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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
P dilip_k
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#

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