The original title (20 12 Tianjin senior high school entrance examination math problem) is shown in the figure. The specific answer steps of the third question of this question are rewarded (the official answer has been given)! ! ! Folding transformation (folding problem);

Coordinate and graphic attributes; Congruent triangles's judgment and nature; Pythagorean theorem; Similar triangles's judgment and nature.

Special topic: geometry synthesis problem.

Analysis: (1) According to the meaning of the question, ∠ OBP = 90, OB=6, at Rt△OBP, ∠ BOP = 30, BP=t, OP=2t, and then using Pythagorean theorem, the equation can be obtained, and the answer can be obtained by solving this equation;

(2) △ OB ′ P and △ QC ′ P are obtained by folding △OBP and △ QCP respectively, so it can be seen that △ OB ′ P △ OBP is easy to prove, and then △OBP∽△PCQ is proportional to the corresponding side of similar triangles.

(iii) Let P be PE⊥OA in E, and it is easy to prove △ PC ′ e ∽△ C ′ QA. The length of c ′ q can be obtained from Pythagorean theorem, and then the corresponding edge of similar triangles is proportional to m= 1.

six

t2- 1 1

six

T+6, you can get the value of t. Solution: Solution: (1) According to the meaning of the question, ∠ OBP = 90, OB=6.

At Rt△OBP, from ∠ BOP = 30, BP=t, OP = 2t.

∫OP2 = OB2+BP2，

That is, (2t)2=62+t2,

Solution: t1= 2 3, T2 =-2 3 (discarded).

The coordinate of point p is (23,6).

(ii) ∫ OB ′ p and △ QC ′ p are obtained by folding △OBP and △QCP respectively.

∴△ob′p≌△obp,△qc′p≌△qcp，

∴∠opb′=∠opb,∠qpc′=∠qpc，

∠∠OPB′+∠OPB+∠QPC′+∠QPC = 180，

∴∠OPB+∠QPC=90，

∠∠BOP+∠OPB = 90，

∴∠BOP=∠CPQ.

∠∠OBP =∠C = 90，

∴△OBP∽△PCQ，

∴OB PC =BP CQ，

Let BP=t, AQ=m, BC= 1 1, AC=6, then PC= 1 1-t, CQ = 6-m.

∴6 1 1-t =t 6-m。

∴m= 1 6 T2- 1 1 6t+6(0 < t < 1 1)。

(iii) using point P as PE⊥OA in E,

∴∠pea=∠qac′=90，

∴∠pc′e+∠epc′=90，

∵∠PC′E+∠QC′A = 90，

∴∠epc′=∠qc′a，

∴△pc′e∽△c′qa，

∴pe AC′= PC′c′q，

∫PC′= PC = 1 1-t，PE=OB=6，AQ=m，C′Q = CQ = 6-m，

∴ac′= c’Q2-aq2 = 36- 12m，

∴6 36- 12m = 1 1-T6-m，

∫m = 1 6 T2- 1 1 6t+6，

Solution: t1=1-133, t2= 1 1+ 13 3,

The coordinates of point P are (11-133,6) or (11+133,6).

Comments: This topic examines the nature of folding, the nature of rectangle and similar triangles's judgment and nature. This problem is difficult, so we should pay attention to the corresponding relationship between the figures before and after folding, and the application of the idea of combining numbers with shapes and the idea of equations.