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.