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For the rectangle \(ABCD\) and any point \(P\), we have \(PA^2 + PC^2 = PB^2+ PD^2\).
Proof. \(\v{PA} + \v{PC} = 2\v{PO}, \v{PA} - \v{PC} = \v{CA}\), hence \[ PA^2 + PC^2 = 4PO^2 + CA^2 = 4PO^2 + DB^2 = PB^2 + PD^2 \]
Remark. Since the proof is purely the algebraic computation, the position of \(P\) could be at anywhere in the space, not limited on the plane \(ABCD\).
Example. For \(Rt\triangle ABC\) where \(\angle C = \pi/2\), if some point \(P\) in the plane fits \(\v{PA} + \v{PB} + \lambda\v{PC} = \v 0\), denoted \[ m = \dfrac{PA^2 + PB^2}{PC^2} \] (i) When \(\lambda = 1\), find \(m\).
- Find the minimum value of \(m\).
Solution. (i) For \(\lambda = 1\), we have \(\v{PA} + \v{PB}+ \v{PC} = 1\), in this case \(P\) is the center of gravity for \(\triangle ABC\), \(P\) located on the middle on \(AB\).
- Construct a rectangle \(ACBE\), we have \[ \frac{PA^2 + PB^2}{PC^2} = \frac{PC^2 + PE^2}{PC^2} = 1 + \frac{PE^2}{PC^2} \] Hence the \(m\) is minimal iff \(PE^2/PC^2\) is minimal, that is, \(P=E\) and \(m_\min = 1\).