◆ The landlord should make it clear where point A is, and only say "Answer point A in the first quadrant".

Solution: (1) When point O is the vertex of isosceles ⊿AOP vertex angle:

Draw an arc with point O as the center and the length of OA as the radius, and intersect on the three coordinate axes of P 1, P2, P3 and P4 respectively.

Then: P 1 is (2,0), P2 is (0,2), P3 is (-2,0) and P4 is (0,2);

(2) When point P is the vertex of isosceles ⊿AOP:

The perpendicular line, such as AO, passes through two coordinate axes at P2 and P5 respectively, so it is easy to know that ⊿AOP2 is an equilateral triangle.

∴op2=ao=2； ∠OP2P5=30，P2P5=2OP5，OP2？ =P2P5？ -OP5？ =3OP5？ .

That is 4=3OP5? , OP5=2√3/3, then P5 is (2√3/3,? 0).

③ When point A is the vertex of isosceles ⊿AOP vertex angle:

Draw an arc with A as the center and AO as the radius, and the two axes intersect at O, P2 and P6 respectively, then AP6 = AO = 2；;

Let AB be perpendicular to the X axis in B and ∠ AOB = 30, then AB = ao/2 = 1, OB = √ 3, OP6 = 2ob = 2 √ 3, that is, P6 is (2 √ 3,0).

To sum up, the coordinates of point P are (2,0), (0,2), (-2,0), (0,2), (2 √ 3/3,0) and (2 √ 3,0) respectively.

(2) When the angle between OA and X axis is 45 degrees, the similarity can be obtained in the same way: P 1 is (0,2), P2 is (-2,0), P3 is (0,2), P4 is (2,0),

P5 is (0,2 √2), P6 is (2 √ 2,0), P7 is (0,2) and P8 is (√ 2,0).