liu2008

甲、乙两人分别从A、B两地出发，甲从A走到B，然后再从B走到A;乙从B走到A，然后再从A走到B.
他们第一次相遇(face to face)在C点，第2次相遇(face to face)在D点，求证：
AD-AC=2(BC-AC)

Tianna

很简单，AD+AC = 2BC,

两边同减2AC 即得AD -AC = 2(BC -AC)

liu2008

这道题源于一道四年级题目，原题是这样的AC=36，AD=54，问AB是多少？四年级没学方程

colo

assume:

l = AB
x = AC
y = BD

so,
AD-AC=2(BC-AC) <=>
(l - y) - x = 2 ( l -x -x) <=>
l - x - y = 2l - 4x
<=> 4x = l + x + y
<=> 3x = l + y


speed of 甲 is s1
speed of 乙 is s2

first met face to face
x/s1 = (l-x) / s2

second met face to face
甲 walked (l+y)
乙 walked (2l - y)
so, (l + y)/s1 = (2l - y) /s2

(1) x/s1 = (l-x) / s2
(2) (l + y)/s1 = (2l - y) /s2
===>

x / s1                 (l-x) / s2
--------       =   ----------------
(l + y)/s1           (2l - y) /s2

====>
  x                  (l-x)
-------        =  --------------
(l + y )            (2l - y)

==========>

x * (2l - y) = (l-x) * (l + y )

===>

2lx - xy = l^2 - xy - lx + ly
==>
2lx + lx  = l^2 + ly
===>
3lx = l^2 + ly
===>
3x = l + y

liu2008

anybody?