The strange ratio

Geometry Level 3

12 circles of same radius are situated on a plane in such manner that the 1st circle externally touches the 2nd circle , the 2nd circle externally touches the 3rd circle ....... following the way , ....the 12th circle externally touches the 1st circle and all those 12 circles externally touches a bigger circle . If 'r' is the radius of those 12 smaller circles , 'R' is the radius of the bigger circle and r:R can be expressed as such 1: [{√(m-1)}+ (√n)− 1] where m and n are relatively prime natural numbers , find m×n .


The answer is 14.

Sudipto Podder by Sudipto Podder , 16, Bangladesh

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1 solution

Hongqi Wang
Jan 12, 2021

r = ( R + r ) sin ⁡ 360 ° 12 × 2 = ( R + r ) sin ⁡ 15 ° r R + r = sin ⁡ 15 ° = 6 − 2 4 = 1 6 + 2 r R = 1 6 + 2 − 1 ⟹ m = 7 , n = 2 \begin{aligned} r &= (R+r) \sin {\dfrac {360 \degree}{12 \times 2}} \\ &= (R+r)\sin{15\degree} \\ \dfrac r{R+r} &= \sin{15\degree} \\ &= \dfrac {\sqrt 6 - \sqrt 2}4 \\ &= \dfrac 1{\sqrt 6 + \sqrt 2} \\ \dfrac rR &= \dfrac 1{\sqrt 6 + \sqrt 2 - 1} \\ \implies &m = 7, n = 2 \end{aligned}

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Nice solution !!!

Sudipto Podder - 5 months ago

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