(1) According to Pythagorean theorem, AB = √ (AC 2+BC 2) = 6 √ 3 = 2ac.
∴∠B=30，∠A=60
∴∠PRQ=∠CRQ=∠B=30
(2) When point P is on AB,
∫QR∨AB
∴∠APQ=∠PQR=∠CQR=∠A=60
∴△APQ is an equilateral triangle.
∴x=AQ=PQ=CQ= 1/2AC=3√3/2
(3) As shown in the left figure, we can get that △AQE is an equilateral triangle by imitating (2).
∴y=BE=AB-AE=6√3-AQ=6√3-x
The domain is 0 < x < 3 √ 3/2.