In the triangle APQ, the base is AP=AH-PH=(2 equations 3-t). The vertical line from point q to AP is the height h of triangle APQ. Because AQ=t,
The angle DAH is 60 degrees, so the height h is (3t)/2. S△APQ= 1/2* (root number 3t)/2 * (root number 3-t).
That is, S=- (root number 3/4)t? +3t/2 .
2. There is only one case where △APM is an isosceles triangle, that is, AM=PM, because angle DAC= angle HAC= angle APQ=30 degrees. So the AQP angle is a right angle, that is, PQ is perpendicular to AD. That is, there are two radicals 3-t=2t and t=(2 radicals 3)/3. At this time, S=(2 radicals 3)/3.
In the above case, if the isosceles triangle is formed by AP=PM, the angle APM needs to be 120 degrees since the angle CAH is 30 degrees, and there is no such point P since the angle DAH is 60 degrees. ..
3.A, P and M must form a triangle. When AH and QP are perpendicular to each other, AM will get the maximum value. The maximum value is (2 root numbers 3)-t.