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Question Number 125491 by ajfour last updated on 11/Dec/20

Commented by ajfour last updated on 11/Dec/20

$$Ifoutermosttriangleisequilateral$$$$andradiusofcircleisunity,find$$$$maximumbrownshadedarea.$$

Answered by mr W last updated on 11/Dec/20

Commented by mr W last updated on 11/Dec/20

$$BC=2\cos \theta$$$$AB=\sqrt{7}$$$$\tan \beta =\frac{2{\sin }^2\theta }{\sqrt{3}+2\sin \theta \cos \theta }$$$$DE=\sqrt{3}\tan \beta =\frac{2\sqrt{3}{\sin }^2\theta }{\sqrt{3}+2\sin \theta \cos \theta }$$$$BD=2-\frac{2\sqrt{3}{\sin }^2\theta }{\sqrt{3}+2\sin \theta \cos \theta }$$$$A={\mathrm{\Delta }}_{BCD}=\frac{1}{2}\times 2\cos \theta \times (2-\frac{2\sqrt{3}{\sin }^2\theta }{\sqrt{3}+2\sin \theta \cos \theta })\sin \theta$$$$\Rightarrow A=\sin 2\theta (1-\frac{\sqrt{3}{\sin }^2\theta }{\sqrt{3}+\sin 2\theta })$$$$$$$$A_{max}=0.7393at\theta =35.2354°$$

Commented by ajfour last updated on 11/Dec/20

Commented by mr W last updated on 11/Dec/20

Commented by ajfour last updated on 11/Dec/20

$$\tan \alpha =\frac{\sqrt{3}}{2}$$$$LetbottommostpointofcircleF.$$$$\mathrm{\angle }ADF=\varphi ,centerofcircleO.$$$$letOD=y$$$$\frac{\sin 2\theta }{CD}=\frac{\sin \varphi }{1}=\frac{\sin (\varphi +2\theta )}{y}$$$$\Rightarrow y=\cos 2\theta +\frac{\sin 2\theta }{\tan \varphi }$$$$\tan \varphi =\frac{\sqrt{3}}{1-OD}=\frac{\sqrt{3}}{1-y}$$$$\Rightarrow y=\cos 2\theta +\frac{(1-y)\sin 2\theta }{\sqrt{3}}$$$$\Rightarrow y=\frac{\sqrt{3}\cos 2\theta +\sin 2\theta }{\sqrt{3}+\sin 2\theta }$$$$Areas=2△=(1+y)\sin 2\theta$$$$s=(1+y)\sin 2\theta$$$$=(1+\frac{\sqrt{3}\cos 2\theta +\sin 2\theta }{\sqrt{3}+\sin 2\theta })\sin 2\theta$$$$let\tan \theta =m$$$$s=\sin 2\theta (2-\frac{2\sqrt{3}\sin {}^2\theta }{\sqrt{3}+2\sin \theta \cos \theta })$$$$=\frac{4m}{(1+m^2)}[1-\frac{\sqrt{3}{m}^2}{\sqrt{3}(1+m^2)+2m}]$$$$\Rightarrow {△}_{max}=\frac{s_{max}}{2}\approx 0.7393$$$$\theta \approx {\tan }^{-1}(0.7063)\approx 35.234°$$

Commented by ajfour last updated on 11/Dec/20

$$s=f\left(m\right)$$

Commented by ajfour last updated on 11/Dec/20

$$Excellent,Sir,Thanksalot!$$

Commented by mr W last updated on 12/Dec/20

$$s=\sin 2\theta (2-\frac{2\sqrt{3}\sin {}^2\theta }{\sqrt{3}+2\sin \theta \cos \theta })$$$$\Rightarrow s=2\sin 2\theta (1-\frac{\sqrt{3}\sin {}^2\theta }{\sqrt{3}+\sin 2\theta })$$$$wegotthesameequationinterms$$$$of\theta .$$

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