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Mathematical finale circle
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:

â‘  When point P is on the right side of point B, as shown in Figure 2,

If ∠ BCP = 15 and ∠ PCO = 30,

So PO=CO? Tan30 =, at this time t = 4+;

â‘¡ When point P is to the left of point B, as shown in Figure 3,

From ∠ BCP = 15, ∠ PCO = 60,

So op = cotan 60 = 3,

At this point, t=4+3,

The value of t is 4+ or 4+3;

(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.

Interviewee: Teacher 083

1。 Let c coordinate be (0, x), then CO=BO=3, so x=3 C coordinate is (0, 3).

2。 The angle BCP= 15 degrees, so when the point p is between OBs, the angles PCO=30 degrees, CO=3 and OP=t-4 are easy to know.

So t-4 = root number 3, which means t=4+ root number 3.

When the point P is between AB, it is easy to know the angles ACO=60 degrees, CO=3, OP=t-4,

So t-4 =3 times the root number 3, that is, t=4+3 times the root number 3.

Interviewee: Teacher 077

(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.