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A circle of radius R touches externally a set of 12 circles each of radius r surrounding it. Each one of the smaller circles touches two circles of the set. Then \mathrm{\frac{R}{r}=\sqrt{m}+\sqrt{n}-1 \text {, where } m, n \in N\text { and } m+n \text { is }}

Option: 1

7


Option: 2

8


Option: 3

9


Option: 4

5


Answers (1)

best_answer

 Let the small circles of centres  A  and B meet at  M.

\mathrm{\begin{aligned} & \frac{A M}{O A}=\sin \frac{2 \pi}{24}=\sin 15^{\circ} \\ & \Rightarrow(R+r) \frac{\sqrt{3}-1}{2 \sqrt{2}}=r, \frac{R}{r}=\frac{2 \sqrt{2}}{\sqrt{3}-1}\left(1-\frac{\sqrt{3}-1}{2 \sqrt{2}}\right) \\ & =\frac{(\sqrt{3}+1)}{2}(2 \sqrt{2}-\sqrt{3}+1) \\ & =\sqrt{6}+\sqrt{2}-1 \Rightarrow m=6, n=2, m+n=8 \end{aligned}}

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