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Question Number 186410 by mr W last updated on 04/Feb/23

Commented by mr W last updated on 04/Feb/23

$$seeQ186302$$

Answered by mr W last updated on 04/Feb/23

$$R=\frac{\sqrt{3}}{6}$$$$CO=2R$$$$\frac{\sin \gamma }{\sin \alpha }=\frac{CO}{R}=2$$$$\Rightarrow \sin \gamma =2\sin \alpha$$$$similarly$$$$\sin \delta =2\sin \beta$$$$\frac{CA}{\sin (\gamma +\alpha )}=\frac{R}{\sin \alpha }$$$$CA=\frac{R\sin (\gamma +\alpha )}{\sin \alpha }=R(\cos \gamma +\frac{\sin \gamma }{\tan \alpha })$$$$\Rightarrow CA=R(\sqrt{1-4{\sin }^2\alpha }+2\cos \alpha )$$$$AB=CA\times \sin (\alpha +\beta )=R(\sqrt{1-4{\sin }^2\alpha }+2\cos \alpha )\sin (\alpha +\beta )$$$$BA=2R\cos \varphi =2R\sin \delta =4R\sin \beta$$$$R(\sqrt{1-4{\sin }^2\alpha }+2\cos \alpha )\sin (\alpha +\beta )=4R\sin \beta$$$$(\sqrt{1-4{\sin }^2\alpha }+2\cos \alpha )(\frac{\sin \alpha }{\tan \beta }+\cos \alpha )=4$$$$\Rightarrow \tan \beta =\frac{\sin \alpha }{\frac{4}{\sqrt{1-4{\sin }^2\alpha }+2\cos \alpha }-\cos \alpha }$$$$\mathrm{\Delta }=\frac{{CA}^2\sin (\alpha +\beta )\cos (\alpha +\beta )}{2}$$$$\mathrm{\Delta }=\frac{R^2\sin 2(\alpha +\beta ){(\sqrt{1-4{\sin }^2\alpha }+2\cos \alpha )}^2}{4}$$$$\Rightarrow \mathrm{\Delta }=\frac{\sin 2(\alpha +\beta ){(\sqrt{1-4{\sin }^2\alpha }+2\cos \alpha )}^2}{48}$$$${\mathrm{\Delta }}_{max}=\frac{1}{6}at\alpha \approx 0.2111$$

Commented by mr W last updated on 04/Feb/23

Commented by mr W last updated on 04/Feb/23

$$itisnotobvioustomethatthe$$$$maximumtriangleshouldbeatthis$$$$position.$$

Commented by ajfour last updated on 04/Feb/23

$$ThanksSir,buttheobviousness$$$$ofthehorizontaldiameteris$$$$bewilderingindeed!$$

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