20 12 penultimate question of Hebei mathematics senior high school entrance examination

Solution: (1) 𕚸 BCO = ∠ CBO = 45

∴OC=OB=3，

Point c is on the positive semi-axis of the y axis,

∴ The coordinate of point C is (0,3);

(2) Consider in two situations:

(1) When point P is on the right side of point B,

If ∠ BCP = 15 and ∠ PCO = 30,

So PO=CO? Tan30 = root number 3, when t=4+ root number 3;

② When point P is to the left of point B,

From ∠ BCP = 15, ∠ PCO = 60,

Therefore, op = cotan 60 = 3 root number 3.

T=3+3 radical number 3

All in all, ...

(3) According to the meaning of the question, if ⊙P is tangent to the side of the quadrilateral ABCD, there are the following three situations:

① When ⊙P and BC are tangent to point C, there is ∠ bcp = 90,

So ∠ OCP = 45, OP=3, where t =1;

(2) When ⊙P and CD are tangent to point C, there is PC⊥CD, that is, point P coincides with point O, and at this time t = 4；;

③ When ⊙P is tangent to AD, ∠ dao = 90,

∴ Point A is the tangent point, as shown in Figure 4, PC2=PA2=(9-t)2, PO2=(t-4)2,

So (9-t)2=(t-4)2+32, that is, 81-18t+T2 = T2-8t+16+9.

Solution: t=5.6,

The value of t is 1 or 4 or 5.6.