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How to comment on the last problem of Suzhou mathematics in 2008
Solution: (1)∫△ABC is a right triangle with AC=8 and BC=6.

∴AB=.

(2)∵PM⊥AC PN⊥BC

∴mp∥bc AC∑pn (two lines perpendicular to the same line are parallel),

AP = x,AB= 10,BC=6,AC=8,BP= 10-x

∴PM=

PN==8-

The ∴ perimeter of the rectangle pmcn = 2 (pm+pn) = 2 (x+8-x) =14.

∴x=5.

(3)∵PM⊥AC,PN⊥BC,

∴AC∥PN.

∴∠A=∠NPB.

∴△AMP∽△PNB.

∴ When P is the midpoint of AB, that is, AP=PB, △ amp △ PNB,

At this time, S△AMP=S△PNB=,

And rectangular PMCN area =PM? MC=3×4= 12,

∴ There is no x value that can make the areas of △PAM and △PBN equal to the area of rectangular PMCN at the same time.