Consider a regular heptagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6} A_{7}$ inscribed in a unit circle.
Let $p_{1} = (A_{1} A_{2}) (A_{1} A_{3}) (A_{1} A_{4}) (A_{1} A_{5}) (A_{1} A_{6}) (A_{1} A_{7})$ ,
$p_{2} = (A_{2} A_{3}) (A_{2} A_{4}) (A_{2} A_{5}) (A_{2} A_{6}) (A_{2} A_{7}),$
$p_{3} = (A_{3} A_{4}) (A_{3} A_{5}) (A_{3} A_{6}) (A_{3} A_{7}),$
$p_{4} = (A_{4} A_{5}) (A_{4} A_{6}) (A_{4} A_{7}),$
$p_{5} = (A_{5} A_{6}) (A_{5} A_{7})$ and $p_{6} = A_{6} A_{7}$ where $A_{i} A_{j}$ is length of the segment joining $A_{i}$ and $A_{j}$ then value of product ${(p_{1} p_{2} p_{3} p_{4} p_{5} p_{6})}^{\frac{2}{7}}$ is

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answer is 7.

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Detailed Solution

Let vertices $A_{1}, A_{2}, .... A_{7}$ are $7^{t h}$ roots of unity. Let $A_{1} (1), A_{2} (α), A_{3} (α^{2}), A_{4} (α^{3})$

$A_{5} = (α^{4}) = \bar{α^{3}}, A_{6} = (α^{5}) = \bar{α^{2}}, A_{7} = (α^{6}) = \bar{α} (1 - α) (1 - α^{2}) (1 - α^{3}) (1 - α^{4}) (1 - α^{5}) (1 - α^{6}) = 7 \Rightarrow {(| 1 - α | | 1 - α^{2} | | 1 - α^{3} |)}^{2} = 7 p_{1} = (A_{1} A_{2}) (A_{1} A_{3}) (A_{1} A_{4}) (A_{1} A_{5}) (A_{1} A_{6}) (A_{1} A_{7}) = {(| 1 - α | | 1 - α^{2} | | 1 - α^{3} |)}^{2} = {(\sqrt{7})}^{2}$

Similarly, $p_{2} = | 1 - α | | 1 - α^{2} | | 1 - α^{3} | | 1 - α^{2} | \sqrt{7} (| 1 - α^{3} | | 1 - a^{2} |) p_{3} = \sqrt{7} (| 1 - α^{2} |) p_{4} = \sqrt{7}, p_{5} = | 1 - α | | 1 - α^{2} | a n d p_{6} = | 1 - α | \Rightarrow p_{1} . p_{2} . p_{3} . p_{4} . p_{5} . p_{6} = {(\sqrt{7})}^{7}$