\(cos\widehat{AOB}=\dfrac{1.0+0.1-1.1}{\sqrt{2}.\sqrt{2}}=-\dfrac{1}{2}\Rightarrow\widehat{AOB}=120^0\)
\(S_{ODE}=\dfrac{1}{2}S_{OAB}\Leftrightarrow\dfrac{1}{2}OD.OE.sin120^0=\dfrac{1}{4}OA.OB.sin120^0\)
\(\Leftrightarrow OD.OE=\dfrac{1}{2}OA.OB=1\)
Định lý hàm cos:
\(DE^2=OD^2+OE^2-2OD.OE.cos120^0\ge2OD.OE-2OD.OE.cos120^0=3\)
\(\Rightarrow DE\ge\sqrt{3}\)
Dấu "=" xảy ra khi \(OD=OE=1\)
M là trung điểm AB \(\Rightarrow M\left(\dfrac{1}{2};\dfrac{1}{2};0\right)\)
N là trung điểm DE \(\Rightarrow\dfrac{ON}{OM}=\dfrac{OD}{OA}=\dfrac{1}{\sqrt{2}}\Rightarrow ON=\dfrac{OM}{\sqrt{2}}\)
\(\Rightarrow N\left(\dfrac{1}{2\sqrt{2}};\dfrac{1}{2\sqrt{2}};0\right)\)
