Friday, April 10, 2015


Well, on my last blog post, I have shown with a strong Olympiad Math competition problem about how useful we have to always bear in mind that when $x$ is an solution to the function, says, $P(x)$, then we need to consider if $-x$ could also be a solution the that function of $P(x)$.

Fortunately that is not something that is hard to check. All that we need to do is to determine if $P(x)=P(-x)$.

I will show with one more challenging, hard and intriguing math competition problem why remembering to check with this fact could save our day from struggling with the problem.

Find all numbers of $m$ such that the equation $|x+|x|+m|+|x-|x|-m|=2$ has exactly three solutions.

If we replace $x$ by $\color{purple}{-x}$ to the given equation, we see that:



$|-[x-|x|-\color{black}{m}]|+|-[x+|x|+\color{black}{m}]|=2$ which is the exact opposite of the original equation $|x+|x|+m|+|x-|x|-m|=2$.

This is a giant discovery and now, we can say if $x$ is a solution to the given equation, then $-x$ is also a solution to that equation.

Thus, in order for the equation to have an odd number (3) of solutions, it is clear that $x=0$ must be one of these.

Substituting $x=0$ into the equation, we see that




$m=\pm 1$

If $m=1$ and $x\ge 0$, then the equation becomes

$|2x+1|+1=2$ and so $x=0$

But $-x$ is a solution iff $x$ is, so there is only one solution in this case.

If $m=-1$ and $x\ge 0$, then the equation becomes

$|2x-1|+1=2$ and so $x=0,\,1$

Thus, in this case, we have three solutions where $x=-1,\,0,\,1$

$\therefore m=-1$ is the unique number for which the equation has exactly three solutions.

See it now? If we have not thought from the angle where both $x$ and $-x$ are the solution for any given equation provided $P(x)=P(-x)$, then we would most likely still be biting our knuckles and pulling our hair out, without knowing how to approach this problem efficient and effectively.

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