ABCD are the vertex of a regular polygon Supposing that 1AB1

A,B,C,D,... are the vertex of a regular polygon. Supposing that 1/AB=(1/AC)+(1/AD), then how many sides has the polygon?

Solution

If n is the number of polygon\'s sides, R is the radius of the circumscribed circle and x=pi/n, then we can write AB=2Rsinx, AC=2Rsin2x, AD=2Rsin3x.

From the expression 1/AB=(1/AC)+(1/AD), we\'ll have:

1/sin x= (1/sin 2x) + (1/sin 3x)

Having the same denominator, both sides of equation, we\\ll have:

sin 2x*sin 3x= sin x* sin 3x + sin x*sin 2x

We\'ll have the common factor sin x and we\'ll write

sin2x=2sinx cos x

2sinx cos x sin 3x=sin x(sin 3x + sin 2x)

We\'ll simplify with sin x

2 cos x sin 3x=(sin 3x + sin 2x)

We\'ll transform the product in sum

2 cos x sin 3x=sin((x+3x)/2)-sin((x-3x)/2)=sin 2x+sin x

sin 2x+sin x=sin 3x + sin 2x

sin 3x - sin x =0

2cos2xsinx=0

x=2kpi

x=2kpi/7 + pi/7

x=pi/n, so n=7