20 1 1 A math problem in Changsha senior high school entrance examination. Can anyone do that?

(1) let b be BM, perpendicular to the x axis, m.

The angle BOM=30 degrees BM=OB/2= 1 OM= root number 3, so the coordinate of b is (root number 3, 1).

(2) Because AB=AO AQ=AP angle PAO= angle QAB=60 degrees-angle O AQ, all triangles PAO are equal to triangle QAB.

So angle ABQ = angle AOP=90 degrees.

(3) If BQ//AO, the angle OAB= the angle ABQ=90 degrees and the angle OAB=60 degrees are different.

If OQ//AB, the angle BOQ=60 degrees and the angle BQO=90 degrees, then BQ= radical number 3=PO, so we get P(- radical number 3,0).