Revenge of the impossible Sangaku Part 2

To make the solution simpler, let's assume that the side length of the big green equilateral triangle is 1, at the end we will multiply the found length of each required element by 12 :

• The big green triangle has a side length of $\boxed{1}$

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• The big red circle circumbscribes a regular hexagone of side length 1, then the diameter of this circle is 2 and the height of the hexagone is $\sqrt{3}$ . We can deduce that the radius of the small orange circle = $\boxed{\frac{2-\sqrt{3}}{4}}$

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• The big light pink circle is inscribed in a rhombus of side length 1. We use trigonometry to calculate the radius of the big pink circle : sin(60°)= $\frac{R}{0,5}$ . $\boxed{R=\frac{\sqrt{3}}{4}}$

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• Next we can calculate the side of the big light blue square inscribed in the big light pink circle. We use the pythagorean theorem to determine the side, knowing the diameter of the circle : s²=2R² <=> $\boxed{s=\frac{\sqrt{6}}{4}}$

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• We determine the side length of the small equilateral triangle now. We denote h its height : h= $\frac{2R-s}{2}$ . Next, we use trigonometry to solve for the side of the triangle which is $\boxed{\frac{1}{2}\sqrt{\frac{3}{2}-\sqrt{2}}}$

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• Now we calculate the radius of the small pink circle inscribed in the circular sector of radius $\frac{1}{2}$ with an angle of 60°. We use trigonometry to calculate the radius : 2r+r= $\frac{1}{2}$ <=> $\boxed{r=\frac{1}{6}}$

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• The small pink square is inscribed in the small pink circle, using the pythagorean theorem, we find the side = $\boxed{\frac{\sqrt{2}}{6}}$

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• At last we calculate the side of the small black triangle. The black triangle and the big green triangle share the same center. We denote $x$ the radius of the circumcircle of the black triangle. The distance between one vertex of the green triangle and the center of the black triangle is equal to its height multiplied by $\frac{2}{3}$ which give $\frac{\sqrt{3}}{3}$ . This same distance is equal to $\frac{1}{2}+x$ <=> $x=\frac{2\sqrt{3}-2}{6}$ . Using this result and basic trigonometry we find that the side length of the black triangle is $\boxed{\frac{2-\sqrt{3}}{2}}$

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• We add the values found and we multiply the result by 12, since the side length of the big green triangle is 12 not 1, we find : $\boxed{38+3\sqrt{6}-6\sqrt{3}-\sqrt{2}}$
• We calculate : $\boxed{k=\sqrt{38+3+6+2}=7}$
• We can finally solve the trigonometric equation now that we know the interval and the answer is $\boxed{\frac{20π}{7}≈9}$ You can find a good explanation here to solve this equation