proportion problem

[nisar_cck]

can any one solve this........
if (2y+2z-x) : a = (2z+2x-y) : b = (2x+2y-z) : c

then prove that

x : (2b+2c-a) = y: (2c+2a-b) = z : (2a+2b-c)

 

[soroban]

Hello, nisar_cck!

I'll do <u>half</u> of it for you . . .

Can any one solve this . . .

If . (-x +2y+2z) : a .= .(2x-y+2x) : b .= .(2x+2y-z) : c

then prove that: .x : (-a+2b+2c) .= .y: (2a-b+2c) .= .z : (2a+2b-c)

. . . . . . . . . - x + 2y + 2x . . . . 2x - y + 2x
We have: . ---------------- . = . ---------------
. . . . . . . . . . . . . .a . . . . . . . . . . . . .b

. . which simplifies to: . (2a + b)x - (a + 2b)y + (2a - 2b)z .= .0 . [1]


. . . . . . . . . .2x - y + 2x . . . . 2x + 2y - z
We have: . -------------- . = . ----------------
. . . . . . . . . . . . . b . . . . . . . . . . . . c

. . which simplifies to: . (2b - 2c)x + (2b + c)y - (b + 2c)z .= .0 . [2]


Multiply [1] by (b + 2c): . . . (b + 2c)(2a + b)x - .(b + 2c)(a + 2b)y + (b + 2c)(2a - 2b)z .= .0
Multiply [2] by (2a - 2b): . (2a - 2b)(2b - 2c)x + (2a - 2b)(2b + c)y - (b + 2c)(2a - 2b)z .= .0

Add: . (6ab + 6bc - 3b²)x + (3ab - 6b² - 6bc)y .= .0

Factor: . 3b(2a + 2c - b)x + 3b(a - 2b - 2c)y .= .0

And we have: . (2a - b + 2c)x . = . (-a + 2b + 2c)y

. . . . . . . . . . . . . . x . . . . . . . . . . . . y
Therefore: . --------------- . = . ---------------
. . . . . . . . . .-a + 2b + 2c . . . . 2a - b + 2c

You can do the other half . . .