Let $A_{0}, A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ ae be regular hexagon inscribed in a circle of unit radius, then the product of the lengths of the line segments $A_{0} A_{1} \cdot A_{0} A_{2} \cdot A_{0} A_{4}$ is _____

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answer is 3.

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Detailed Solution

Given that $A_{0}, A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ are regular hexagon inscribed in a circle of radius $' 1'$

$∠ A_{0} O A = \frac{360}{6} = 60 °$

But in $O A_{0} = O A_{1} = 1$

$∠ O A_{0} A_{1} = ∠ O A_{1} A_{0} 60 °$

Therefore, $Δ A_{0} A_{1}$ is an equilateral triangle

$A_{0} A_{1} = 1 = A_{1} A_{2} = A_{2} A_{3} = A_{3} A_{4} = A_{4} A_{5} = A_{5} A_{0}$

$∠ A_{0} A_{1} A_{2} = 120 °$

Using cosine rule, we get

$\cos 120 ° = \frac{{(A_{0} A_{1})}^{2} + {(A_{1} A_{2})}^{2} - {(A_{0} A_{2})}^{2}}{2 (A_{0} A_{1}) (A_{1} A_{2})}$

$\begin{array}{l} - \frac{1}{2} = \frac{1 + 1 - {(A_{0} A_{2})}^{2}}{2 \times 1 \times 1} \\ - \frac{1}{2} = \frac{2 - {(A_{0} A_{2})}^{2}}{2} \end{array}$

$\begin{array}{l} {(A_{0} A_{2})}^{2} = 3 \\ ∴ A_{0} A_{2} = \sqrt{3} \\ ∴ A_{0} A_{1} . A_{0} A_{2} . A_{0} A_{4} = 1 \times \sqrt{3} \times \sqrt{3} \\ = 3 \end{array}$