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Question Number 214012 by ajfour last updated on 24/Nov/24

Commented by ajfour last updated on 24/Nov/24

$$OutercircleradiusisR.Circlewith$$$$centerAhasradiusr=R/2.$$$$If△ABCisequilateral,findits$$$$edgelength\left(saya\right).$$

Answered by mr W last updated on 24/Nov/24

Commented by mr W last updated on 24/Nov/24

$$c=s-\frac{R}{2}$$$$C(s\sin \alpha ,-\frac{R}{2}+s\cos \alpha )$$$$B(s\sin (\alpha +\frac{\pi }{3}),-\frac{R}{2}+s\cos (\alpha +\frac{\pi }{3}))$$$$s^2{\sin }^2\alpha +{(-\frac{R}{2}+s\cos \alpha )}^2={(R-s+\frac{R}{2})}^2$$$$s^2{\sin }^2(\alpha +\frac{\pi }{3})+{[-\frac{R}{2}+s\cos (\alpha +\frac{\pi }{3})]}^2=R^2$$$$let\lambda =\frac{s}{R}$$$${\lambda }^2{\sin }^2\alpha +{(-\frac{1}{2}+\lambda \cos \alpha )}^2={(\frac{3}{2}-\lambda )}^2$$$$\frac{1}{4}-\lambda \cos \alpha =\frac{9}{4}-3\lambda$$$$\Rightarrow \lambda \cos \alpha =3\lambda -2\Rightarrow \lambda \sin \alpha =2\sqrt{(1-\lambda )(2\lambda -1)}$$$${\lambda }^2{\sin }^2(\alpha +\frac{\pi }{3})+{[-\frac{1}{2}+\lambda \cos (\alpha +\frac{\pi }{3})]}^2=1$$$${\lambda }^2-\lambda \cos (\alpha +\frac{\pi }{3})=\frac{3}{4}$$$${\lambda }^2-\frac{\lambda }{2}(\cos \alpha -\sqrt{3}\sin \alpha )=\frac{3}{4}$$$$2{\lambda }^2-3\lambda +\frac{1}{2}+2\sqrt{3(1-\lambda )(2\lambda -1)}=0$$$$\Rightarrow \lambda \approx 0.52391863,0.97608137$$

Commented by mr W last updated on 24/Nov/24

Commented by mr W last updated on 24/Nov/24

Commented by ajfour last updated on 24/Nov/24

$$YesSir.Itook△side=2s$$$${(8s^2-6s+\frac{1}{2})}^2+12(8s^2-6s+\frac{1}{2})+6=0$$$$\Rightarrow 8s^2+6s+\frac{1}{2}=-6-\sqrt{30}$$$$2s=\frac{3\pm \sqrt{8\sqrt{30}-43}}{4}$$$$\approx 0.97608,0.52392$$

Commented by ajfour last updated on 25/Nov/24

$$Let△sidebe2s.$$$$2s=r+c$$$$AndmidpointofACbe$$$$M(s\sin \alpha ,-r+s\cos \alpha )$$$$BM=s\sqrt{3}$$$$B(s\sin \alpha +s\sqrt{3}\cos \alpha ,-r+s\cos \alpha -s\sqrt{3}\sin \alpha )$$$${OB}^2=R^2$$$$s^2\sin {}^2\alpha +3s^2\cos {}^2\alpha +\cancel{2\sqrt{3}{s}^2\sin \alpha \cos \alpha }$$$$+r^2+s^2\cos {}^2\alpha +3s^2\sin {}^2\alpha -\cancel{2\sqrt{3}{s}^2\sin \alpha \cos \alpha }$$$$-2rs\cos \alpha +2\sqrt{3}rs\sin \alpha =R^2$$$$\Rightarrow$$$$4s^2-2rs\cos \alpha +2\sqrt{3}rs\sin \alpha =R^2-r^2$$$$say\frac{s}{R}=t,and\frac{r}{R}=\frac{1}{2}\left(given\right)$$$$4t^2-t\cos \alpha +\sqrt{3}t\sin \alpha =1-\frac{1}{4}.......\left(i\right)$$$$Now$$$$C(2s\sin \alpha ,-r+2s\cos \alpha )$$$${OC}^2={(R-c)}^2={(R+r-2s)}^2$$$$4s^2-4rs\cos \alpha =(R+2r-2s)(R-2s)$$$$2t^2-t\cos \alpha =(1-t)(1-2t).....\left(ii\right)$$$$\Rightarrow t\cos \alpha =3t-1$$$$from..\left(i\right)$$$$\sqrt{3}t\sin \alpha =1-\frac{1}{4}-4t^2+3t-1$$$$\Rightarrow 3{(3t-1)}^2+{(3t-4t^2-\frac{1}{4})}^2=3t^2$$$$\Rightarrow {(3t-4t^2-\frac{1}{4})}^2+24t^2-18t+3=0$$$${(3t-4t^2-\frac{1}{4})}^2-6(3t-4t^2-\frac{1}{4})+\frac{3}{2}=0$$$$\Rightarrow 3t-4t^2-\frac{1}{4}=3-\sqrt{\frac{15}{2}}$$$$\Rightarrow 16t^2-12t+13-\sqrt{120}=0$$$$\frac{s}{R}=t=\frac{3}{8}\pm \sqrt{\frac{9}{64}-\frac{52}{64}+\frac{8\sqrt{30}}{64}}$$$$2s=sidea=\left(\frac{3\pm \sqrt{8\sqrt{30}-43}}{4}\right)R✓$$$$★$$

Commented by mr W last updated on 25/Nov/24

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