In the figure below OACB is a parallelogram in which M is the midpoint of AC. OM is produced to meet BC which is also produced to meet at point X. Given that OA = a and OB = b, Find the values of r and s such that OX = r OM and CX = s BC. Hence determine the ratio BC: BX.

Given $\vec{OA} = \mathbf{a}$ and $\vec{OB} = \mathbf{b}$.
Since OACB is a parallelogram:
$\vec{AC} = \vec{OB} = \mathbf{b}$
$\vec{BC} = \vec{OA} = \mathbf{a}$
The position vector of C is $\vec{OC} = \vec{OA} + \vec{AC} = \mathbf{a} + \mathbf{b}$.
M is the midpoint of AC, so $\vec{AM} = \frac{1}{2}\vec{AC} = \frac{1}{2}\mathbf{b}$.
The position vector of M is $\vec{OM} = \vec{OA} + \vec{AM} = \mathbf{a} + \frac{1}{2}\mathbf{b}$.

Substitute the expression for $\vec{OM}$:

We know $\vec{BC} = \mathbf{a}$, so $\vec{CX} = s\mathbf{a}$.
Using vector addition, $\vec{OX} = \vec{OC} + \vec{CX}$.
Substitute the expressions for $\vec{OC}$ and $\vec{CX}$:

Equating $(*)$ and $(**)$:

Since $\mathbf{a}$ and $\mathbf{b}$ are non-parallel vectors, we can equate their coefficients:
For $\mathbf{a}$: $r = 1+s$ (Equation 1)
For $\mathbf{b}$: $\frac{r}{2} = 1$ (Equation 2)
From Equation 2, we find $r$:

Substitute $r=2$ into Equation 1:

The values are $\boxed{r=2, s=1}$.

From $\vec{CX} = s \vec{BC}$ and $s=1$, we have $\vec{CX} = \vec{BC}$.
This means that C lies on the line segment BX, and the length of CX is equal to the length of BC.
Therefore, B, C, and X are collinear, and C is the midpoint of BX.
We can write $\vec{BX} = \vec{BC} + \vec{CX}$.
Since $\vec{CX} = \vec{BC}$,

The ratio BC:BX is the ratio of their magnitudes:

The ratio BC:BX is $\boxed{1:2}$.