2015/004) Find x,y when $x \cos \dfrac{\pi}{y} = \sqrt{8 + \sqrt{32 + \sqrt{768}}}$

we have

$768 = 1024 * \dfrac{3}{4}= (32 * \dfrac{\sqrt{3}}{2})^2 = (32 *\cos(\dfrac{\pi}{6}))^2$

so $\sqrt{768} = 32 *\cos(\dfrac{\pi}{6})$
so
 $32 + \sqrt{768} = 32 *(1+\cos(\dfrac{\pi}{6})$
= $64 *\cos^2(\dfrac{\pi}{12}$ using $1+\cos 2x = 2 cos^2 x$

so
 $\sqrt{32 + \sqrt{768}} = 8 \cos(\dfrac{\pi}{12})$
so
 $8 + \sqrt{32 + \sqrt{768}} = 8 ( 1 + \cos(\dfrac{\pi}{12})) = 16 \cos^2 \dfrac{\pi}{24}$

so $\sqrt{8 + \sqrt{32 + \sqrt{768}}} = 4 \cos \dfrac{\pi}{24}$

so x = 4, y = 24 or (x,y) = (4,24)

Note:
I have solved this as a problem of the week at http://mathhelpboards.com/potw-secondary-school-high-school-students-35/problem-week-145-january-5th-2015-a-13891.html