I Was Very Amazed At The Solution 7

Note that $\cos(18108 ^{\circ}) = \cos (108^{\circ})$

Construct a regular pentagon $ABCDE$ and lines $\overline{AD} \ , \ \overline{AC} \ , \ \overline{BD}$ .

Let $\overline{AB} \ = \ \overline{BC} \ = \ \overline{CD} = s \ , \ \overline{AD} \ = \ \overline{AC} \ = \ \overline{BD} = y$

By law of cosine in $\Delta BCD$ , $\implies s^2 + s^2 - 2(s)(s)(\cos(108^{\circ})) = y^2$

As $ABCD$ is a cyclic quadrilateral, we have:

$\begin{aligned} & s(y) + s(s) = y(y) \\ & \implies y^2 - sy - s^2 = 0 \\ & \implies y = \dfrac{s \pm \sqrt{s^2 + 4s^2}}{2} = \dfrac{s \pm s\sqrt{5}}{2} \\ & \implies y^2 = \dfrac{3s^2 \pm s^2 \sqrt{5}}{2} \end{aligned}$

Then,

$\begin{aligned} & s^2 + s^2 - 2(s)(s)(\cos(108^{\circ})) = \dfrac{3s^2 \pm s^2 \sqrt{5}}{2} \\ & \implies 2s^2 - \dfrac{3s^2 \pm s^2 \sqrt{5}}{2} = 2(s)(s)(\cos(108^{\circ})) \\ & \implies \dfrac{s^2 \pm s^2 \sqrt{5}}{2} = 2(s)(s)(\cos(108^{\circ})) \\ & \implies \cos(108^{\circ}) = \dfrac{1 - \sqrt{5}}{4} \end{aligned}$

as $\cos(108^{\circ}) < 0$ .

Then, $a = 1, b = 5, c = 4$

$\therefore 16(a + \sqrt{b + c}) = 16(1 + \sqrt{5 + 4}) = \boxed{64}$