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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

see Q186302

$${see}\:{Q}\mathrm{186302} \\ $$

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

R=((√3)/6) CO=2R ((sin γ)/(sin α))=((CO)/R)=2 ⇒sin γ=2 sin α similarly sin δ=2 sin β ((CA)/(sin (γ+α)))=(R/(sin α)) CA=((R sin (γ+α))/(sin α))=R(cos γ+((sin γ)/(tan α))) ⇒CA=R((√(1−4 sin^2 α))+2 cos α) AB=CA×sin (α+β)=R((√(1−4 sin^2 α))+2 cos α) sin (α+β) BA=2R cos φ=2R sin δ=4R sin β R((√(1−4 sin^2 α))+2 cos α) sin (α+β)=4R sin β ((√(1−4 sin^2 α))+2 cos α)(((sin α)/(tan β))+cos α)=4 ⇒tan β=((sin α)/((4/( (√(1−4 sin^2 α))+2 cos α))−cos α)) Δ=((CA^2 sin (α+β) cos (α+β))/2) Δ=((R^2 sin 2(α+β)((√(1−4 sin^2 α))+2 cos α)^2 )/4) ⇒Δ=(( sin 2(α+β)((√(1−4 sin^2 α))+2 cos α)^2 )/(48)) Δ_(max) =(1/6) at α≈0.2111

$${R}=\frac{\sqrt{\mathrm{3}}}{\mathrm{6}} \\ $$$${CO}=\mathrm{2}{R} \\ $$$$\frac{\mathrm{sin}\:\gamma}{\mathrm{sin}\:\alpha}=\frac{{CO}}{{R}}=\mathrm{2} \\ $$$$\Rightarrow\mathrm{sin}\:\gamma=\mathrm{2}\:\mathrm{sin}\:\alpha \\ $$$${similarly} \\ $$$$\mathrm{sin}\:\delta=\mathrm{2}\:\mathrm{sin}\:\beta \\ $$$$\frac{{CA}}{\mathrm{sin}\:\left(\gamma+\alpha\right)}=\frac{{R}}{\mathrm{sin}\:\alpha} \\ $$$${CA}=\frac{{R}\:\mathrm{sin}\:\left(\gamma+\alpha\right)}{\mathrm{sin}\:\alpha}={R}\left(\mathrm{cos}\:\gamma+\frac{\mathrm{sin}\:\gamma}{\mathrm{tan}\:\alpha}\right) \\ $$$$\Rightarrow{CA}={R}\left(\sqrt{\mathrm{1}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\alpha}+\mathrm{2}\:\mathrm{cos}\:\alpha\right) \\ $$$${AB}={CA}×\mathrm{sin}\:\left(\alpha+\beta\right)={R}\left(\sqrt{\mathrm{1}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\alpha}+\mathrm{2}\:\mathrm{cos}\:\alpha\right)\:\mathrm{sin}\:\left(\alpha+\beta\right) \\ $$$${BA}=\mathrm{2}{R}\:\mathrm{cos}\:\phi=\mathrm{2}{R}\:\mathrm{sin}\:\delta=\mathrm{4}{R}\:\mathrm{sin}\:\beta \\ $$$${R}\left(\sqrt{\mathrm{1}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\alpha}+\mathrm{2}\:\mathrm{cos}\:\alpha\right)\:\mathrm{sin}\:\left(\alpha+\beta\right)=\mathrm{4}{R}\:\mathrm{sin}\:\beta \\ $$$$\left(\sqrt{\mathrm{1}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\alpha}+\mathrm{2}\:\mathrm{cos}\:\alpha\right)\left(\frac{\mathrm{sin}\:\alpha}{\mathrm{tan}\:\beta}+\mathrm{cos}\:\alpha\right)=\mathrm{4} \\ $$$$\Rightarrow\mathrm{tan}\:\beta=\frac{\mathrm{sin}\:\alpha}{\frac{\mathrm{4}}{\:\sqrt{\mathrm{1}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\alpha}+\mathrm{2}\:\mathrm{cos}\:\alpha}−\mathrm{cos}\:\alpha} \\ $$$$\Delta=\frac{{CA}^{\mathrm{2}} \:\mathrm{sin}\:\left(\alpha+\beta\right)\:\mathrm{cos}\:\left(\alpha+\beta\right)}{\mathrm{2}} \\ $$$$\Delta=\frac{{R}^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{2}\left(\alpha+\beta\right)\left(\sqrt{\mathrm{1}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\alpha}+\mathrm{2}\:\mathrm{cos}\:\alpha\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow\Delta=\frac{\:\mathrm{sin}\:\mathrm{2}\left(\alpha+\beta\right)\left(\sqrt{\mathrm{1}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\alpha}+\mathrm{2}\:\mathrm{cos}\:\alpha\right)^{\mathrm{2}} }{\mathrm{48}} \\ $$$$\Delta_{{max}} =\frac{\mathrm{1}}{\mathrm{6}}\:{at}\:\alpha\approx\mathrm{0}.\mathrm{2111} \\ $$

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

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

it is not obvious to me that the maximum triangle should be at this position.

$${it}\:{is}\:{not}\:{obvious}\:{to}\:{me}\:{that}\:{the} \\ $$$${maximum}\:{triangle}\:{should}\:{be}\:{at}\:{this} \\ $$$${position}. \\ $$

Commented by ajfour last updated on 04/Feb/23

Thanks Sir, but the obviousness of the horizontal diameter is bewildering indeed!

$${Thanks}\:{Sir},\:{but}\:{the}\:{obviousness} \\ $$$${of}\:{the}\:{horizontal}\:{diameter}\:{is} \\ $$$$\:{bewildering}\:{indeed}! \\ $$

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