Question

The question is below?

Let `A_1,A_2,....A_n` be the vertices of n-sided regular polygon such that `1/(A_1A_2)=1/(A_1A_3)+1/(A_1A_4)`. Find the value of n.

Answer 1

Let each side of the regular n-sided polygon be `a` and the radius of the circumcircle of the polygon be R. Each side of the polygon will subtend angle `theta=(2pi)/n` at the center `O` of the circumcircle.

The perpendicular dropped from `O` to `A_1A_2` will bisect the `angleA_1OA_2=theta` as well as side `A_1A_2`

So `(A_1A_2)/2=Rsin(theta/2)`

Similarly the perpendicular dropped from `O` to `A_1A_3` will bisect the `angleA_1OA_3=2theta` as well as side `A_1A_3`

So `(A_1A_3)/2=Rsin((2theta)/2)`

And also the perpendicular dropped from `O` to `A_1A_4` will bisect the `angleA_1OA_4=3theta` as well as side `A_1A_4`

So `(A_1A_4)/2=Rsin((3theta)/2)`

Now given condition is

`1/(A_1A_2)=1/(A_1A_3)+1/(A_1A_4)`

Substituting the values of
`A_1A_2,A_1A_3andA_1A_4` we have

`1/(2Rsin(theta/2))=1/(2Rsin((2theta)/2))+1/(2Rsin((3theta)/2))`

`>1/(sin((2theta)/2))=1/(sin(theta/2))-1/(sin((3theta)/2))`

`=>1/(sin((2theta)/2))=(sin((3theta)/2)-sin(theta/2))/(sin((3theta)/2)sin(theta/2))`

`=>1/(sin((2theta)/2))=(2cos((2theta)/2)sin(theta/2))/(sin((3theta)/2)sin(theta/2))`

`=>sin((3theta)/2)=2cos((2theta)/2)sin((2theta)/2)`

`=>sin((3theta)/2)=sin((4theta)/2)`

`=>sin(pi-(3theta)/2)=sin((4theta)/2)`

`=>pi-(3theta)/2=(4theta)/2`

`=>(7theta)/2=pi`

`=>7/2*(2pi)/n=pi`

`=>n=7`