A surprising expression with nested square roots

 Simplify 

$$\frac{\sqrt{10 + \sqrt{1}} + \sqrt{10 + \sqrt{2}} + \dots + \sqrt{10 + \sqrt{99}}}{\sqrt{10 - \sqrt{1}} + \sqrt{10 - \sqrt{2}} + \dots + \sqrt{10 - \sqrt{99}}}$$

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The expression is equal to $\sqrt{2} + 1$.  Surprising!

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Let $a + b$ = 100. 

Now

$$\sqrt{10 - \sqrt{a}} \sqrt{10 + \sqrt{a}} = \sqrt{100 - a}  = \sqrt{b}$$

Let $u = \sqrt{10 + \sqrt{a}}$ and $v = \sqrt{10 - \sqrt{a}}$

So $20 = u^2 + v^2$

and $\sqrt{b} = uv$

Thus

$$\sqrt{20 + 2\sqrt{b}} = \sqrt{u^2 + v^2 + 2uv} = u+v$$

Similarly

$$\sqrt{20 - 2\sqrt{b}} = \sqrt{u^2 + v^2 - 2uv} = u-v$$

Thus if 

$$P = \sum_{a=1}^{99} \sqrt{10 + \sqrt{a}}$$

and

$$Q = \sum_{a=1}^{99} \sqrt{10 - \sqrt{a}}$$

We get 

$$\sqrt{2} P = P + Q$$ and

$$\sqrt{2} Q = P - Q$$

This gives us $$\frac{P}{Q} = \sqrt{2} + 1$$