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Question Number 162490 by amin96 last updated on 29/Dec/21

Answered by mr W last updated on 30/Dec/21

Commented by mr W last updated on 30/Dec/21

$$R=4=radius$$$$OA=a=1$$$$s=sidelengthofequilateral$$$$OC=\sqrt{s^2-a^2}$$$$DC=s\cos (\frac{\pi }{3}+\alpha )=\frac{1}{2}(s\cos \alpha -\sqrt{3}s\sin \alpha )$$$$=\frac{1}{2}(\sqrt{s^2-a^2}-\sqrt{3}a)$$$$DB=s\sin (\frac{\pi }{3}+\alpha )=\frac{1}{2}(\sqrt{3}s\cos \alpha +s\sin \alpha )$$$$=\frac{1}{2}(\sqrt{3(s^2-a^2)}+a)$$$$OD=\sqrt{s^2-a^2}-\frac{1}{2}(\sqrt{s^2-a^2}-\sqrt{3}a)$$$$=\frac{1}{2}(\sqrt{s^2-a^2}+\sqrt{3}a)$$$${OD}^2+{DB}^2={OB}^2$$$$\frac{1}{4}{(\sqrt{s^2-a^2}+\sqrt{3}a)}^2+\frac{1}{4}{(\sqrt{3(s^2-a^2)}+a)}^2=R^2$$$$s^2-a^2+3a^2+2a\sqrt{3(s^2-a^2)}+3(s^2-a^2)+a^2+2a\sqrt{3(s^2-a^2)}=4R^2$$$$a\sqrt{3(s^2-a^2)}=R^2-s^2$$$$3a^2(s^2-a^2)=s^4+R^4-2R^2{s}^2$$$$s^4-(2R^2+3a^2)s^2+R^4+3a^4=0$$$$s^2=\frac{2R^2+3a^2-\sqrt{{(2R^2+3a^2)}^2-4(R^4+3a^4)}}{2}$$$$=\frac{2R^2+3a^2-a\sqrt{3(4R^2-a^2)}}{2}$$$$withR=4,a=1:$$$$s^2=\frac{2\times 4^2+3\times 1^2-1\sqrt{3(4\times 4^2-1^2)}}{2}=\frac{35-3\sqrt{21}}{2}$$$$areaofblueequilateral$$$$A_{blue}=\frac{\sqrt{3}{s}^2}{4}=\frac{\sqrt{3}(35-3\sqrt{21})}{8}=\frac{35\sqrt{3}-9\sqrt{7}}{8}\approx 4.601$$$$$$$$===================$$$$let\xi =\frac{a}{R}$$$${\left(\frac{s}{R}\right)}^2=\frac{2+3{\xi }^2-\xi \sqrt{3(4-{\xi }^2)}}{2}$$$${\left(\frac{s}{R}\right)}_{max}=1at\xi =0$$$${\left(\frac{s}{R}\right)}_{min}=\sqrt{3}-1at\xi =\frac{\sqrt{3}-1}{\sqrt{2}}$$

Commented by mr W last updated on 30/Dec/21

Commented by amin96 last updated on 30/Dec/21

$$\mathit{g}\mathit{r}\mathit{e}\mathit{a}\mathit{a}\mathit{t}\mathit{s}\mathit{i}\mathit{r}$$

Commented by Tawa11 last updated on 30/Dec/21

$$\mathrm{Great}\mathrm{sir}$$

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