A Pre-RMO question! -5

Given that $x = \dfrac {4ab}{a+b} \implies \left(\dfrac 1a+\dfrac 1b\right)x = 4$ . Then we have:

$\begin{aligned} \left(\frac {x+2a}{x-2a} + \frac {x+2b}{x-2b} \right)^2 & = \left(\frac {\left(\frac 1a+\frac 1b\right)(x+2a)}{\left(\frac 1a+\frac 1b\right)(x-2a)} + \frac {\left(\frac 1a+\frac 1b\right)(x+2b)}{\left(\frac 1a+\frac 1b\right)(x-2b)} \right)^2 \\ & = \left(\frac {4+2+\frac {2a}b}{4-2-\frac {2a}b} + \frac {4+2+\frac {2b}a}{4-2-\frac {2b}a} \right)^2 \\ & = \left(\frac {6b+2a}{2b-2a} + \frac {6a+2b}{2a-2b} \right)^2 \\ & = \left(\frac {3b+a-3a-b}{b-a} \right)^2 \\ & = \left(\frac {2b-2a}{b-a} \right)^2 = 2^2 = \boxed 4 \end{aligned}$

I suppose it is cheating, but for the purposes of this test we know the answer is a numerical one independent of x,a, and b. Take a=1, b=2. Then, x=3 and simple arithmetic shows the desired quantity is 4

$x = \frac{4ab}{a + b}$

Above expression can be rearranged in 2-ways

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$1^{st}$ way

$\frac{x}{2a} = \frac{2b}{a + b}$

Applying componendo and dividendo

$\frac{x + 2a}{x - 2a} = \frac{a + 3b}{b - a}$ ........ (1)

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$2^{nd}$ way

$\frac{x}{2b} = \frac{2a}{a + b}$

Applying componendo and dividendo

$\frac{x + 2b}{x - 2b} = \frac{3a + b}{a - b}$ ........ (2)

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Adding (1) and (2)

$\frac{x + 2a}{x - 2a} + \frac{x + 2b}{x - 2b} = \frac{a + 3b}{b - a} + \frac{3a + b}{a - b}$

$\frac{x + 2a}{x - 2a} + \frac{x + 2b}{x - 2b} = 2$

$\boxed{(\frac{x + 2a}{x - 2a} + \frac{x + 2b}{x - 2b}) ^ 2 = 4}$

We can write $\frac{x}{2a}=\frac{2b}{a+b}\implies \frac{x+2a}{x-2a}=\frac{a+3b}{b-a}$ .

Also $\frac{x}{2b}=\frac{2a}{a+b}\implies \frac{x+2b}{x-2b}=\frac{3a+b}{a-b}$ .

So, the given expression is equal to

$\left (\frac{3a+b-a-3b}{a-b}\right) ^2=\boxed 4$ .