Two polygons in a circle

Geometry Level 5

Two regular polygons are inscribed in the same circle. First polygon has 2016 sides and the second polygon has n ( < 2016 ) n (<2016) sides.

If the polygons have only one vertex in common, find the number of possible values of n n .


The answer is 575.

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1 solution

Harsh Shrivastava
Nov 24, 2016

Call the polygon with number of sides 2016 2016 A A and call the other polygon B B .

In the argand plane, sides of a A A are represented by the 2016 t h 2016th roots of unity and sides of B B are represented by the n t h nth roots of unity.

Thus the number of common vertices is given by the number of common roots of z 2016 1 = 0 z^{2016}-1=0 and z n 1 = 0 z^{n}-1=0 which implies number of common vertices = g c d ( 2016 , n ) =gcd(2016,n) .

It is given in the question that this value is one.

Therefore, number of possible values of n n is all the integers less than or equal 2016 2016 which are co-prime with it.

Thus our answer is ϕ ( 2016 ) 1 = 575 \phi (2016) - 1 = 575 . Since n is not 1. \text{Since n is not 1.}

ϕ ( x ) \phi (x) represents the Euler Totient Function .


Generalization:

For two regular polygons with number of sides m m and n n ,number of common vertices is gcd ( m , n ) \gcd(m,n) .

nice question

initially forgot to subtract 1 and thought something went wrong!but yeah (+1) :)

Rohith M.Athreya - 4 years, 6 months ago

I think that you need to add a condition that n < 2016 n<2016 otherwise any prime fits.

Maria Kozlowska - 4 years, 6 months ago

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Well yes thanks for pointing that error out.

I have edited the problem.

Harsh Shrivastava - 4 years, 6 months ago

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