To solve this problem, let the coordinate of Bn be (xn, yn), with 2/3x12 = √ 3x1= √ 3/2.

So ~ for the convenience of comparison:

( 1)2/3 x 1^2=√3/3 x 1

(2) 2/3 x2^2-4/3 x 1^2=√3/3 x2

2/3 x3^2-4/3 x2^2+4/3 x 1^2=√3/3 x3

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2/3 x2008^2-4/3 x2007^2+4/3 x2006^2-...-4/3 x 1^2=√3/3 x2008

Then you can see the above formula, followed by the front, for example:

( 1)+(2)=2/3(x2^2-x 1^2)=√3/3(x 1+x2)

That is x2-x 1=√3/2.

Similarly, it can be calculated as x2008-x2007=√3/2.

x 1=√3/2

Arithmetic progression, you can calculate the results yourself ~