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By Cauchy-Schwarz inequality,

$(a^2 \cot 9^\circ + b^2 \cot 27^\circ + c^2 \cot 63^\circ + d^2 \cot 81^\circ)(\frac{1}{\cot 9^\circ}+\frac{1}{\cot 27^\circ}+\frac{1}{\cot 63^\circ}+\frac{1}{\cot 81^\circ}) \geq (a+b+c+d)^2 = 25$ (Since $a+b+c+d = 5$

Now, let $(a^2 \cot 9^\circ + b^2 \cot 27^\circ + c^2 \cot 63^\circ + d^2 \cot 81^\circ) = N$ Then, dividing both sides by $\frac{1}{\cot 9^\circ}+\frac{1}{\cot 27^\circ}+\frac{1}{\cot 63^\circ}+\frac{1}{\cot 81^\circ}$ ,

$\large N \geq \frac{25}{\frac{1}{\cot 9^\circ}+\frac{1}{\cot 27^\circ}+\frac{1}{\cot 63^\circ}+\frac{1}{\cot 81^\circ}}= \frac{25}{\tan 9^\circ +\tan 81^\circ +\tan 27^\circ +\tan 63^\circ}$

Now, by the identity $\tan x^\circ + \tan(90-x)^\circ = 2 \csc (2x)$ , we see that $\tan 9^\circ +\tan 81^\circ +\tan 27^\circ +\tan 63^\circ = 2(\csc 18^\circ + \csc 54 ^\circ) = 4 \sqrt{5}$

Therefore, $\large N \geq \frac{25}{\tan 9^\circ +\tan 81^\circ +\tan 27^\circ +\tan 63^\circ} = \frac{25}{4 \sqrt{5}} = \boxed{\frac{\sqrt{125}}{4}}$

Our answer is $125+4 = \boxed {129}$

Let us take a general case where $x+y=k$ and $p=ax^{2}+by^{2}$ which is to be minimised.

$\Rightarrow$ $p=(a+b)x^{2}-(2kb)x+bk^{2}$ or $p=(a+b)y^{2}-(2kb)y+bk^{2}$

So, $x_{min}=\frac{b}{a+b}k$ and $y_{min}=\frac{a}{a+b}k$

$\therefore$ $p_{min}=\frac{ab}{a+b}k^{2}$ - - - $(1)$

So, $p=a^2 \cot 9^\circ + b^2 \cot 27^\circ + c^2 \cot 63^\circ + d^2 \cot 81^\circ$ can be minimised as follows:

$p_1=a^2 \cot 9^\circ+ d^2 \cot 81^\circ$

$p_2=b^2 \cot 27^\circ + c^2 \cot 63^\circ$

$p_{min}=p_{1min}+p_{2min}$ where $p_1=\frac{sin18^{\circ}}{2}k_1^{2}$ ; $p_2=\frac{cos36^{\circ}}{2}k_2^{2}$ , $a+d=k_1$ ; $b+c=k_2$ and $k_1+k_2=5$

$\therefore$ $p_{min}=\frac{\sqrt{125}}{4}$ (using $(1)$ )