## Tuesday, April 4, 2006

### Where Are They? Topic: Algebra. Level: Olympiad.

Problem: (Math Olympiad Treasures - 1.1.10) Find the locus of points $(x,y)$ for which

$x^3+y^3+3xy = 1$.

Solution: Recall the useful factorization

$a^3+b^3+c^3-3abc = \frac{1}{2}(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]$.

Setting $a = x$, $b = y$, and $c = -1$, we obtain

$x^3+y^3-1+3xy = \frac{1}{2}(x+y-1)[(x-y)^2+(y+1)^2+(-1-x)^2]$.

But from the given condition this expression is zero, so either

$x+y-1 = 0$,

which gives us the line $x+y = 1$ or

$(x-y)^2+(y+1)^2+(-1-x)^2 = 0$,

which gives us the point $(-1,-1)$.

So the locus of points is the line $x+y = 1$ and $(-1,-1)$. QED.

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Comment: The factorization basically finished the problem; it's nice to remember them because when they apply, they are extremely useful.

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Practice Problem: (Math Olympiad Treasures - 1.1.6) Let $a,b,c$ be distinct real numbers. Prove that the following equality cannot hold:

$\sqrt{a-b}+\sqrt{b-c}+\sqrt{c-a} = 0$.

1. Set x = cbrt(a-b), y = cbrt(b-c), z = cbrt(c-a). Then x^3 + y^3 + z^3 = 0. Moreover, from the factorization we have

-3xyz = 1/2(x + y + z)( (x-y)^2 + (y-z)^2 + (z-x)^2 ). The LHS cannot be zero because a, b, c are distinct, hence x + y + z \neq 0. QED.

2. How do you know all these factorizations? I've never even heard of that one.

3. Well I learned this one from the book that I got the problem from, but it's shown up a few times (there was a USAMTS question related to it last year?).