- Thread starter
- Admin
- #1
- Feb 14, 2012
- 3,967
For the integers $a, b, c$, we have $\dfrac{2a-b}{c}=\dfrac{2b+c}{a}=\dfrac{-2a-c}{b}$.
If $a+b=2004$, find $a+b+c$.
If $a+b=2004$, find $a+b+c$.
For the integers $a, b, c$, we have $\dfrac{2a-b}{c}=\dfrac{2b+c}{a}=\dfrac{-2a-c}{b}$.
If $a+b=2004$, find $a+b+c$.
$\dfrac{2a-b}{c}=\dfrac{2b+c}{a}$ | $\dfrac{2a-b}{c}=\dfrac{-2a-c}{b}$ | $\dfrac{2b+c}{a}=\dfrac{-2a-c}{b}$ |
$2a^2-ab=2bc+c^2$ | $(b+c)(2a-b+c)=0$ | |
But notice that $b+c\ne 0$ because if $b+c=0$, $2a^2-ab=2bc+c^2$ becomes $2a^2-ac=-2c^2+c^2$ $2a^2-ac+c^2=0$ $2a^2-ac+c^2=0$ $2(a-\frac{c}{4})+\frac{7c^2}{8}\ne 0$ for all integers $a, c$. Hence, $2a-b+c=0$ | $\dfrac{2b+c}{a}=\dfrac{-(2a+c)}{b}$ $\dfrac{2b+c}{a}=\dfrac{-b}{b}$ $2b+c=-a$ $a+2b+c=0$ |
let :For the integers $a, b, c$, we have $\dfrac{2a-b}{c}=\dfrac{2b+c}{a}=\dfrac{-2a-c}{b}$.
If $a+b=2004$, find $a+b+c$.