## Wednesday, May 27, 2015

### IMO Problem: If $\dfrac{(x-y)(y-z)(z-x)}{(x+y)(y+z)(z+x)}=\dfrac{2014}{2015}$, evaluate $\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}$.

If $\dfrac{(x-y)(y-z)(z-x)}{(x+y)(y+z)(z+x)}=\dfrac{2014}{2015}$, evaluate $\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}$.

We all know in solving this problem correctly, our method should focus on turning the given equation so that the LHS of the equation could be represented by the wanted expression. That is, we hope to turn $\dfrac{(x-y)(y-z)(z-x)}{(x+y)(y+z)(z+x)}=\dfrac{2014}{2015}$ so the LHS can take the form $\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}$.

Let us try that out now:

$\dfrac{(x-y)(y-z)(z-x)}{(x+y)(y+z)(z+x)}=\dfrac{2014}{2015}$

$\dfrac{x^2(z-y)+y^2(x-z)+z^2(y-x)}{(x+y)(y+z)(z+x)}=\dfrac{2014}{2015}$

$\dfrac{x^2(z-y)}{(x+y)(y+z)(z+x)}+\dfrac{y^2(x-z)}{(x+y)(y+z)(z+x)}+\dfrac{z^2(y-x)}{(x+y)(y+z)(z+x)}=\dfrac{2014}{2015}$

This looks messy, we should stop proceeding with this approach at this point.

Okay, why not we tried to cross multiply the given equation and see where that leads us?

$\dfrac{(x-y)(y-z)(z-x)}{(x+y)(y+z)(z+x)}=\dfrac{2014}{2015}$

$2015(x-y)(y-z)(z-x)=2014(x+y)(y+z)(z+x)$

$2015(x^2(z-y)+y^2(x-z)+z^2(y-x))=2014(x^2(z+y)+y^2(x+z)+z^2(y+x))$

$x^2(2015z-2015y-2014z-2014y)+y^2(2015x-2015z-2014x-2014z)+z^2(2015y-2015x-2014y-2014x)=0$

$x^2(z-4029y)+y^2(x-4029z)+z^2(y-4029x)=0$

Not only does that not sound promising, it probably will give us more headache than ever if we continue working with the equation $x^2(z-4029y)+y^2(x-4029z)+z^2(y-4029x)=0$.

What does the expression want exactly? We might want to combine adding the three fractions:

$\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}$

$=\dfrac{x(y+z)(z+x)+y(x+y)(z+x)+z(x+y)(y+z)}{(x+y)(y+z)(z+x)}$

$=\dfrac{2xyz+x^2y+x^2z+xy^2+xz^2+y^2z+yz^2}{(x+y)(y+z)(z+x)}$

Unfortunately this looks necessarily complicated...

We should try some other approach instead. Substitution method is a great heuristic skill when we have encountered the right challenge contest problem.

If we let $a=x+y,\,b=y+z,\,c=z+x$, we should aware that:

1.

[MATH]\color{yellow}\bbox[5px,purple]{\begin{align*}a-c&=x+y-z-x\\&=y-z\end{align*}}[/MATH]

[MATH]\color{yellow}\bbox[5px,purple]{\begin{align*}b-a&=y+z-x-y\\&=z-x\end{align*}}[/MATH]

[MATH]\color{yellow}\bbox[5px,purple]{\begin{align*}c-b&=z+x-y-z\\&=x-y\end{align*}}[/MATH]

and

2.

[MATH]\color{yellow}\bbox[5px,green]{\begin{align*}a-c+b&=x+y-z-x+y+z\\&=2y\end{align*}}[/MATH]

[MATH]\color{yellow}\bbox[5px,green]{\begin{align*}b-a+c&=y+z-x-y+z+x\\&=2z\end{align*}}[/MATH]

[MATH]\color{yellow}\bbox[5px,green]{\begin{align*}c-b+a&=z+x-y-z+x+y\\&=2x\end{align*}}[/MATH]

Therefore, the given equation becomes:

$\dfrac{(x-y)(y-z)(z-x)}{(x+y)(y+z)(z+x)}=\dfrac{2014}{2015}$

$\dfrac{(c-b)(a-c)(b-a)}{(a)(b)(c)}=\dfrac{2014}{2015}$

$\dfrac{(c-b)(a-c)(b-a)}{(a)(b)(c)}=\dfrac{2014}{2015}$

And our target expression becomes:

$\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}$

$=\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}$

$=\dfrac{c-b+a}{2a}+\dfrac{a-c+b}{2b}+\dfrac{b-a+c}{2c}$

$=\dfrac{a}{2a}+\dfrac{b}{2b}+\dfrac{c}{2c}+\dfrac{c-b}{2a}+\dfrac{a-c}{2b}+\dfrac{b-a}{2c}$

$=\dfrac{3}{2}+\dfrac{1}{2}\left(\dfrac{c-b}{a}+\dfrac{a-c}{b}+\dfrac{b-a}{c}\right)$

$=\dfrac{3}{2}+\dfrac{1}{2}\left(-\dfrac{(c-b)(a-c)(b-a)}{abc}\right)$

$=\dfrac{3}{2}-\dfrac{1}{2}\left(\dfrac{2014}{2015}\right)$

$=\dfrac{4031}{4030}$

I know many of you have one question that is bothering to the greatest, how do I know

$\dfrac{c-b}{a}+\dfrac{a-c}{b}+\dfrac{b-a}{c}=-\dfrac{(c-b)(a-c)(b-a)}{abc}$ instead?

I memorized it, hehehe...

It's worth to mention here that

[MATH]\color{yellow}\bbox[5px,orange]{(c-b)(a-c)(b-a)=a^2b-a^2c-ab^2+ac^2+b^2c-bc^2}[/MATH]

and

[MATH]\color{yellow}\bbox[5px,blue]{\begin{align*}(c-b)bc+(a-c)ac+(b-a)ab &=-a^2b+a^2c+ab^2-ac^2-b^2c+bc^2\\&=-(a^2b-a^2c-ab^2+ac^2+b^2c-bc^2)\\&=-(c-b)(a-c)(b-a)\end{align*}}[/MATH]

Therefore, if we have

$\dfrac{c-b}{a}+\dfrac{a-c}{b}+\dfrac{b-a}{c}$

It can be added so to get

$\dfrac{c-b}{a}+\dfrac{a-c}{b}+\dfrac{b-a}{c}$

$=\dfrac{(c-b)bc+(a-c)ac+(b-a)ab}{abc}$

$=\dfrac{-(c-b)(a-c)(b-a)}{abc}$