Question
Let $${A_0}{A_1}{A_2}{A_3}{A_4}{A_5}$$ be a regular hexagon inscribed in a circle of unit radius. Then the product of the lengths of the line segments $${A_0}{A_1},{A_0}{A_2}\,{\text{and }}{A_0}{A_4}\,{\text{is}}$$
A.
$$\frac{3}{4}$$
B.
$${3\sqrt 3 }$$
C.
$$3$$
D.
$$\frac{{3\sqrt 3 }}{2}$$
Answer :
$$3$$
Solution :
Let the vertices be $${z_0},{z_1},.....,{z_5}$$ w.r.t centre $$O$$ as origin $$\left| {{z_0}} \right| = 1,$$
$$\eqalign{
& {A_0}{A_1} = \left| {{z_1} - {z_0}} \right| = \left| {{z_0}{e^{i\theta }} - {z_0}} \right| \cr
& \therefore {A_0}{A_1} = \left| {{z_0}} \right|\left| {\cos \theta + i\sin \theta - 1} \right| \cr
& = 1 \cdot \sqrt {{{\left( {\cos \theta - 1} \right)}^2} + {{\sin }^2}\theta } = \sqrt {2\left( {1 - \cos \theta } \right)} \cr
& \therefore {A_0}{A_1} = \sqrt {2.2{{\sin }^2}\frac{\theta }{2}} = 2\sin \frac{\theta }{2} \cr} $$
Where $$\theta = \frac{{2\pi }}{6} = \frac{\pi }{3}.$$ Replacing $$\theta $$ by $$2\theta $$ and $$4\theta ,$$
we get, $${A_0}{A_2} = 2\sin \frac{{2\theta }}{2} = 2\sin \theta \,\,\& \,\,{A_0}{A_4} = 2\sin \frac{{4\theta }}{2} = 2\sin 2\theta $$
$$\eqalign{
& \therefore \left( {{A_0}{A_1}} \right)\left( {{A_0}{A_2}} \right)\left( {{A_0}{A_4}} \right) \cr
& = 8\sin \frac{\pi }{6}\sin \frac{\pi }{3}\sin \frac{{2\pi }}{3} \cr
& = 8\left( {\frac{1}{2}} \right)\left( {\frac{{\sqrt 3 }}{2}} \right)\left( {\frac{{\sqrt 3 }}{2}} \right) = 3 \cr} $$