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Question Number 75570 by naka3546 last updated on 12/Dec/19

Commented by naka3546 last updated on 12/Dec/19

$$\frac{\left[bluearea\right]}{\left[squarearea\right]}=?$$

Answered by mr W last updated on 13/Dec/19

Commented by mr W last updated on 13/Dec/19

$$METHOD1$$$$saysidelengthofsquare=2$$$${(2-r\cos \theta )}^2+{(r\sin \theta -1)}^2=1^2$$$$r^2-2(\sin \theta +2\cos \theta )r+4=0$$$$$$$$\sin \theta +2\cos \theta =\sqrt{5}\cos (\theta -\alpha )$$$$with\alpha ={\tan }^{-1}\frac{1}{2}$$$$r^2-2\sqrt{5}\cos (\theta -\alpha )r+4=0$$$$\Rightarrow r=\sqrt{5}\cos (\theta -\alpha )-\sqrt{5{\cos }^2(\theta -\alpha )-4}$$$$r^2=10{\cos }^2(\theta -\alpha )-2\sqrt{5}\cos (\theta -\alpha )\sqrt{5{\cos }^2(\theta -\alpha )-4}-4$$$$$$$$2^2-2\sqrt{5}\cos ({\theta }_1-\alpha )2+4=0$$$$\Rightarrow \cos ({\theta }_1-\alpha )=\frac{2}{\sqrt{5}}=\cos \alpha$$$$\Rightarrow {\theta }_1=2\alpha =2{\tan }^{-1}\frac{1}{2}$$$$$$$$A_{blue}={\int }_{\frac{\pi }{4}}^{{\theta }_1}\frac{1}{2}(2^2-r^2)d\theta$$$$=\frac{1}{2}{\int }_{\frac{\pi }{4}}^{2\alpha }(8-10{\cos }^2(\theta -\alpha )+2\sqrt{5}\cos (\theta -\alpha )\sqrt{5{\cos }^2(\theta -\alpha )-4})d\theta$$$$={\int }_{\frac{\pi }{4}}^{2\alpha }(4-5{\cos }^2(\theta -\alpha )+\sqrt{5}\cos (\theta -\alpha )\sqrt{5{\cos }^2(\theta -\alpha )-4})d(\theta -\alpha )$$$$={\int }_{\frac{\pi }{4}-\alpha }^{\alpha }(4-5{\cos }^2\phi +\sqrt{5}\cos \phi \sqrt{5{\cos }^2\phi -4})d\phi$$$$={\int }_{\frac{\pi }{4}-\alpha }^{\alpha }(\frac{3}{2}-\frac{5}{2}\cos 2\phi +\sqrt{5}\cos \phi \sqrt{1-5{\sin }^2\phi })d\phi$$$$={[\frac{3}{2}\phi -\frac{5}{4}\sin 2\phi ]}_{\frac{\pi }{4}-\alpha }^{\alpha }+{\int }_{\frac{\pi }{4}-\alpha }^{\alpha }\sqrt{1-{\left(\sqrt{5}\sin \phi \right)}^2}d\left(\sqrt{5}\sin \phi \right)$$$$={[\frac{3}{2}\phi -\frac{5}{4}\sin 2\phi +\frac{1}{2}{\sin }^{-1}\left(\sqrt{5}\sin \phi \right)+\frac{\sqrt{5}\sin \phi \sqrt{1-5{\sin }^2\phi }}{2}]}_{\frac{\pi }{4}-\alpha }^{\alpha }$$$$=\frac{1}{2}[3(2{\tan }^{-1}\frac{1}{2}-\frac{\pi }{4})-\frac{5}{2}(\frac{4}{5}-\frac{3}{5})+\frac{\pi }{2}-\frac{\pi }{4}+0-\frac{1}{2}]$$$$=3{\tan }^{-1}\frac{1}{2}-\frac{\pi +2}{4}$$$$A_{square}=2^2=4$$$$\frac{A_{blue}}{A_{square}}=\frac{3}{4}{\tan }^{-1}\frac{1}{2}-\frac{\pi +2}{16}\approx 0.026386$$

Answered by mr W last updated on 13/Dec/19

Commented by mr W last updated on 13/Dec/19

$$thanksforencouraging,sir!$$

Commented by mr W last updated on 13/Dec/19

$$METHOD2$$$$saysquaresidelength=2$$$$2\sin \frac{\alpha }{2}=1\sin \frac{\beta }{2}=1\sin (\frac{\pi }{2}-\frac{\alpha }{2})=1\cos \frac{\alpha }{2}$$$$\Rightarrow \tan \frac{\alpha }{2}=\frac{1}{2}$$$$\Rightarrow \alpha =2{\tan }^{-1}\frac{1}{2}$$$$A_{blue}=\frac{2^2}{2}(\alpha -\sin \alpha )+\frac{1^2}{2}(\beta -\sin \beta )$$$$-\frac{\pi \times 2^2}{8}+\frac{{\left(\sqrt{2}\right)}^2}{2}-\frac{1^2}{2}(\frac{\pi }{2}-\sin \frac{\pi }{2})$$$$A_{blue}=\frac{2^2}{2}(2{\tan }^{-1}\frac{1}{2}-\frac{4}{5})+\frac{1^2}{2}(\pi -2{\tan }^{-1}\frac{1}{2}-\frac{4}{5})-\frac{\pi }{2}+1-\frac{\pi }{4}+\frac{1}{2}$$$$A_{blue}=4{\tan }^{-1}\frac{1}{2}-\frac{8}{5}+\frac{\pi }{2}-{\tan }^{-1}\frac{1}{2}-\frac{2}{5}-\frac{\pi }{2}+1-\frac{\pi }{4}+\frac{1}{2}$$$$A_{blue}=3{\tan }^{-1}\frac{1}{2}-\frac{\pi +2}{4}$$$$A_{square}=2^2=4$$$$\Rightarrow \frac{A_{blue}}{A_{square}}=\frac{3}{4}{\tan }^{-1}\frac{1}{2}-\frac{\pi +2}{16}\approx 0.026386$$

Commented by behi83417@gmail.com last updated on 13/Dec/19

$$\mathrm{sir}\mathrm{mrW}!\mathrm{you}\mathrm{are}\mathrm{amazing}!$$$$\mathrm{thank}\mathrm{you}\mathrm{for}\mathrm{being}\mathrm{at}\mathrm{this}\mathrm{forum}.$$

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