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Quadratic function method for solving the math final problem of Luzhou 20 12 middle school entrance examination.
y=-? (x﹣m)? +(? m? +m+? ﹚

D﹙m? m? +m﹢? ﹚

When y=0, x = m (m+ 1)

X =- 1 or x=2m+ 1.

Therefore, a is a constant (﹙ 1, 0 ﹚, b ﹙ 2m+ 1, 0).

1)m=3/2,D(3/2,25/8)

tanADH = AH/DH =( 1+ 1.5)/(25/8)= 4/5

2) When ∠ ADB = 60, an equilateral triangle

√3﹙ 1+m﹚=? m? +m﹢?

M =-1+2 √ 3 or-1.

Because B is on the right of A, M =- 1+2 √ 3.

When ∠ ADB = 90, isosceles right triangle

1+m=? m? +m﹢?

Similarly, M= 1.

So 1 ≤ m ≤- 1+2 √ 3.

3)AB=2m+2,OC=m+? ,S 1=(2m+2)(m+? )/2,

DH intersects BC at F, CT is perpendicular to DH and DG is perpendicular to CB.

FH/CO=BH/BO

(m+ 1)/(2m+ 1)= FH/( 1/2+ 1)

DF=DH-FH=(m? +2m+ 1)/2-(m? + 1.5m+0.5)/(2m+ 1)=[(m+ 1)? (2m+ 1)-(2m? +3m+ 1)]⊙[2(2m+ 1)]

DF=(m+ 1)m/2

S2=S triangle DCF+S triangle DFB=BO×DF=[(2m+ 1)×(m? +m)/2]/2=S 1

m=2,- 1,- 1/2

So m= 1.

BO=3,CO= 1.5,BC=(3√5)/2

S2= 15/2

So DG=2√5,

So the distance is 2√5.

I made it myself. It should be right.