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.

1 Answer
Jun 6, 2018

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