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Question Number 73737 by ajfour last updated on 15/Nov/19

Commented by ajfour last updated on 15/Nov/19

$$FindR/r.$$

Answered by ajfour last updated on 15/Nov/19

$$Igotx=\frac{R}{r}=5+2\sqrt{6}(isitright?)$$

Commented by mr W last updated on 15/Nov/19

$$correctsir!$$

Answered by mr W last updated on 15/Nov/19

Commented by mr W last updated on 15/Nov/19

$$[\frac{2}{3}\sqrt{3}r+\sqrt{{(R+r)}^2-R^2}]\times 2=\frac{1}{3}\sqrt{3}r+\sqrt{{(R+r)}^2-r^2}$$$$\sqrt{3}r+2\sqrt{r(2R+r)}=\sqrt{R(R+2Rr)}$$$$4r\sqrt{3r(2R+r)}=(R-7r)(R+r)$$$$R^4-12R^{3}r+22R^2{r}^2-12{Rr}^3+r^2=0$$$${(R-r)}^2(R^2-10Rr+r^2)=0$$$$\Rightarrow R=(5+2\sqrt{6})r$$

Commented by ajfour last updated on 15/Nov/19

$$\tan 30°=\frac{\sqrt{{(R+r)}^2-R^2}+\frac{2r}{\sqrt{3}}}{R}$$$$let\frac{r}{R}=x$$$$\Rightarrow \frac{1}{\sqrt{3}}=\sqrt{x^2+2x}+\frac{2x}{\sqrt{3}}$$$${(1-2x)}^2=3(x^2+2x)$$$$\Rightarrow x^2-10x+1=0$$$$\Rightarrow x=5-2\sqrt{6}$$$$\frac{R}{r}=\frac{1}{x}=5+2\sqrt{6}.$$

Commented by ajfour last updated on 15/Nov/19

$$ThankyouSir.Ihavealsoposted$$$$howisolvedit,Sir.$$

Commented by mr W last updated on 15/Nov/19

$$verynice!$$$$yourpathisbetter,sinceityields$$$$aquadraticequation.$$

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