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.