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$\vv{AB}·\vv{BC}=\vv{BC}·\vv{CA}=\vv{CA}·\vv{AB}$则三角形ABC是等边三角形
$\vv{AB}·\vv{BC}=\vv{BC}·\vv{CA}=\vv{CD}·\vv{DA}=\vv{DA}·\vv{AB}$则四边形ABCD或ACBD或ABDC是矩形
凸四边形的情况很好证明,若
$\vv{AB}·\vv{BC}=\vv{BC}·\vv{CA}=\vv{CD}·\vv{DA}=\vv{DA}·\vv{AB}$>0,则四个内角都是钝角
$\vv{AB}·\vv{BC}=\vv{BC}·\vv{CA}=\vv{CD}·\vv{DA}=\vv{DA}·\vv{AB}$<0,则四个内角都是锐角,矛盾.所以四边形ABCD是矩形
那么五边形情况怎样?(首先五边形及五角星满足)
设点列{$A_n$},满足$A_nA_{n+1}-A_{n+2}A_{n+3}⊥A_{n+1}A_{n+2}$
这时候可以看出来实际上是俩数列
确定这样的复数列{$A_n$},使$A_{n+1}=A_1,A_{n+2}=A_2,A_{n+3}=A_3$,且存在实数列{$k_n$},$A_{n+3}=A_{n+2}-A_{n+1}+ik_n(A_{n+1}-A_{n})$ |
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