hEatLove wrote: $ \frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+...+\sqrt{10+\sqrt{99}}}{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+...+\sqrt{10-\sqrt{99}}}=? $ The result is $1+\sqrt 2$ and it's a more general result : $\frac{\sum_{k=1}^{n-1}\sqrt{1+\sqrt{\frac kn}}}{\sum_{k=1}^{n-1}\sqrt{1-\sqrt{\frac kn}}}$ $=1+\sqrt 2$ But I dont know how to prove it up to now

Wow, that is pretty amazing! I have no idea how to prove it either, but it's very cool.

Got it ! $1+\sqrt{\frac kn}\in(1,2)$ and let then $x_k\in(0,\frac{\pi}4)$ such that $\cos x_k=\sqrt{\frac{1+\sqrt{\frac kn}}2}$ We get $\cos 2x_k=2\cos^2x_k-1=\sqrt{\frac kn}$ And so $\cos^2 2x_k+\cos^2 2x_{n-k}=1$ and so $2x_{n-k}=\frac{\pi}2-2x_k$ and so $x_{n-k}=\frac{\pi}4-x_k$ So $2\sum_{k=1}^{n-1}\cos x_k$ $=\sum_{k=1}^{n-1}(\cos x_k+\cos x_{n-k})$ $=\sum_{k=1}^{n-1}(\cos x_k+\cos(\frac{\pi}4-x_k))$ $=2\cos\frac{\pi}8\sum_{k=1}^{n-1}\cos(x_k-\frac{\pi}8)$ Hence a first result : $\sum_{k=1}^{n-1}\sqrt{\frac{1+\sqrt{\frac kn}}2}$ $=\cos\frac{\pi}8\sum_{k=1}^{n-1}\cos(x_k-\frac{\pi}8)$ Then $\sin x_k=\sqrt{\frac{1-\sqrt{\frac kn}}2}$ and so : $2\sum_{k=1}^{n-1}\sin x_k$ $=\sum_{k=1}^{n-1}(\sin x_k+\sin x_{n-k})$ $=\sum_{k=1}^{n-1}(\sin x_k+\sin(\frac{\pi}4-x_k))$ $=2\sin\frac{\pi}8\sum_{k=1}^{n-1}\cos(x_k-\frac{\pi}8)$ Hence a second result : $\sum_{k=1}^{n-1}\sqrt{\frac{1-\sqrt{\frac kn}}2}$ $=\sin\frac{\pi}8\sum_{k=1}^{n-1}\cos(x_k-\frac{\pi}8)$ And now, maaaagic : $\frac{\sum_{k=1}^{n-1}\sqrt{\frac{1+\sqrt{\frac kn}}2}}{\sum_{k=1}^{n-1}\sqrt{\frac{1-\sqrt{\frac kn}}2}}$ $=\cot\frac{\pi}8$ And so $\frac{\sum_{k=1}^{n-1}\sqrt{1+\sqrt{\frac kn}}}{\sum_{k=1}^{n-1}\sqrt{1-\sqrt{\frac kn}}}$ $=\cot\frac{\pi}8$ And it's easy to compute $\cot\frac{\pi}8=1+\sqrt 2$ And so $\boxed{\frac{\sum_{k=1}^{n-1}\sqrt{1+\sqrt{\frac kn}}}{\sum_{k=1}^{n-1}\sqrt{1-\sqrt{\frac kn}}}=1+\sqrt 2}$

Nice solution pco! For those interested in a solution without using trigonometry, it boils down to using the identity $\sqrt{10+\sqrt{k}} + \sqrt{10-\sqrt{k}} = \sqrt{2} \cdot \sqrt{10+\sqrt{100-k}}$. This can be verified by squaring each side. The question, of course, is how would we recognize this identity without trigonometry... I don't have a good answer to that, other than to think that since conjugates multiply nicely they might also add nicely. And in this case they do! Once the identity is verified, we can take the sum of each side as k ranges from 1 to 99. This gives $N + D = \sqrt{2} \cdot N$, where N is the sum in the numerator and D is the sum in the denominator of the original problem. Solving for $\frac{N}{D}$ gives $\sqrt{2}+1$.

Say $x= \frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+...+\sqrt{10+\sqrt{99}}}{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+...+\sqrt{10-\sqrt{99}}}$ $\sqrt{a+\sqrt{k}}+\sqrt{a-\sqrt{k}}=\sqrt{2}\sqrt{a+\sqrt{a^2-k}}$ $\sqrt{a+\sqrt{k}}-\sqrt{a-\sqrt{k}}=\sqrt{2}\sqrt{a-\sqrt{a^2-k}}$ Therefore we get $\frac{x+1}{x-1}=x\implies x=1\pm \sqrt{2}$ Clearly $x$ must be positive $\implies x=1+\sqrt{2}$ I see that gauss202 has already given the ans