解:延长EC到F,使CF=CE
∵tan∠BPD=1/3,CE=x,∴CP=3x,EF=2x,∵∠ACB=90°∴EP=根号10x.
∵△ADE∽△PEF∴AD:EP=DE:EF,即1:根号10x=DE:2x ∴DE=根号10/5
在AF上截取AM=AB,∵AD=AE∴BM‖DE ∴AD:AB=DE:BM,
设CM=m,则BC=3m,BM=根号10m ∴1:(m+x+1)=根号10/5:根号10m
得:m=(x+1)/4 ∴AB=5(x+1)/4 BC=3(x+1)/4
∴△ABC的周长为y=AB+BC+AC=5(x+1)/4+3(x+1)/4+x+1=3x+3(x>0)
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