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Question Number 139957 by cherokeesay last updated on 02/May/21

Commented by mr W last updated on 02/May/21

$$R=r+\frac{2}{3}\times \sqrt{3}r$$$$\Rightarrow \frac{R}{r}=1+\frac{2\sqrt{3}}{3}$$

Commented by Dwaipayan Shikari last updated on 02/May/21

$$Ihavedoneinthesameway.Butwhattheoremyouused$$$$belowsir?KissingCircleTheorem?$$

Commented by mr W last updated on 02/May/21

$$Descartestheorem$$

Commented by cherokeesay last updated on 03/May/21

$$thankyousir!$$

Answered by mr W last updated on 02/May/21

$${(\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}-\frac{1}{R})}^2=2(\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{R^2})$$$${(\frac{3}{r}-\frac{1}{R})}^2=2(\frac{3}{r^2}+\frac{1}{R^2})$$$$\frac{9}{r^2}+\frac{1}{R^2}-\frac{6}{rR}=\frac{6}{r^2}+\frac{2}{R^2}$$$$\frac{3R^2}{r^2}-\frac{6R}{r}-1=0$$$$\Rightarrow \frac{R}{r}=1+\frac{2\sqrt{3}}{3}\approx 2.1547$$

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