## Monday, October 5, 2015

### IMO Algebra Problem: Find maximum of a₁+a₂+a₃+a₄-a₁a₂-a₁a₃-a₁a₄-a₂a₃-a₂a₄-a₃a₄+a₁a₂a₃+a₁a₂a₄+a₁a₃a₄+a₂a₃a₄-a₁a₂a₃a₄

Find the maximum of

$a_1+a_2+a_3+a_4-a_1a_2-a_1a_3-a_1a_4-a_2a_3-a_2a_4-a_3a_4+a_1a_2a_3+a_1a_2a_4+a_1a_3a_4+a_2a_3a_4-a_1a_2a_3a_4$

where $|a_i|\le 1,\,i=1,\,2,\,3,\,4$.

Whop! We're not given anything and we're required to determine the minimum value of the given expression that contains as much as 4 variables in it! That's outrageous, you might reason.

Perhaps the test creators wanted to see if we're bold enough to tell them back this problem couldn't be solved, you whisper. Oh, you can point it out and whisper all you want, but, this is an authentic Mathematics Olympiad problem, so, it is a valid problem and could be solved neatly with some algebraic skills.

But, you would stamp your foot in frustration, saying you're not "math" person, and so you're not born with knowing how to algebraically manipulating what you have, and are not given to assist in your solution.

Hey, I already mention thousand times before that you don't have to be a born "genius" to solve hard and intriguing math problems effectively, I strongly and firmly believe that everyone is capable of learning math. Learning how to become an efficient and effective problem solver is something that we can learn, just like we learn how to type on keyboards!

There is however one criteria that helps you to get out of the problem. It is about your determination and persistence, never giving up and sticking with it that count.

Let's not beat about the bush and cut to the chase.

We first let

$P=a_1+a_2+a_3+a_4-a_1a_2-a_1a_3-a_1a_4-a_2a_3-a_2a_4-a_3a_4+a_1a_2a_3$

$\,\,\,\,\,\,\,\,\,\,\,+a_1a_2a_4+a_1a_3a_4+a_2a_3a_4-a_1a_2a_3a_4$

We then try to see if we could factor $P$. Well, that we are less sure if after factorization (if only it could be done), what benefit and good can come of it, but, that is what we could do now, try to factor instead of staring at the problem mindlessly and hopefully we could gain some perspective from there.

Speaking of factorization, the first and foremost thing is to factor out the common factor after appropriate groupings have been done to the expression that we wanted to factor.

Look at what we have now,

$a_1+a_2+a_3+a_4-a_1a_2-a_1a_3-a_1a_4-a_2a_3-a_2a_4-a_3a_4+a_1a_2a_3+a_1a_2a_4+a_1a_3a_4+a_2a_3a_4-a_1a_2a_3a_4$

how can we group "appropriately" in this case, we've never encountered expression such as this before. It's trivial to factor $a^2+2ab+b^2=a^2+ab+ab+b^2=a(a+b)+b(a+b)=(a+b)^2$, but it seems virtually impossible to factor $P$.

Shh...didn't I tell you not to gracefully say No in all circumstances when it comes to IMO problems? Most of the time, nothing creative, productive, or worthwhile happens after "NO". No discourages thinking and when you stop thinking, you will develop a rusty mind and you eventually become a mindless nattering yes man to the suggestions that thrown at you. You lose the ability to judge.

I agree that this is a very hard task to factor something you are so unfamiliar with, and you could try anything that you wanted to group, sometimes, goofing around with silly attempts grant us brilliant ideas.

$P$

$=a_1+a_2+a_3+a_4-a_1a_2-a_1a_3-a_1a_4-a_2a_3-a_2a_4-a_3a_4+a_1a_2a_3$
$\,\,\,\,\,\,a_1a_2a_4+a_1a_3a_4+a_2a_3a_4-a_1a_2a_3a_4$

$=a_1(1-a_2)+a_2(1-a_4)+a_3(1-a_2)+a_4(1-a_1)-a_1a_3-a_2a_4+a_1a_2a_3$
$\,\,\,\,\,\,+a_1a_2a_4+a_1a_3a_4+a_2a_3a_4-a_1a_2a_3a_4$

$=a_1(1-a_2)+a_2(1-a_4)+a_3(1-a_2)+a_4(1-a_1)-a_1a_3(1-a_2-a_4)$
$\,\,\,\,\,\,-a_2a_4(1-a_1-a_3)-a_1a_2a_3a_4$

$=(1-a_2)(a_1+a_3-a_1a_3)+(1-a_4)(a_2)+(1-a_1)(a_4-a_2a_4)+a_1a_3a_4$
$\,\,\,\,\,\,+a_2a_4a_3-a_1a_2a_3a_4$

Aww, what we have gotten so far is so ugly, but we never say no, we try again for good!

$P$

$=a_1+a_2+a_3+a_4-a_1a_2-a_1a_3-a_1a_4-a_2a_3-a_2a_4-a_3a_4+a_1a_2a_3$
$\,\,\,\,\,\,+a_1a_2a_4+a_1a_3a_4+a_2a_3a_4-a_1a_2a_3a_4$

$=a_1(1-a_2-a_3-a_4+a_2a_3+a_2a_4+a_3a_4)+a_2(1-a_3-a_4+a_3a_4)$
$\,\,\,\,\,\,+a_3(1-a_4)+a_4-a_1a_2a_3a_4$

$=a_1((1-a_2)(1-a_3)-a_4+a_2a_4+a_3a_4)+a_2(1-a_3)(1-a_4)+a_3(1-a_4)$
$\,\,\,\,\,\,+a_4-a_1a_2a_3a_4$

$=a_1(1-a_2)(1-a_3)-a_1a_4(1-a_2-a_3)+a_2(1-a_3)(1-a_4)+a_3(1-a_4)+a_4$
$\,\,\,\,\,\,-a_1a_2a_3a_4$

$=a_1(1-a_2)(1-a_3)-a_1a_4(1-a_2-a_3+a_2a_3)+a_2(1-a_3)(1-a_4)$
$\,\,\,\,\,\,+a_3(1-a_4)+a_4$

$=a_1(1-a_2)(1-a_3)-a_1a_4(1-a_2)(1-a_3)+a_2(1-a_3)(1-a_4)+a_3(1-a_4)$
$\,\,\,\,\,\,+a_4$

$=a_1(1-a_2)(1-a_3)-a_1a_4(1-a_2)(1-a_3)+a_2(1-a_3)(1-a_4)+a_3-a_3a_4$
$\,\,\,\,\,\,+a_4$

$=a_1(1-a_2)(1-a_3)-a_1a_4(1-a_2)(1-a_3)+a_2(1-a_3)(1-a_4)$
$\,\,\,\,\,\,-(a_3a_4-a_3-a_4+1)+1$

$=a_1(1-a_2)(1-a_3)-a_1a_4(1-a_2)(1-a_3)+a_2(1-a_3)(1-a_4)$
$\,\,\,\,\,\,-(1-a_3)(1-a_4)+1$

$=a_1(1-a_2)(1-a_3)-a_1a_4(1-a_2)(1-a_3)-(1-a_3)(1-a_4)(1-a_2)+1$

$=a_1(1-a_2)(1-a_3)-(1-a_2)(1-a_3)(a_1a_4+1-a_4)+1$

$=a_1(1-a_2)(1-a_3)-(1-a_2)(1-a_3)(a_1a_4+1-a_4-a_3+a_3)+1$

$=a_1(1-a_2)(1-a_3)-(1-a_2)(1-a_3)(a_1a_4+1-a_4-a_1)$
$\,\,\,\,\,\,-a_1(1-a_2)(1-a_3)+1$

$=\cancel{a_1(1-a_2)(1-a_3)}-(1-a_2)(1-a_3)(1-a_1)(1-a_4)$
$\,\,\,\,\,\,-\cancel{a_1(1-a_2)(1-a_3)}+1$

$=1-(1-a_1)(1-a_2)(1-a_3)(1-a_4)$

Wow, this is incredibly beautiful form of $P$ that we know it is always less than or equal to $1$.

Since we're told $|a_i|\le1,\,i=1,\,2,\,3,\,4$, and from the newest and beautiful form of $P=1-(1-a_1)(1-a_2)(1-a_3)(1-a_4)$, we won't ramble and say this problem is unapproachable, instead, we could conclude it now the maximum of $P$ is $1$ occurs at $a_1=a_2=a_3=a_4=1$.