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ABCD is a unit square. There are 2 circles in picture. The center of one is A and passing from F. The center of another is D and passing from C.
If $\alpha=2\beta$ then I got the following results

$|AF|=x$

$\cos(2\beta)=\frac{1}{x}$

and from isoscale triangle (ADG): $\sin(3\beta)=\frac{x}{2}$

$\sin(3\beta)\cos(2\beta)=\frac{1}{2}$

$\sin(5\beta)+\sin(\beta)=1$

$\sin(\beta)=P$

my first question: Is it possible to find a solution of P via radicals?

Second question: During my drawing the figure that Area of (CGH region) is equal about Area of (HBF region) (just I noticed with geometrical drawing tool).I do not claim that they are equal Because I know very well that it is impossible to Squaring a circle.(more info is in Wiki link)

I try to find approximation of $\pi$ in radicals via the figure if possible. Note:If you know interesting $\pi$ approximations via Squaring a circle, I would like to know them. Thanks for advices and answers

EDIT:

I asked the question to wolfram alpha. I got fifth degree polinom with real coefficients.

$\sin(5\beta)+\sin(\beta)=1$

$\sin^5(\beta)+5\sin(\beta)\cos^4(\beta)-10\sin^3(\beta)\cos^2(\beta)+\sin(\beta)=1$

$\sin^5(\beta)+5\sin(\beta)(1-\sin^2(\beta))^2-10\sin^3(\beta)(1-\sin^2(\beta))+\sin(\beta)=1$

$\sin(\beta)=P$

$16P^5-20P^3+6P=1$

Really $P=\sin(\pi /6)=\frac{1}{2}$ is a solution of the polinom.

All roots are in the link. What a pity that wolfram did not offer me radical solutions for 4 degree polinom either. I will check other tools to find radical solutions of the roots. For my second question UPDATE:

I decided to see if really Areas are about eqaul to each other.

the root is for my first drawing: I will take the solution about $P\approx0.188286$ from wolfram solution.

$\sin(\beta)\approx0.188286$

$\beta\approx0.189416$

$|AF|=x$

$\cos(2\beta)=\frac{1}{x}$

$x=\frac{1}{\cos(2\beta)}\approx1.076314$

Area of (CGH region)= (wiki link) enter image description here

Area of (HBF region)= (Wiki Link) enter image description here

RESULT: I wonder if I use more digits of the root what Areas will be. it seems that numerical solution of areas are very near each other. I will update if I find radical solution of polinom and I will update if I find some other results. Your helps will be appreciated very much to analyze the results.

Thanks a lot to Chris K. Caldwel for his contribution. I drew the solution he offered . enter image description here

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To answer your first question: one solution to $\sin(5\beta)+\sin(\beta)=1$ is $\pi/6$, so since $sin(5\beta)$ can be written as fifth degree polynomial in terms of $\sin\beta$, the remaining four solutions to $\sin(5\beta)+\sin(\beta)=1$ are the zeros of a fourth degree polynomial, hence can all be expressed in terms of radicals. –  Chris K. Caldwell May 2 '12 at 16:05
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1 Answer

up vote 2 down vote accepted

I don't think that this answer meaningfull but if you need this roots, here they are $$ x_1=\frac{1}{2} $$ $$ x_2=-\frac{1}{8}-\frac{1}{8 \sqrt{\frac{3}{35+2 \sqrt[3]{3160-24 \sqrt{1713}}+4 \sqrt[3]{395+3 \sqrt{1713}}}}}-\frac{1}{2} \sqrt{\frac{35}{24}-\frac{1}{24} \sqrt[3]{3160-24 \sqrt{1713}}-\frac{1}{12} \sqrt[3]{395+3 \sqrt{1713}}-\frac{15}{8} \sqrt{\frac{3}{35+2 \sqrt[3]{3160-24 \sqrt{1713}}+4 \sqrt[3]{395+3 \sqrt{1713}}}}} $$ $$ x_3=-\frac{1}{8}-\frac{1}{8 \sqrt{\frac{3}{35+2 \sqrt[3]{3160-24 \sqrt{1713}}+4 \sqrt[3]{395+3 \sqrt{1713}}}}}+\frac{1}{2} \sqrt{\frac{35}{24}-\frac{1}{24} \sqrt[3]{3160-24 \sqrt{1713}}-\frac{1}{12} \sqrt[3]{395+3 \sqrt{1713}}-\frac{15}{8} \sqrt{\frac{3}{35+2 \sqrt[3]{3160-24 \sqrt{1713}}+4 \sqrt[3]{395+3 \sqrt{1713}}}}} $$ $$ x_4=-\frac{1}{8}+\frac{1}{8 \sqrt{\frac{3}{35+2 \sqrt[3]{3160-24 \sqrt{1713}}+4 \sqrt[3]{395+3 \sqrt{1713}}}}}-\frac{1}{2} \sqrt{\frac{35}{24}-\frac{1}{24} \sqrt[3]{3160-24 \sqrt{1713}}-\frac{1}{12} \sqrt[3]{395+3 \sqrt{1713}}+\frac{15}{8} \sqrt{\frac{3}{35+2 \sqrt[3]{3160-24 \sqrt{1713}}+4 \sqrt[3]{395+3 \sqrt{1713}}}}} $$ $$ x_5=-\frac{1}{8}+\frac{1}{8 \sqrt{\frac{3}{35+2 \sqrt[3]{3160-24 \sqrt{1713}}+4 \sqrt[3]{395+3 \sqrt{1713}}}}}+\frac{1}{2} \sqrt{\frac{35}{24}-\frac{1}{24} \sqrt[3]{3160-24 \sqrt{1713}}-\frac{1}{12} \sqrt[3]{395+3 \sqrt{1713}}+\frac{15}{8} \sqrt{\frac{3}{35+2 \sqrt[3]{3160-24 \sqrt{1713}}+4 \sqrt[3]{395+3 \sqrt{1713}}}}} $$ Numerical values $$ x_1=0.500000000000000000000000000000, $$ $$ x_2=-0.821076869248865871679368938530-0.147733318623280702329856547465i, $$ $$ x_3=-0.821076869248865871679368938530+0.147733318623280702329856547465i, $$ $$ x_4=0.188285649881385672907404476142, $$ $$ x_5=0.953868088616346070451333400918 $$ The respecitive angles $\beta=\arcsin(x)$ in radians are $$ \beta_1=0.523598775598298873077107230547, $$ $$ \beta_2=-0.923163176894719154945644411954-0.242491674789456740938340067472i, $$ $$ \beta_3=-0.923163176894719154945644411954+0.242491674789456740938340067472i, $$ $$ \beta_4=0.189416282605918466874902490444, $$ $$ \beta_5=1.26586671937652178290413890330 $$ The areas of domains $CGH$, $HBF$ are given by the formulas $$ S_{CGH}=3 \beta -\frac{1}{2} \sin (6 \beta )-\frac{1}{2} \beta \sec ^2(2 \beta )-\frac{1}{2}\sqrt{\sec ^2(2 \beta )-1}-\frac{\pi }{4}+1 $$ $$ S_{HBF}=\beta \sec ^2(2 \beta )-\frac{1}{2} \sqrt{\sec ^2(2 \beta )-1} $$ The respecitve numericals values for $CGH$ are $$ S_{CGH,1}=-0.127824791583588083302276786026, $$ $$ S_{CGH,2}=-3.99683686501392644407946565939-0.54938014174210294554189328651i, $$ $$ S_{CGH,3}=-3.99683686501392644407946565939+0.54938014174210294554189328651i, $$ $$ S_{CGH,4}=0.0205233661564834188106917673615, $$ $$ S_{CGH,5}=2.23756796644691039586047775118 $$ for $HFB$ are $$ S_{HFB,1}=1.22836969860875684554470575143, $$ $$ S_{HFB,2}=0.15747480940566025178679372899+3.71420167984539571383021593893i, $$ $$ S_{HFB,3}=0.15747480940566025178679372899-3.71420167984539571383021593893i, $$ $$ S_{HFB,4}=0.0203997399628726939700469522529, $$ $$ S_{HFB,5}=1.53450150774254892235003332376 $$ Computations were made via Mathematica $8$ with $30$ digit precision.

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Really quite huge radical results. I did not expect to see such complex results. What about Areas . Is it possible to show how they are near each other via using Mathematica 8. Pity that I do not have mathematica 8. Thanks a lot for help and answer. very appreciated –  Mathlover May 2 '12 at 22:40
    
Just give the formulas for this areas, i will compute them and extend the answer –  Norbert May 2 '12 at 22:47
    
They are in my question. I calculated for P≈0.188286. Is not it clear for you? –  Mathlover May 2 '12 at 22:51
1  
As far as I understand they are $$1-\left(\frac{\beta x^2}{2}+\frac{\sqrt{x^2-1}}{2}+\frac{1}{2}\sin(6\beta)+\left(\frac{\pi}{4}-3\bet‌​a\right)\right)$$ and $$\beta x^2-\frac{\sqrt{x^2-1}}{2}$$ Right? –  Norbert May 2 '12 at 22:56
    
Yes. Correct results –  Mathlover May 2 '12 at 22:59
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