Circles in the Circle

Geometry Level 3

Let n 3 n \ge 3 be an integer. Assume that inside a big circle, exactly n n small circles of radius r r can be drawn so that each small circle touches the big circle and also touches both its adjacent small circles. Then, the radius of the big circle is:

None of the others r ( 1 + csc 2 π n ) r \left(1 + \csc\frac {2\pi} n \right) r csc π n r \csc\frac \pi n r ( 1 + csc π 2 n ) r \left(1 + \csc\frac \pi {2n} \right) r ( 1 + csc π n ) r \left(1 + \csc\frac \pi n \right)

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2 solutions

Chew-Seong Cheong
Feb 21, 2020

Consider a general case of n n small circles of radius r r touching the circumference of a large circle internally. The the distance between centers of two adjacent small circles is 2 r 2r and the angle extended by the two centers and the center O O of the large circle is (\frac {2\pi}n}. Then the radius of the large circle is given by:

O Q = P Q + O P = r + r sin π n = r ( 1 + cosec π n ) OQ = PQ + OP = r + \frac r{\sin \frac \pi n} = \boxed{r\left(1+\cosec \frac \pi n\right)}

The centers of n n small circles form a regular n n -gon whose each internal angle is ( 2 n 4 ) π 2 n (2n-4)\dfrac{π}{2n} . The distance of any vertex of this n n -gon from it's centre is r sec ( π 2 π n ) = r cosec π n r\sec (\dfrac{π}{2}-\dfrac{π}{n})=r\cosec \dfrac{π}{n} . Hence, the radius of the big circle is r + r cosec π n = r ( 1 + cosec π n ) r+r\cosec \dfrac{π}{n}=\boxed {r(1+\cosec \dfrac{π}{n})}

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