Solution: (1) Draw a straight line parallel to BC through point M and intersect with AC at point F.

Because △ABC is a regular triangle, AF = AM = X, DF = CD = Y.

Cf = AC-af = 4-X。

CF=2y

So 2y=4-x gives 2y+x = 4 (0

(2)S△ABC = 4 * 2√3/2 = 4√3

S△AMF=x*x√3/2=x？ √3

S△DCF = 1/5(S△ABC-S△AMF)= 1/5(4√3-x？ √3)

∫S△DCF = x * 1/2 * y√3/2。

∴ 1/5(4√3-x？ √3)=x* 1/2*y√3/2

2y=4-x

∴3x？ -20x+32=0

X = 8/3 or x=4

X≠4 again.

∴x=8/3

(3) When crossing M, it is ME vertical AC, and the vertical foot is E. ..

∴DE=AE-AE

=AC-CD-AE

=4-y- 1/2x

2y=4-x

∴DE=4-y- 1/2(4-2y)

=4-y-2+y

=2

When m moves on AB, the length of DE will not change.