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(method1)
(1) vectorOA+vectorOC= k (vector OB),  k=/= 2  (in fact k=2 sin(18 deg))
(2) vectorOB+vectorOD= k (vector OC)
(3) vectorOC+vectorOE= k (vector OD)
(4) vectorOD+vectorOA= k (vector OE)
(5) vectorOE+vectorOB= k (vector OA)
(1)+(2)+...+(5) then
2 (OA+ OB+OC+OD+OE)= k (OA+OB+OC+OD+OE)
but k =/= 2, so OA+OB+OC+OD+OE= vector(0)
 
(method2)
1. Let O(0,0), A(1,0), B(cost, sint), C(cos2t, sin2t), D(cos3t, sin3t), 
    E(cos4t, sin4t) ,where t= 2pi/5.
2. The solutions of x^5-1=0 are 1+0i, cost+i sint, ..., cos4t+i sin4t.
3. The sum of roots of x^5-1=0 equals 0, so
    (1+0i)+(cost+i sint)+...+(cos4t+i sin4t)= 0+0 i
    so, vector OA+OB+...+OE = (0, 0)
 

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