A geometry problem by Rocco Dalto

$\begin{aligned} \cos \frac \pi C & = \frac 12 \sqrt{2+\color{#3D99F6}{\sqrt{2+\sqrt 1}}} \\ \cos^2 \frac \pi C & = \frac 12 + \frac {\color{#3D99F6}{\sqrt 3}}4 \\ 2\cos^2 \frac \pi C & = 1 + \frac {\color{#3D99F6}{\sqrt 3}}2 \\ 2\cos^2 \frac \pi C - 1 & = \frac {\color{#3D99F6}{\sqrt 3}}2 \\ \cos \frac {2\pi}C & = \frac {\color{#3D99F6}{\sqrt 3}}2 & \color{#3D99F6}{...(1)} \\ \frac {2\pi}C & = \frac \pi 6 \\ \implies C & = 12 \end{aligned}$

Similarly,

$\begin{aligned} \color{#3D99F6}{(1)}: \quad \cos \frac {2\pi}B & = \frac {\sqrt{2+\color{#D61F06}{\sqrt 2}}}2 \\ \cos^2 \frac {2\pi}B & = \frac 12 + \frac {\color{#D61F06}{\sqrt 2}}4 \\ 2\cos^2 \frac {2\pi}B - 1 & = \frac {\color{#D61F06}{\sqrt 2}}2 \\ \cos \frac {4\pi}B & = \frac {\color{#D61F06}{\sqrt 2}}2 & \color{#D61F06} {...(2)} \\ \frac {4\pi}B & = \frac \pi 4 \\ \implies B & = 16 \end{aligned}$

Again,

$\begin{aligned} \color{#D61F06}{(2)}: \quad \cos \frac {4\pi}A & = \frac {\sqrt 3}2 \\ \frac {4\pi}A & = \frac \pi 6 \\ \implies A & = 24 \end{aligned}$

Therefore, $A+B+C = 24+16+12 = \boxed{52}$

$\large \cos \dfrac \pi A =\dfrac12 \sqrt{2 + \sqrt{2 + \sqrt3}} \implies$

${\bf 4cos^2(\frac{\pi}{A}) = }$ ${\bf 2 + \sqrt{2 + \sqrt{3}} \implies }$

${\bf 4cos^2(\frac{\pi}{A}) - 2 = \sqrt{2 + \sqrt{3}} \implies }$

${\bf 2(2cos^2(\frac{\pi}{A}) - 1) = \sqrt{2 + \sqrt{3}} \implies }$

${\bf 2cos(\frac{2\pi}{A}) = \sqrt{2 + \sqrt{3}} \implies }$

${\bf 4cos^2(\frac{2\pi}{A}) = 2 + \sqrt{3} \implies}$

${\bf 4cos^2(\frac{2\pi}{A}) - 2 = \sqrt{3} \implies}$

${\bf 2(2cos^2(\frac{2\pi}{A}) - 1) = \sqrt{3} \implies }$

${\bf 2cos(\frac{4\pi}{A}) = \sqrt{3} \implies }$

${\bf cos(\frac{4\pi}{A}) = \frac{\sqrt{3}}{2} \implies }$

${\bf \frac{4\pi}{A} = \frac{pi}{6} \implies }$

${\bf A = 24 }$

Similarly, $\large \cos \dfrac \pi B =\dfrac12 \sqrt{2 + \sqrt{2 + \sqrt2}} \implies$

${\bf 2cos(\frac{4\pi}{B}) = \sqrt{2} \implies }$

${\bf cos(\frac{4\pi}{B}) = \frac{1}{\sqrt{2}} \implies}$

${\bf \frac{4\pi}{B} = \frac{\pi}{4} \implies}$

${\bf B = 16 }$

and

$\large \cos \dfrac \pi C =\dfrac12 \sqrt{2 + \sqrt{2 + \sqrt1}} \implies$

${\bf 2cos(\frac{4\pi}{C}) = 1 \implies }$

${\bf cos(\frac{4\pi}{C}) = \frac{1}{{2}} \implies}$

${\bf \frac{4\pi}{C} = \frac{\pi}{3} \implies}$

${\bf C = 12 \implies }$

${\bf A + B + C = 52 }$