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Wednesday, May 27, 2015

IMO Problem: If (xy)(yz)(zx)(x+y)(y+z)(z+x)=20142015, evaluate xx+y+yy+z+zz+x.

If (xy)(yz)(zx)(x+y)(y+z)(z+x)=20142015, evaluate xx+y+yy+z+zz+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 (xy)(yz)(zx)(x+y)(y+z)(z+x)=20142015 so the LHS can take the form xx+y+yy+z+zz+x.

Let us try that out now:

(xy)(yz)(zx)(x+y)(y+z)(z+x)=20142015

x2(zy)+y2(xz)+z2(yx)(x+y)(y+z)(z+x)=20142015

x2(zy)(x+y)(y+z)(z+x)+y2(xz)(x+y)(y+z)(z+x)+z2(yx)(x+y)(y+z)(z+x)=20142015

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?

(xy)(yz)(zx)(x+y)(y+z)(z+x)=20142015

2015(xy)(yz)(zx)=2014(x+y)(y+z)(z+x)

2015(x2(zy)+y2(xz)+z2(yx))=2014(x2(z+y)+y2(x+z)+z2(y+x))

x2(2015z2015y2014z2014y)+y2(2015x2015z2014x2014z)+z2(2015y2015x2014y2014x)=0

x2(z4029y)+y2(x4029z)+z2(y4029x)=0

Not only does that not sound promising, it probably will give us more headache than ever if we continue working with the equation x2(z4029y)+y2(x4029z)+z2(y4029x)=0.

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

xx+y+yy+z+zz+x

=x(y+z)(z+x)+y(x+y)(z+x)+z(x+y)(y+z)(x+y)(y+z)(z+x)

=2xyz+x2y+x2z+xy2+xz2+y2z+yz2(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.

ac=x+yzx=yz

ba=y+zxy=zx

cb=z+xyz=xy

and

2.

ac+b=x+yzx+y+z=2y

ba+c=y+zxy+z+x=2z

cb+a=z+xyz+x+y=2x

Therefore, the given equation becomes:

(xy)(yz)(zx)(x+y)(y+z)(z+x)=20142015

(cb)(ac)(ba)(a)(b)(c)=20142015

(cb)(ac)(ba)(a)(b)(c)=20142015

And our target expression becomes:

xx+y+yy+z+zz+x

=xa+yb+zc

=cb+a2a+ac+b2b+ba+c2c

=a2a+b2b+c2c+cb2a+ac2b+ba2c

=32+12(cba+acb+bac)

=32+12((cb)(ac)(ba)abc)

=3212(20142015)

=40314030

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

cba+acb+bac=(cb)(ac)(ba)abc instead?

I memorized it, hehehe...

It's worth to mention here that

(cb)(ac)(ba)=a2ba2cab2+ac2+b2cbc2

and

(cb)bc+(ac)ac+(ba)ab=a2b+a2c+ab2ac2b2c+bc2=(a2ba2cab2+ac2+b2cbc2)=(cb)(ac)(ba)

Therefore, if we have

cba+acb+bac

It can be added so to get

cba+acb+bac

=(cb)bc+(ac)ac+(ba)ababc

=(cb)(ac)(ba)abc

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