AM= 1, then

ME= 1.5，MB=4，

Because MN=5 and BN=3.

And then you can get

OM/MN=MH/MB=OH/BN ( 1)

OE/DE=HE/AE=OH/DA (2)

Let OM be x, and substitute Me = 1.5, BN = 3, Mn = 5, MB = 4 (1).

Then MH = 4x/5 and OH = 3x/5.

Substitute (2)

He =3x/ 10.

Because MH+ he = me

So 4x/5+3x/ 10= 1.5.

X= 15/ 1 1.

And so on = 5-15/11= 40/1/.

(2) Obviously, when m passes through the midpoint e, the part that falls in the bright area is the whole MN with a length of 5.

Therefore, when x≤2.5,

Let OM=m

[Then the method is the same as (1)]

OH/BN=MH/MB=MO/MN, that is, OH/t=MH/(5-x)=MO/5.

And OH=2HE.

The simultaneous equations 2HE=MH*t/(5-x) and HE+MH=2.5-x are obtained.

Get the expression,

MH=(5-2x)(5-x)/(t+ 10-2x)

So MO/5=MH/(5-x)

MO =(25- 10x)/(t+ 10-2x)

So y = 5-(25-10x)/(t+10-2x) = (25+5t)/(10+t-2x) (0 ≤ x ≤ 2.5).

y = 5(2.5 & lt； x≤5)

(3) When 2.5

When 0≤x≤2.5,

5/y = 1+(25- 10x)/(25+5t)

Let 0 ≤ x 1

Then 5/y1-5/y2 = (25-10x1)/(25+5t1)-(25-10x2)/(25+5t2).

Because MB 2+BN 2 = Mn 2

So t 2 = 10x-x 2.

T is in the range of 0≤x≤2.5, and increases with the increase of x.

So 25+5t 1

And 25-10x1>; 25- 10x2

Denominator increases, numerator decreases, and the whole decreases.

so 5/y 1-5/y2 >； 0

Get y 1

Therefore, in the range of 0≤x≤2.5, y increases with the increase of x;

When 2.5

References:

Junior high school+senior high school learning experience ~ this topic is still a classic question type, and it is all the same.