MHB Prove: (PC)·(QE) + (PD)·(QF) < 8

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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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Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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