TrigoNOPEmetry

Let $\space C_n (x) = \displaystyle \sum _{i=0} ^n {\cos{((2i+1)x)}} \space$ and $\space S_n (x) = \displaystyle \sum _{i=0} ^n {\sin{((2i+1)x)}}$

We note that:

$\begin{aligned} C_3 \left( \frac{\pi}{8} \right) & = \cos{\frac{\pi}{8}} + \cos{\frac{3\pi}{8}} + \cos{\frac{5\pi}{8}} + \cos{\frac{9\pi}{8}} \\ & = \cos{\frac{\pi}{8}} + \cos{\frac{3\pi}{8}} - \cos{\frac{3\pi}{8}} - \cos{\frac{\pi}{8}} \\ & = 0 \end{aligned}$

$\Rightarrow C_n \left( \frac{\pi}{8} \right) = 0\space$ for $\space n = 3, 7, 11, ... 4k+1, ...$ where $k = 0,1,2...$

$\Rightarrow C_n \left( \frac{\pi}{8} \right) = C_{n \text{ mod } 4} \left( \frac{\pi}{8} \right) \quad \Rightarrow C_{200} \left( \frac{\pi}{8} \right) = C_0 \left( \frac{\pi}{8} \right) = \cos{\frac{\pi}{8}}$

Similarly, $S_n \left( \frac{\pi}{8} \right) = S_{n \text{ mod } 8} \left( \frac{\pi}{8} \right) \quad \Rightarrow S_{200} \left( \frac{\pi}{8} \right) = S_0 \left( \frac{\pi}{8} \right) = \sin{\frac{\pi}{8}}$

Therefore,

$\Omega_{200} \left( \frac{\pi}{8} \right) = \dfrac {C_{200} \left( \frac{\pi}{8} \right)} {S_{200} \left( \frac{\pi}{8} \right)} = \dfrac {\cos{\frac{\pi}{8}}}{\sin{\frac{\pi}{8}}} = \dfrac {2\cos^2 {\frac{\pi}{8}}}{\sin{\frac{\pi}{8}} \cos{\frac{\pi}{8}}} = \dfrac {\cos{\frac{\pi}{4}}+1}{\sin{\frac{\pi}{4}}} \\ \quad \quad \quad \quad = \dfrac {\frac{1}{\sqrt{2}}+1}{\frac{1}{\sqrt{2}}} = 1 + \sqrt{2}$

$\Rightarrow \begin{aligned} \sqrt{a+b\sqrt{c}} & = 1 + \sqrt{2} \\ a + b\sqrt{c} & = (1+\sqrt{2})^2 \\ & = 1+2\sqrt{2}+2 \\ & = 3 + 2\sqrt{3} \end{aligned} \quad \Rightarrow a +b+c = 3+2+2 = \boxed{7}$

${ \Omega }_{ n }(x)=\frac { \sum _{ i=0 }^{ n }{ \cos { \{ (2i+1)x\} } } }{ \sum _{ i=0 }^{ n }{ \sin { \{ (2i+1)x\} } } }$

Multiplying and dividing with $2\sin { x }$ ,

${ \Omega }_{ n }(x)=\frac { \sum _{ i=0 }^{ n }{ 2\sin { x } \cos { \{ (2i+1)x\} } } }{ \sum _{ i=0 }^{ n }{ 2\sin { x } \sin { \{ (2i+1)x\} } } }$

Applying required identities,

${ \Omega }_{ n }(x)=\frac { \sum _{ i=0 }^{ n }{ \sin { \{ (2i+2)x\} -\sin { 2ix } } } }{ \sum _{ i=0 }^{ n }{ \cos { 2ix } -\cos { \{ (2i+2)x\} } } }$

These are telescoping series. All middle terms will cancel out to yield,

${ \Omega }_{ n }(x)=\frac { \sin { \{ (2n+2)x\} } }{ 1-\cos { \{ (2n+2)x\} } }$

Applying double angle formulas,

${ \Omega }_{ n }(x)=\frac { 2\sin { \{ (n+1)x\} } \cos { \{ (n+1)x\} } }{ 2\sin ^{ 2 }{ \{ (n+1)x\} } }$

We have simplified the tedious expression and got this result.

${ \Omega }_{ n }(x)=\cot { \{ (n+1)x\} }$

We now put values $n=200$ and $x=\frac { \pi }{ 8 }$

${ \Omega }_{ 200 }(\frac { \pi }{ 8 } )=\cot { \frac { 201\pi }{ 8 } }$

Simplifying Angle,

${ \Omega }_{ 200 }(\frac { \pi }{ 8 } )=\cot { 25\pi +\frac { \pi }{ 8 } }$

Further simplifying,

${ \Omega }_{ 200 }(\frac { \pi }{ 8 } )=\cot { \frac { \pi }{ 8 } }$

Putting $\cot { \frac { \pi }{ 8 } }=\sqrt { 2 } +1$ ,

${ \Omega }_{ 200 }(\frac { \pi }{ 8 } )=\sqrt { 2 } +1$

Manipulating the above expression to get answer,

${ \Omega }_{ 200 }(\frac { \pi }{ 8 } )=\sqrt { { (\sqrt { 2 } +1) }^{ 2 } }$

Simplifying,

${ \Omega }_{ 200 }(\frac { \pi }{ 8 } )=\sqrt { 3+2\sqrt { 2 } }$

Therefore the answer is $\boxed { 7 }$ .

Notice that $\displaystyle S = \Omega_{200}\left(\frac{\pi}{8}\right)$ is, after expansion:

$\displaystyle \frac{cos(x) + cos(3x) + ... + cos(201x) + ... + cos(399x) + cos(401x)}{sin(x) + sin(3x) + ... + sin(201x) + ... + sin(399x) + sin(401x)}$

Grouping in Gaussian fashion, we get:

$\displaystyle \frac { cos(x)+cos(401x)+cos(3x)+cos(399x)+...+cos(201x) }{ sin(x)+sin(401x)+sin(3x)+sin(399x)+...+sin(201x) }$

Using the Sum-to-Product formulae, we get :

$\displaystyle \Rightarrow \quad \frac { 2cos(201x)cos(200x)+2cos(201x)cos(198x)+...+cos(201x) }{ 2sin(201x)cos(200x)+2sin(201x)cos(198x)+...+sin(201x) }$

Which conveniently reduces to :

$\displaystyle \Rightarrow \quad \frac { cos(201x)(\displaystyle \sum _{ i=0 }^{ 100 }{ cos((200-2i)x) } ) }{ sin(201x)(\displaystyle\sum _{ i=0 }^{ 100 }{ cos((200-2i)x) } ) } \quad$

Therefore, $\displaystyle S = \Omega_{200}\left(\frac{\pi}{8}\right) = cot\left(\frac{201\pi}{8}\right)$

Now, $cot\left(\frac{201\pi}{8}\right) = cot\left(24\pi + \frac{9\pi}{8}\right) = cot\left(\frac{9\pi}{8}\right) = cot\left(\frac{\pi}{8}\right)$

Also, using the Double-Angle formula for cotangents and using $y= cot\left(\frac{\pi}{8}\right)$ :

$\displaystyle cot\left(2*\frac{\pi}{8}\right) = \frac{y^2 - 1}{2y} = 1$

Solving which gives $y= cot\left(\frac{\pi}{8}\right) = S = 1 + \sqrt{2}$

I'm a little bit cross at the problem-setter for making the problem so twisted, but assume :

$\displaystyle \sqrt{a+b\sqrt{c}} = 1 + \sqrt{2} \Rightarrow a + b\sqrt{c} = (1+\sqrt{2})^2 = 3+2\sqrt{2}$

Hence (finally!)

$\displaystyle a + b + c = 3 + 2 + 2 = \boxed{7}$