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Question Number 121793 by ajfour last updated on 11/Nov/20

Commented by ajfour last updated on 13/Nov/20

$$IfO\stackrel{⌢}{DE}isasector,and△ABC$$$$isequilateralwithminimumarea$$$$hasanareaequalto△OEF,find\theta .$$$$$$

Commented by ajfour last updated on 14/Nov/20

$$mrWSirkindlyhelpwiththisQ.$$

Answered by ajfour last updated on 13/Nov/20

$$letOD=OE=1,Fbeorigin.$$$$OF=p=\cos \theta ,FB=y,FC=x$$$$2p\sqrt{1-p^2}=\frac{\sqrt{3}{s}^2}{4}$$$$\Rightarrow 64\left(\frac{p^2}{s^2}\right)(\frac{1}{s^2}-\frac{p^2}{s^2})=3$$$$\Rightarrow say{\left(\frac{p}{s}\right)}^2=t$$$$\Rightarrow t^2-\frac{t}{s^2}+\frac{3}{64}=0$$$$\Rightarrow t=\frac{p^2}{s^2}=\frac{1}{2s^2}\pm \sqrt{\frac{1}{4s^4}-\frac{3}{64}}...\left(I\right)$$$$or{p}^2=\frac{1}{2}\pm \sqrt{\frac{1}{4}-\frac{3s^4}{64}}$$$${BC}^2=s^2=x^2+y^2$$$$letMbemidpointofBC.$$$$M(\frac{x}{2},\frac{y}{2})$$$$slopeofAMm=\frac{x}{y}$$$$A(p+\frac{x}{2}+\frac{s\sqrt{3}}{2}\frac{y}{s},\frac{y}{2}+\frac{s\sqrt{3}}{2}\frac{x}{s})$$$$AsOA=1$$$${(p+\frac{x}{2}+\frac{y\sqrt{3}}{2})}^2+{(\frac{y}{2}+\frac{x\sqrt{3}}{2})}^2=1$$$$\Rightarrow$$$$p^2+p(x+y\sqrt{3})+x^2+y^2+\sqrt{3}xy=1$$$$with{s}^2=x^2+y^{2}aminimum$$$$letx=s\cos \varphi ,y=s\sin \varphi$$$$\Rightarrow$$$$p^2+sp(\cos \varphi +\sqrt{3}\sin \varphi )+$$$$s^2+\sqrt{3}{s}^2\sin \varphi \cos \varphi =1$$$$\Rightarrow \frac{p^2}{s^2}+\frac{p}{s}(\cos \varphi +\sqrt{3}\sin \varphi )+1$$$$+\sqrt{3}\sin \varphi \cos \varphi -\frac{1}{s^2}=0$$$$Andfrom\left(I\right)wehave$$$$\frac{p^2}{s^2}=\frac{1}{2s^2}\pm \sqrt{\frac{1}{4s^4}-\frac{3}{64}}$$$$......$$$$$$

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