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Prove: (PC)·(QE) + (PD)·(QF) < 8

Albert

Well-known member
Jan 25, 2013
1,225
PC.QE+PD.QF.jpg
 

Albert

Well-known member
Jan 25, 2013
1,225

construct OM $\perp CD$
point M is the midpoint of CD
$PC\times QE \leq \dfrac {PC^2+QE^2}{2}----(1)$
$PD\times QF \leq \dfrac {PD^2+QF^2}{2}----(2)$
BUT $PC^2+PD^2$=$(CM-PM)^2+(DM+PM)^2=2(CM^2+PM^2)$
=$2(CM^2+OM^2)=2OC^2=8$
also $QE^2+QF^2=8$
$\therefore (1)+(2)\leq \dfrac {(8+8)}{2}=8$
and the proof is finished
 
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