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Math problems only write solutions.
This question is a typical comprehensive question of Grade Three, in which there are many knowledge points, and it involves the concept of cycle. You haven't learned it, so we discuss the ideas in categories:

The maximum value of 2△2△FGH, then the area of point P increases when it moves to point C, and then point F is on BC, that is, the area decreases. So you must make point C CM vertical AB, and point M the maximum point, which is also the demarcation point. List the function expressions and discuss them in sections.

3)2 It's a little difficult to ask questions, and discuss them in categories.

1) is easy to handle, that is, I is the inscribed circle of a triangle, that is, when point P coincides with point B,

2) Mainly discuss the positional relationship of I, establish a rectangular coordinate system with point A as the origin, and then delete other graphics and redraw. In fact, I have been in it all the time, simply solving linear AC. BC equation. When the coordinates of point I are brought in, the linear AC or BC equation should be greater than the ordinate of I.

Now let's discuss the position of I:

1, when 0 < t ≤ 2, the coordinate of point Q is (2√3-√3t, 0), then the coordinate of point P is (2√3+√3t, 0), then PQ = 2 √ 3+√ 3t-2 √ 3 = 2 √ 3t, then the coordinate of point I.

So I click the coordinates (2√3, t),

2) When 2 < t ≤ 4, the coordinates of point Q (t-2) √ 3,0 and point P (2√3+√3t) are at this time, so PQ=4√3, and the midpoint of PQ (√3t,2) at this time, you can contact the I coordinate yourself.

3) When 4 < t ≤ 6, at this time, the coordinates of Q point (2 √ 3-(t-4) √ 3,0) and P point (2√3+√3t) are solved by themselves as above.

4) When 6 < t ≤ 8, the coordinates of point Q (t-6) √ 3,0 and point P (2√3+√3t) should be solved as above.