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Question Number 184441 by ajfour last updated on 06/Jan/23

Commented by mr W last updated on 07/Jan/23

$$a=\frac{\sqrt{3}Rr}{\sqrt{R^2+r^2-Rr}}$$

Answered by ajfour last updated on 07/Jan/23

$$a=2r\cos \theta =2R\cos \varphi$$$$\theta +\varphi =\frac{\pi }{3}$$$$\Rightarrow {(\cos \theta \cos \varphi -\frac{1}{2})}^2$$$$=(1-\cos {}^2\theta )(1-\cos {}^2\varphi )$$$$\Rightarrow {(\frac{a^2}{4rR}-\frac{1}{2})}^2=(1-\frac{a^2}{4r^2})(1-\frac{a^2}{4R^2})$$$$\Rightarrow \frac{a^2}{4}(\frac{1}{r^2}+\frac{1}{R^2}-\frac{1}{rR})=\frac{3}{4}$$$$\Rightarrow a^2=\frac{3r^2{R}^2}{r^2+R^2-rR}$$$$\Rightarrow a=\frac{\sqrt{3}rR}{\sqrt{r^2+R^2-rR}}$$

Commented by mr W last updated on 07/Jan/23

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Answered by mr W last updated on 07/Jan/23

$$\cos \alpha =\frac{a}{2r}$$$$\cos \beta =\frac{a}{2R}$$$$\alpha +\beta =\frac{\pi }{3}$$$$\cos (\alpha +\beta )=\frac{a^2}{4Rr}-\frac{\sqrt{(4R^2-a^2)(4r^2-a^2)}}{4Rr}=\frac{1}{2}$$$$a^2-2Rr=\sqrt{(4R^2-a^2)(4r^2-a^2)}$$$$(R^2+r^2-Rr)a^2=3R^2{r}^2$$$$\Rightarrow a=Rr\sqrt{\frac{3}{R^2+r^2-Rr}}$$

Commented by mr W last updated on 07/Jan/23

Commented by mr W last updated on 07/Jan/23

Commented by ajfour last updated on 08/Jan/23

$$Thankssir,{we}^{\prime }vegotsamewayshere!$$

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