Suzhou senior high school entrance examination 20 10 math problem 18.

Solution: connect AP and BP, and transfer P when PQ⊥x axis is in Q;

∫∠AOB = 90，

∴AB is⊙ O, then∠ ∠APB = 90；;

In Rt△AOB, OB=2, OA=2, root number 3? According to Pythagorean theorem, AB = 4；;

∫op shares ∠AOB, ∴? Arc BP= arc AP;

So △ABP is isosceles Rt△, AP=2. Equation 2? ;

At Rt△POQ, ∠ POQ = 45, PQ = OQ；;

Let PQ=OQ=x, then the root number 3 of AQ=2? -x；

In Rt△APQ, from Pythagorean theorem:

AP = AQ+PQ, that is, (2? Root number 3-x)+x = 8;

The solution is x=? Root number 3+ 1, x= root number 3? - 1;

Because ∠ POA > ∠ OAB, PQ > OB, that is, X > 2；;

∴PQ=OQ=x=？ Root number 3+1;

That is, the coordinate of point P is (root number 3? +1, root number 3? + 1).

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