MHB What is the ordered pair (x,y) for the given equation?

[anemone]

If $x$ and $y$ are positive integers such that $\sqrt{8+\sqrt{{32}+\sqrt{768}}}=x\cos \dfrac{\pi}{y}$,

compute the ordered pair $(x,\,y)$.

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[anemone]

Hi MHB,

I want to apologize for I just discovered that I made a typo in this very first high school POTW for the new year and I was in fact being advised by both valuable members (Opalg and lfdahl) of the matter.

My heartfelt thanks and appreciation to Opalg and lfdahl for advising me of the matter.

The problem has been edited and I am looking forward to your participation to this POTW.

Thank you for reading.

 

[anemone]

Congratulations to the following members for their correct solutions::)

1. kaliprasad
2. lfdahl

Solution from kaliprasad:

Spoiler

We have

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

so $\sqrt{768} = 32 *\cos(\dfrac{\pi}{6})$

and

$\begin{align*}32 + \sqrt{768}&= 32 *(1+\cos(\dfrac{\pi}{6}))---\text{using}\,\,1+\cos 2x = 2 cos^2 x\\&=64 *\cos^2(\dfrac{\pi}{12})\end{align*}$

Taking square root on both sides gives

$\sqrt{32 + \sqrt{768}} = 8 \cos(\dfrac{\pi}{12})$

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

So $\sqrt{8 + \sqrt{32 + \sqrt{768}}} = 4 \cos \dfrac{\pi}{24}$$\therefore x=4,\,y=24$ or $(x,\,y)=(4,\,24)$

Solution from lfdahl:

Spoiler

$x$ and $y$ are positive integers, so I am looking for a cosine value with an exact expression of the form:

\[cos(\frac{\pi }{y})=\frac{\sqrt{8+\sqrt{32+\sqrt{768}}}}{x}=2\frac{\sqrt{2+\sqrt{2+\sqrt{3}}}}{x} \;\;\;\;\;\;\; (1).\]

I know, that $cos(\frac{\pi}{3}) = \frac{1}{2}$. By using the well known relation:

\[cos(\frac{\alpha }{2})=\sqrt{\frac{cos(\alpha )+1}{2}}\] I get successively:

\[cos(\frac{\pi }{6})=\sqrt{\frac{cos(\frac{\pi }{3})+1}{2}}=\frac{\sqrt{3}}{2}
\\\\
cos(\frac{\pi }{12})=\sqrt{\frac{cos(\frac{\pi }{6})+1}{2}}=\frac{\sqrt{2+\sqrt{3}}}{2}
\\\\
cos(\frac{\pi }{24})=\sqrt{\frac{cos(\frac{\pi }{12})+1}{2}}=\frac{\sqrt{2+\sqrt{2+\sqrt{3}}}}{2}\]

Therefore, the ordered pair $(x,y)=(4,24)$ is a solution to (1).