that $x + y = 10$.

1. Uh, x^3+y^3 = (x+y)^3 - 3xy(x+y)

then just treat it as a polynomial in (x+y), and you're doneee.

2. Viewing the problem as a polynomial in t is a nice trick, I should learn to see those kinds of solutions XD

Anonymous: Your solution method can be generalized. In the general case we are presented with a condition symmetric in all of its variables, so we may substitute the symmetric polynomials u = x + y, v = xy. In this case, we are left with

u^3 - 3uv + u^3 + 30v = 2000
2u^3 + 3v(10-u) = 2000
u(2u^2 + 27v) = 2000

(Incidentally, I don't see how it's immediately obvious from here that u = 10 is the only real solution. Possibly I'm tired.) Let w = 10-u. Then

2(1000 - 300w + 30w^2 - w^3) + 3vw = 2000
2w^3 - 60w^2 + (600 - 3v) w = 0

w = 0 is an obvious root, and it's easy to show the other two are complex.

3. Actually, your expression factors into

(u - 10)(2u^2 +20u + 200 - 3v) = 0,

from which it isn't so difficult to show that the second term is always positive. Completing squares is one way to do it.

4. Whoa, okay, the third line of that first part is completely wrong. Not sure what I did there.