Problem - 3669
Let $a$, $b$, and $c$ be three distinct numbers such that $$\frac{a+b}{a-b}=\frac{b+c}{2(b-c)}=\frac{c+a}{3(c-a)}$$
Prove that $8a + 9b + 5c = 0$.
Let's first solve the following system
$$
\left\{
\begin{array}{ccc}
x+z&=8\\
x+y&=9\\
y+z&=5
\end{array}
\right.
\implies
\left\{
\begin{array}{c}
x=6\\
y=3\\
z=2
\end{array}
\right.
$$
Then, from the given conditions, we have
$$\frac{6(a+b)}{6(a-b)}=\frac{3(b+c)}{6(b-c)}=\frac{2(c+a)}{6(c-a)}=k$$
where $k$ is a constant.
\begin{align*}
\therefore\quad &8a + 9b+5c \\
&= 6(a+b)+3(b+c)+2(c+a)\\
&= k\Big(6(a-b)+6(b-c)+6(c-a)\Big)\\
&= \boxed{0}
\end{align*}