(1)由△P1OA1是等腰直角三角形,得y1=x1,则有x12=4,故x1=±2(负舍),点P1(2,2).
(2)过P1作P1B⊥OA1于B,过P2作P2C⊥A1A2于C, ∵△OP1A1、△A1P1A2是等腰直角三角形, ∴OB=BP1=BA1=x1=y1 ∴y2=A1C=OC-A1B-OB=x2-x1-y1, 同理可得:y3=x3-x2-y2,y4=x4-x3-y3,…,y10=x10-x9-y9, 又yn=,则:x2-4=,解得,x2=2+2. ∴y2=2-2, ∴x3-4=,x3=2+2,y3=2-2, 同理,依次得x2=2+2,y2=2-2, x3=2+2,y3=2-2, x4=2+2,y4=2-2, x5=2+2,y5=2-2, … x9=2+2,y9=2-2, x10=2+2,y10=2-2, ∴y1+y2+y3+…+y10=2+2-2+2-2+2-2+…+2-2+2-2=2. |