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24 lines
806 B
24 lines
806 B
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根据条件，知道$OA=\sqrt{3},OB=1$
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$\therefore AB=2,\angle OAB=30^{\circ},\angle OBA=60^{\circ}$
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### 第一问
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当$A'B \perp OB$时,$A'B=\sqrt{2},OA'=OA=\sqrt{3}$
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$\therefore OB=\sqrt{\sqrt{3}^2=\sqrt{2}^2}=1$
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$\therefore A'(\sqrt{2},1)$
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### 第二问
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$P$是中点，则$OP$是直角三角形$AOB$的斜边中线，
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$\therefore OP=AP=A'P=BP=1$
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$\because \angle OBA=60^{\circ}$
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$\therefore \triangle OPB$是等边三角形
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$\therefore \angle OPB=60^{\circ}$
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$\therefore \angle OPA=120^{\circ}$
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由于翻折，所以$\angle OPA=\angle OPA'$
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$\therefore \angle BPA'=60^{\circ}$
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$\therefore \triangle A'BP$是等边三角形
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$\therefore A'B=1$
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### 第三问
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