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Question Number 128944 by ajfour last updated on 11/Jan/21

Commented by ajfour last updated on 11/Jan/21

$$Findradiusratio\frac{R}{r}.$$$$Thepolygonisaregular$$$$pentagon.$$

Answered by mr W last updated on 11/Jan/21

Commented by mr W last updated on 11/Jan/21

$$sidelengthofpentagon=a$$$$a=\frac{2r}{\tan 36°}$$$$AE=\frac{r}{\tan 36°}\times \tan 72°=\frac{2r}{1-{\tan }^{2}36°}$$$$AC=\frac{2r}{1-{\tan }^{2}36°}-2r$$$$AB=\frac{2r}{\tan 36°}\times \sin 36°=2r\cos 36°$$$$BC=2r(\frac{1}{1-{\tan }^{2}36°}-1-\cos 36°)$$$$GB=\frac{2r}{\tan 36°}\times \cos 36°$$$${BG}^2=(2R-BC)BC$$$${\left(\frac{2r}{\tan 36°}\times \cos 36°\right)}^2=(2R-2r(\frac{1}{1-{\tan }^{2}36°}-1-\cos 36°))2r(\frac{1}{1-{\tan }^{2}36°}-1-\cos 36°)$$$$let\lambda =\frac{R}{r}$$$${\left(\frac{\cos 36°}{\tan 36°}\right)}^2=(\lambda -\frac{1}{1-{\tan }^{2}36°}+1+\cos 36°)(\frac{1}{1-{\tan }^{2}36°}-1-\cos 36°)$$$$\Rightarrow \lambda =\frac{{\cos }^{2}36°}{{\tan }^{2}36°(\frac{1}{1-{\tan }^{2}36°}-1-\cos 36°)}+\frac{1}{1-{\tan }^{2}36°}-1-\cos 36°$$$$=\frac{35+23\sqrt{5}}{20}$$$$\approx 4.321478$$

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