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Question Number 187689 by Mingma last updated on 20/Feb/23

Answered by mr W last updated on 21/Feb/23

Commented by mr W last updated on 26/Feb/23

$$\alpha =30°-\gamma$$$$2\delta +\gamma =90°\Rightarrow \delta =45°-\frac{\gamma }{2}$$$$\frac{1}{\tan \delta }=1+4\sin \alpha$$$$\frac{1}{\tan (45°-\frac{\gamma }{2})}=1+4\sin (30°-\gamma )$$$$\frac{1+\tan \frac{\gamma }{2}}{1-\tan \frac{\gamma }{2}}=1+2\cos \gamma -2\sqrt{3}\sin \gamma$$$$lett=\tan \frac{\gamma }{2}$$$$\frac{1+t}{1-t}=1+\frac{2(1-t^2)}{1+t^2}-\frac{4\sqrt{3}t}{1-t^2}$$$$\frac{1+t}{1-t}=\frac{(3-t^2)(1-t^2)-4\sqrt{3}t(1+t^2)}{(1+t^2)(1-t^2)}$$$${(1+t)}^2(1+t^2)=(3-t^2)(1-t^2)-4\sqrt{3}t(1+t^2)$$$$(1+2\sqrt{3})t^3+3t^2+(1+2\sqrt{3})t-1=0$$$$\Rightarrow t\approx 0.192115=\tan \frac{\gamma }{2}$$$$(\gamma \approx 21.749760°)$$$$$$$$a=1+4\cos \alpha +1=2+4\cos (30°-\gamma )$$$$(a\approx 5.958603)$$$$$$$$\beta =90°-60°-\alpha =30°-30°+\gamma =\gamma$$$$b=1+4\cos \beta +1=2+4\cos \gamma$$$$(b\approx 5.715244)$$$$A_{green}=ab-6\times \pi \times 1^2$$$$=(2+4\cos (30-\gamma ))(2+4\cos \gamma )-6\pi$$$$\approx 34.054870-6\pi$$

Answered by a.lgnaoui last updated on 21/Feb/23

$$SoitQuadrilatetere\left[MNPQ\right]$$$$\measuredangle FNQ=\frac{\measuredangle YNQ}{2}=\frac{\theta }{2}$$$$\tan \frac{\theta }{2}=\frac{BS}{NS}=\frac{R}{NS}\Rightarrow NS=\frac{R}{\tan \frac{\theta }{2}}$$$$NQ=NS+BK+KT$$$$=\frac{R}{\tan \frac{\theta }{2}}+4R\cos \frac{\theta }{2}+R$$$$NQ=\cot \frac{\theta }{2}+4\cos \frac{\theta }{2}+1\left(1\right)$$$$dautrepartMP=NQ$$$$MP=IA+AH+HL$$$$=R+4R\cos \phi +R\left(2\right)$$$$\phi +\measuredangle ACB+(\frac{\pi }{2}-\frac{\theta }{2})=\pi$$$$AB=BC=AC=3R\Rightarrow \measuredangle ACB=\frac{\pi }{3}$$$$\Rightarrow \phi =\frac{\pi }{3}+\frac{\theta }{2}$$$$\left(2\right)MP=2+4\sin (\frac{\pi }{3}+\frac{\theta }{2})$$$$=2+2\sin \frac{\theta }{2}+2\sqrt{3}\cos \frac{\theta }{2}$$$$$$$$\left(1\right)et\left(2\right)\cot \frac{\theta }{2}+4\cos \frac{\theta }{2}+1=$$$$=2+2\sin \frac{\theta }{2}+2\sqrt{3}\cos \frac{\theta }{2}$$$$\cot \frac{\theta }{2}=(2\sqrt{3}-4)\cos \frac{\theta }{2}+2\sin \frac{\theta }{2}+1$$$$\cos \frac{\theta }{2}=(2\sqrt{3}-4)\sin \frac{\theta }{2}\cos \frac{\theta }{2}+{2\sin }^2\frac{\theta }{2}+\sin \frac{\theta }{2}$$$$\cos \frac{\theta }{2}(1+(4-2\sqrt{3})\sin \frac{\theta }{2})=2\sin {}^2\frac{\theta }{2}+\sin \frac{\theta }{2}$$$$\cos \frac{\theta }{2}=\frac{2\sin {}^2\frac{\theta }{2}+\sin \frac{\theta }{2}}{1+(4-2\sqrt{3})\sin \frac{\theta }{2}}$$$$(1-\sin {}^2\frac{\theta }{2})=\frac{{\sin }^2\frac{\theta }{2}{(2\sin \frac{\theta }{2}+1)}^2}{{(1+(4-2\sqrt{3})\sin \frac{\theta }{2})}^2}$$$$posonsy=\sin \frac{\theta }{2}$$$$1-y^2=\frac{y^2{(2y+1)}^2}{{[1+(4-2\sqrt{3})y]}^2}$$$$4y^4+4y^3+2y^2=-{(4-2\sqrt{3})}^2{y}^4-2(4-2\sqrt{3})y^3+(4-2\sqrt{3})y^2+2(4-2\sqrt{3})y+1$$$$4(4-\sqrt{3})y^4+2(3-\sqrt{3})y^3+(\sqrt{3}-1)y^2-2(2-\sqrt{3})y-\frac{1}{2}=0$$$${\mathit{y}}^4+\frac{2}{13}(9-\sqrt{3}){\mathit{y}}^3+\frac{1}{13}(3\sqrt{3}-1){\mathit{y}}^2-\frac{5-2\sqrt{3}}{13}\mathit{y}-\frac{4-\sqrt{3}}{26}=0$$$$y=0,48533919$$$$$$$$\frac{\theta }{2}={\sin }^{-1}\left(y\right)\cong 30°\alpha =\measuredangle VNM=30$$$$$$$$\mathrm{x}=5+1=6$$$$GreenArea={\mathbf{x}}^2-6\mathit{\pi }{\mathit{R}}^2(R=1)$$$$$$$$Area\left(green\right)=\mathrm{x}.\mathrm{z}-6\pi R^2$$$$Area=\left(PQ\times NQ\right)-6\pi R^2$$$$\mathbf{z}=PQ=PL+TQ+4R[\sin \frac{\pi }{6}+\cos (\frac{\pi }{6}+\frac{\pi }{3})]$$$$PQ=4Area=\mathrm{x}\times \mathrm{z}=24$$$$$$$$\mathit{A}\mathit{r}\mathit{e}\mathit{a}\left(green\right)=24-6\pi$$

Commented by a.lgnaoui last updated on 21/Feb/23

Commented by Rupesh123 last updated on 21/Feb/23

Good job, bro!

Commented by mr W last updated on 22/Feb/23

$$clearlywrong!$$$$PQ=4?howcanthenthreecirclesbe$$$$placedsidebyside?$$$$besidesclearlyPQ>NQ!$$

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