## Friday, June 16, 2006

### Mission Accomplished! Topic: Algebra. Level: AIME.

Problem: (1991 India National Olympiad - #7) Solve the following system for real $x, y, z$

$x+y-z = 4$

$x^2-y^2+z^2 = -4$

$xyz = 6$.

Solution: Well, we simply go about solving this system the regular way - look for something to eliminate. We try using the first and second equations and find

$(x-z)^2 = (4-y)^2 \Rightarrow x^2-2xz+z^2 = 16-8y+y^2$,

which we subtract from the second equation to get

$2xz = 8y-20$.

Using the third equation, we have $2xz = \frac{12}{y}$ so we substitute, multiply through by $y$ (it can't be zero), and divide by $4$ to get

$0 = 2y^2-5y-3 = (2y+1)(y-3)$.

So $y = -\frac{1}{2}, 3$. But from the second equation, we see that $y = -\frac{1}{2}$ gives $x^2+z^2 < 0$ but they are real so this is impossible. Hence $y = 3$. It remains to solve for $x$ and $z$ using

$x-z = 1$

$xz = 2$.

Substitute $z = \frac{2}{x}$ into the first one and again multiply through by $x$ to find

$x^2-x-2 = (x-2)(x+1) = 0$.

We then have $x = 2, -1$ with corresponding $z = 1, -2$. Checking, we see that both solutions work, so our final solution set is

$(2, 3, 1)$; $(-1, 3, -2)$.

QED.

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Comment: Not bad for an olympiad question - just requires basic algebraic manipulation and solving quadratics. Looking for the right things to square and substitute made this problem considerably easier.

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Practice Problem: (1994 India National Olympiad - #2) If $x^5-x^3+x = a$, prove that $x^6 \ge 2a-1$.

#### 1 comment:

1. We wish to show

x^6 - 2x^5 + 2x^3 - 2x + 1 \ge 0

This factors as

(x-1)^2 (x^4 - x^2 + 1) \ge (x-1)^2 (x^4 - 2x^2 + 1) \ge (x-1)^2 (x^2 - 1)^2 \ge 0

QED.

(India MO problems are a little on the easy side, as far as MO problems go...)