Let `A_(1), A_(2), A_(3),…,A_(n)` be the vertices of an n-sided regular polygon such that `(1)/(A_(1)A_(2))=(1)/(A_(1)A_(3))+(1)/(A_(1)A_(4)).` Find the value of n.

Explanation

Inscribe the regular polygon in a circle of radius R. Since `A_(1)A_(2), A_(1)A_(3)` and `A_(1)A_(4)` subtends arcs of measures `(2pi)/(n), (4pi)/n, (6pi)/n`,
`therefore A_(1)A_(2) = 2R sin(pi/n)`
`A_(1)A_(2) = 2Rsin(2pi/n)`
`A_(1)A_(2) = 2Rsin(3pi)/n`
`therefore 1/(A_(1)A_(2)) = 1/(A_(1)A_(3)) + 1/(A_(1)A_(4))`
`rArr 1/(2Rsin(pi//r)) = 1/(2Rsin(2pi//n)) + 1/(2Rsin(3pi//n))`
`rArr 1/(sin(pi/n)) =1/(sin(2pi)/n) + 1/(sin(3pi)/n)`
`rArr 1/(sinpi/n) -1/(sin(3pi)/n) = 1/(sin(2pi)/n)`
`(sin(3pi)/n-sin(pi/n))/(sin(pi/n)sin(3pi)/n)= 1(sin(2-pi)/(n))` `{therefore sin(pi//4) ne 0}`
`((2sin(pi/n) cos(2pi)/n)/(sin(pi/n)sin(3pi)/n))= 1/(sin(2pi)/n)`
`2sin(2pi)/b cos(2pi)/n = sin(3pi)/n`

`sin(4pi)/n = sin(3pi)/n`
`sin(4pi)/n - sin(3pi)/n=0`
`2sin(pi/(2n))cos(7pi)/(2n)=0`
`rArr cos(7pi)/(2n)=0` `[therefore sin(pi/(2n)) ne 0]`
`(7pi)/(2n) = 2m+1 pi/2, m int Z`
`n=7/(2m+1) rArr 2m+1` must divides 7
`rArr m=7, m=3`
All these steps, that we carried out are invertible, hence the converse also follows
In regular heptogen, `A_(1), A_(2), ............. , A_(7)` we have
`1/(A_(1)A_(2)) = 1/(A_(1)A_(3)) + 1/(A_(1)A_(4))`