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Question Number 78443 by Maclaurin Stickker last updated on 17/Jan/20

Answered by mr W last updated on 18/Jan/20

Commented by mr W last updated on 18/Jan/20

$$sidelengthofsquare=a=x_0$$$$EF=2r+\frac{2r}{\sqrt{2}}=(2+\sqrt{2})r$$$$EB=r+\frac{r}{\tan \frac{\theta }{2}}=a\cos \theta \Rightarrow \frac{a}{r}=\frac{1+\frac{1}{\tan \frac{\theta }{2}}}{\cos \theta }$$$$EC=r+\frac{r}{\tan (\frac{\pi }{4}-\frac{\theta }{2})}=FB$$$$EB=EF+FB$$$$r+\frac{r}{\tan \frac{\theta }{2}}=(2+\sqrt{2})r+r+\frac{r}{\tan (\frac{\pi }{4}-\frac{\theta }{2})}$$$$\frac{1}{\tan \frac{\theta }{2}}=2+\sqrt{2}+\frac{1}{\tan (\frac{\pi }{4}-\frac{\theta }{2})}$$$$\frac{1}{\tan \frac{\theta }{2}}=2+\sqrt{2}+\frac{1+\tan \frac{\theta }{2}}{1-\tan \frac{\theta }{2}}$$$$lett=\tan \frac{\theta }{2}$$$$\frac{1}{t}-\frac{1+t}{1-t}=2+\sqrt{2}$$$$(1+\sqrt{2})t^2-(4+\sqrt{2})t+1=0$$$$t=\frac{4+\sqrt{2}-\sqrt{{(4+\sqrt{2})}^2-4(1+\sqrt{2})}}{2(1+\sqrt{2})}$$$$t=\tan \frac{\theta }{2}=\frac{3\sqrt{2}-2-\sqrt{2(13-8\sqrt{2})}}{2}\Rightarrow \theta =22.96°$$$$\frac{x_0}{r}=\frac{a}{r}=\frac{1+\frac{1}{\tan \frac{\theta }{2}}}{\cos \theta }=\frac{1+t^2}{(1-t)t}=6.4336$$

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