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Question

Let A1,A2,,An be the vertices of a regular polygon of n sides in a circle of radius unity and
a=|A1A2|2+|A1A3|2+|A1An|2,
b=|A1A2||A1A3||A1An|, then ab=

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Solution

Let us assume that O is the centre of the regular polygon and z0,z1,,zn1 represent the affixes of A1,A2,,An, such that
z0=1,z1=α,z2=α2,,zn1=αn1, where α=ei2π/n
(Since all nth roots of unity can be represented as verticies of regular polygon inscribed in a circle of unit radius)
Now, |A1Ar|2=|αr1|2=|1αr|2
=1cos2rπnisin2rπn2
=(1cos2rπn)2+(sin2rπn)2
=22cos2rπn
nr=2|A1A2|2=nr=1(22cosrπn)
=2(n1)2[cos2πn+cos4πn+cos2(n1)πn]
=2(n1)2 Real part of (α+α2++αn1)
=2(n1)2(1)=2n
[1+α+α2++αn1=0]
|A1A2|2+|A1A3|2+|A1An|2=2n

Also, let E=|A1A2||A1A3||A1An|
=|1α||1α2||1α3||1αn1|
=|(1α)(1α2)(1α3)(1αn1)|
Since, 1,α,α2,,αn1 are the roots of zn1=0
(z1)(zα)(zα2)(zαn1)=zn1
(zα)(zα2)(zαn1)=zn1z1
=1+z+z2++zn1
Substituting z=1, we have
=(1α)(1α2)(1α3)(1αn1)=n
|1α||1α2||1α3||1αn1|=n
Hence, the value of ab=2nn=2

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