Wednesday, May 27, 2015

2015 IMO contest problem: If$x,\,y,\,z$ are real numbers such that $x+2y+3z=6$ and $x^2+4y^2+9z^2=12$, evaluate $xyz$.

If$x,\,y,\,z$ are real numbers such that $x+2y+3z=6$ and $x^2+4y^2+9z^2=12$, evaluate $xyz$.

There is something that this Olympiad problem might trick us because it's obvious that $x^2,\,4y^2,\,9z^2$ are squares of $x,\,2y,\,3z$ and hence, one has reason to believe that the proper first step in solving this problem is to square the first given equation:

$x+2y+3z=6$

$(x+2y+3z)^2=(6)^2$

$x^2+(2y)^2+(3z)^2+2((x)(2y)+(x)(3z)+(2y)(3z))=36$

$x^2+4y^2+9z^2+4xy+6xz+12yz=36$

$12+4xy+6xz+12yz=36$

$2xy+3xz+6yz=12$

Ops...how could we get the value for $xyz$ from the equation above?

We were happy initially that we thought our first step led us in the right direction but it appears like we have met a situation that there is no immediate best next step to proceed...but we don't give up easily, we might want to continue because we hold fast to the principle that any given mathematical Olympiad problem has more than one way of solving, so we continue with what we have started:

$x(2xy+3xz+6yz)=12x$

$2x^2y+3x^2z+6xyz=12x$-(1)

$y(2xy+3xz+6yz)=12y$

$2xy^2+6y^2z+3xyz=12y$-(2)

$z(2xy+3xz+6yz)=12z$

$3xz^2+6z^2y+2xyz=12z$-(3)

Adding the three equations, we get:

$2x^2y+3x^2z+2xy^2+6y^2z+3xz^2+6z^2y+(6xyz+3xyz+2xyz)=12x$

$2x^2y+3x^2z+2xy^2+6y^2z+3xz^2+6z^2y+11xyz=12x$

Sigh...this approach seems unproductive. As a critical thinker, we have to have the ability when to quit when we have encountered with the wrong approach. We need to change our direction so we head in the right path.

Looking back to the given equations, something is amiss, we have only two equations when there are three variables in the equations, albeit we're asked to find not for the value for each variable, but their product, this might suggest we don't have to solve for the equations for all the three variables.

Ah! We got an idea! Inequality theorem might help!!!

We could use the Cauchy Schwarz inequality on the sum $x+2y+3z$ with some algebraic trick:

Cauchy Schwarz inequality says:

$a_1b_1+a_2b_2\le \sqrt{a_1^2+a_2^2}\sqrt{b_1^2+b_2^2}$

Also, note that we could multiply each of the term on the sum $x+2y+3z$ by $1$ without changing the value given sum since $x(1)+2y(1)+3z(1)=x+2y+3z$,

Therefore we have:

$x(1)+2y(1)+3z(1)\le \sqrt{x^2+(2y)^2+(3z)^2}\sqrt{1^2+1^2+1^2}$

$x+2y+3z\le \sqrt{x^2+4y^2+9z^2}\sqrt{3}$

Since we're told $x+2y+3z=6$ and $x^2+4y^2+9z^2=12$, by replacing the values into the inequality above we have:

$6\le \sqrt{12}\sqrt{3}$

We can conclude that equality must hold.

That means $x=2y=3z=2$ and hence $xyz=2(1)\left(\dfrac{2}{3}\right)=\dfrac{4}{3}$ and we're done!