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Question Number 212802 by Spillover last updated on 24/Oct/24

Answered by mr W last updated on 24/Oct/24

Commented by mr W last updated on 24/Oct/24

$${\left(2R\right)}^2-{\left(2r\right)}^2={(R+r)}^2-{(R-r)}^2$$$${\left(\frac{R}{r}\right)}^2-\left(\frac{R}{r}\right)-1=0$$$$\Rightarrow \frac{R}{r}=\frac{1+\sqrt{5}}{2}=\phi$$

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