## Thursday, April 30, 2015

### Algebraic Method to Tackle the Mock APMO Problem

There exists another way to tackle the previously discussed AMPO mock problem (Asian Pacific Mathematics Olympiad Mock Problem ).

Find [MATH]\sum_{x=0}^{101}\dfrac{\dfrac{2x}{101}-1}{\dfrac{3x^2}{10201}-\dfrac{3x}{101}+1}[/MATH].

In case you are not well prepared to attack the problem analytically, you could still tackle it algebraically, that is purely allowable and no one will ever say algebraic method is not awesome!

For simplicity's sake, we let $x_i=\dfrac{i}{101}$ and $f(x)=\dfrac{\dfrac{2x}{101}-1}{\dfrac{3x^2}{10201}-\dfrac{3x}{101}+1}$, we then have:

$f(x)=\dfrac{\dfrac{2x}{101}-1}{\dfrac{3x^2}{10201}-\dfrac{3x}{101}+1}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{2\left(\dfrac{x}{101}\right)-1}{3\left(\dfrac{x}{101}\right)^2-3\left(\dfrac{x}{101}\right)+1}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{2x_i-1}{3x_i^2-3x_i+1}$

Furthermore, if we replace $x_i$ by $1-x_i$, we see that:

$f(x)=\dfrac{2x_i-1}{3x_i^2-3x_i+1}$

$f(1-x_i)=\dfrac{2(1-x_i)-1}{3(1-x_i)^2-3(1-x_i)+1}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{2-2x_i-1}{3x_i^2-6x_i+3-3+3x_i+1}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{1-2x_i}{3x_i^2-3x_i+1}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,=-\left(-\dfrac{1-2x_i}{3x_i^2-3x_i+1}\right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,=-\left(\dfrac{2x_i-1}{3x_i^2-3x_i+1}\right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,=-f(x_i)$

Thus, what we can conclude by now is that we have proved that $f(x_i)+f(1-x_i)=0$

Our next step is to make full use of this piece of valuable data and also $x_i=\dfrac{i}{101}$:

1.

Observe that $1-\dfrac{i}{101}=\dfrac{101-i}{101}$, this tells us $1-x_i=x_{101-i}$

2.

We let

[MATH]S=\sum_{x=0}^{101}\dfrac{\dfrac{2x}{101}-1}{\dfrac{3x^2}{10201}-\dfrac{3x}{101}+1}=\sum_{x=0}^{101}\dfrac{2x_i-1}{3x_i^2-3x_i+1}=\sum_{x=0}^{101} f(x_i)--(*)[/MATH]

We can also have

[MATH]S=\sum_{x=0}^{101} f(1-x_i)--(**)[/MATH]

3.

By adding the two equations (*) and (**) up yields:

[MATH]2S=\sum_{x=0}^{101} (f(x_i)+f(1-x_i))[/MATH]

[MATH]\,\,\,\,\,\,\,\,\,=\sum_{x=0}^{101} (f(x_i)+f(1-x_i))[/MATH]

[MATH]\,\,\,\,\,\,\,\,\,=\sum_{x=0}^{101} (0)[/MATH]

4.

Therefore,

[MATH]S=\sum_{x=0}^{101}\dfrac{\dfrac{2x}{101}-1}{\dfrac{3x^2}{10201}-\dfrac{3x}{101}+1}=0[/MATH].