Let A0A1A2A3A4A5 be a regular hexagon inscribed in a circle of unit radius. Then the product of the lengths of the line segmentsA0Al,A0A2, and A0A4 is

Let O be the center of the circle. ∠A0OA1=360∘6=60∘Thus, A0OA1  is an equilateral triangle. We get also  A0A1=1A0A2=A0A4=2A0D=2OA0sin⁡60∘=2(1)32=3A0A1A0A2A0A4=(1)(3)(3)=3