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Question Number 54045 by ajfour last updated on 28/Jan/19

Commented by ajfour last updated on 28/Jan/19

$$ABCDisasquare,anditsdiagonal$$$$asasideoftherectangle.FindR/r.$$

Answered by ajfour last updated on 28/Jan/19

$$\frac{R}{r}=2\left(\frac{\frac{\sqrt{5}+1}{2}+\sqrt{2}+1}{\sqrt{5}+\sqrt{2}+3}\right)$$$$=\frac{\sqrt{5}+2\sqrt{2}+3}{\sqrt{5}+\sqrt{2}+3}.$$

Answered by mr W last updated on 28/Jan/19

$$letAB=1$$$$\tan \alpha =\frac{1}{2}\Rightarrow \tan \frac{\alpha }{2}=\sqrt{5}-2$$$$\tan \beta =2\Rightarrow \tan \frac{\beta }{2}=\frac{\sqrt{5}-1}{2}$$$$\tan \frac{45°}{2}=\sqrt{2}-1$$$$$$$$\frac{R}{\tan \frac{\alpha }{2}}+\frac{R}{\tan \frac{45°}{2}}=\sqrt{2}$$$$\frac{R}{\sqrt{5}-2}+\frac{R}{\sqrt{2}-1}=\sqrt{2}$$$$\Rightarrow R=\frac{(2-\sqrt{2})(\sqrt{5}-2)}{\sqrt{5}+\sqrt{2}-3}$$$$\frac{r}{\tan \frac{45°}{2}}+\frac{r}{\tan \frac{\beta }{2}}=\frac{\sqrt{2}}{2}$$$$\frac{r}{\sqrt{2}-1}+\frac{2r}{\sqrt{5}-1}=\frac{\sqrt{2}}{2}$$$$\Rightarrow r=\frac{(2-\sqrt{2})(\sqrt{5}-1)}{2(\sqrt{5}+2\sqrt{2}-3)}$$$$\Rightarrow \frac{R}{r}=\frac{(3-\sqrt{5})(\sqrt{5}+2\sqrt{2}-3)}{2(\sqrt{5}+\sqrt{2}-3)}\approx 1.21265$$

Commented by ajfour last updated on 29/Jan/19

$$thanksSir.$$

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