IMO-2001-6

[Tahmid Hasan]

let $a>b>c>d$ be positive integers and suppose
$ac+bd=(b+d+a-c)(b+d-a+c)$.
prove that $ab+cd$ is composite.[easiest IMO-6 of 21-th century from my point of view ]

[sourav das]

First of all, feeling easy and hard is relative. And i think until you see the solution, the intuitions are really unique. So in my point of view, every IMO problem is quite unique in contest and has a beautiful solution.

The real trick (I'm consenting that I've seen this part from an article)

i)Our given equation becomes $a^2+c^2-ac=b^2+d^2+bd$

A very familiar form is $a^2+c^2-2ac$ $cos60$=$b^2+d^2-2bd$ $cos120$

So, why don't we transform our $N.T.$ problem to a $Geometry$ problem or rather related to geometry?

With the help of this hint, i have solved the problem. So, everyone should work alone after this hint...

[FahimFerdous]

I totally agree with Sourav. I also used the hint that Sourav gave and solved the problem. I got the hint from 'How To Solve It'. But it's true that the official solution is much more beautiful and cruxy (a new adjective maybe) than this one and needs unique ideas.

[Tahmid Hasan]

oops,i forgo post post a reply . actually i used the same lemma as well.
another lemma i used was if $\frac {a^2+b^2-c^2-d^2}{cos x}$ is a natural number for $0^{\circ}<x<180^{\circ}$ then $cos x$ is rational.
@Sourav vaiya,could you please share a link to the article if possible?

[FahimFerdous]

http://www.math.ust.hk/excalibur/v12_n1.pdf