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Question Number 203416 by Spillover last updated on 18/Jan/24

Answered by a.lgnaoui last updated on 19/Jan/24

$$$$$$\mathbf{R}=2\mathbf{r}+\mathbf{BM}\cos \mathit{\alpha }\left(1\right)$$$$\mathbf{BM}\sin \mathit{\alpha }=\mathbf{r}\left(2\right)$$$$$$$$\Rightarrow \{\begin{array}{l}\cos \mathit{\alpha }=\frac{\mathbf{R}-2\mathbf{r}}{\mathbf{BM}}\\ \sin \mathit{\alpha }=\frac{\mathbf{r}}{\mathbf{BM}}\end{array}$$$$\tan \mathit{\alpha }=1\Rightarrow =45\mathbf{BM}=\mathbf{r}\sqrt{2}$$$$$$$$\{\begin{array}{l}\frac{\mathbf{R}-2\mathbf{r}}{\mathbf{BM}}=\frac{\mathbf{R}-2\mathbf{r}}{\mathbf{r}\sqrt{2}}\\ \frac{\mathbf{r}}{\mathbf{BM}}=\frac{1}{\sqrt{2}}\end{array}$$$$$$$$\frac{\left(1\right)}{2\mathbf{r}}\Rightarrow \frac{\mathbf{R}}{2\mathbf{r}}=\frac{2\sqrt{2}+1}{2\sqrt{2}}$$$$$$$$$$

Commented by a.lgnaoui last updated on 19/Jan/24

Answered by mr W last updated on 19/Jan/24

Commented by mr W last updated on 19/Jan/24

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

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