Solution: Solution (1) Let the analytical formula of straight line AC be y=kx+b, and substitute point A (4,0) and point C (1 2) into 14k+b = 02k+b = 2. The solution is k =-2/3b = 8. According to the property of isosceles trapezoid ∴AH= 1, then op = OA-ah-HP = 4-1-bn = 3-t: point q is point PQ =-2/3 (3-t)+8/3 on AC ∴. PQ = 1/2(4-2t)(2/3t+2/3)=-2/3t? +2/3t+4/3; When t= 1 /2, s max =3/ 2.
(3) There are two situations: ①QM=QA, and MP=AP, that is, 3-3t=t+ 1, t=0.5(2 points) ②QM=MA, that is, QM2=MA2, MP2+PQ2=MA2 is obtained from Pythagorean theorem, that is, (3-3t). +(2 /3 t+2 /3)? =(4-2t)? , t 1=59/ 49, t2=- 1 (truncated) ∴ When t=0.5 or t 1=59 49, △AMQ is an isosceles triangle with MQ as its waist.
This topic examines the solution of analytical formula of straight line, the representation of triangle area in coordinate system, the maximum value of quadratic function, and the conditions for finding isosceles triangle.