TITLE: Let X be the point of intersection between CM and AN, where M is the midpoint of AB and N satisfies the condition |CB| = 4|CN| . Find $\overline{CX}$.
QUESTION [0 upvotes]: In the triangle ΔABC, let N be the point on the side BC that satisfies the condition |CB| = 4|CN|, and let X be the point of intersection between CM and AN, where M is the midpoint of AB. If $\overline{CN}$ = $\vec{u}$ and $\overline{BM}$ = $\vec{v}$, express $\overline{CX}$ in terms of $\vec{u}$ and $\vec{v}$. Calculate $\frac{\overline{CX}}{\overline{CM}}$ and $\frac{Area(ΔACX)}{Area(ΔACM)}$.


I've tried to express the sides and the median $\overline{CM}$ in terms of $\vec{u}$ and $\vec{v}$, but I don't see how to express $\overline{CX}$. 
$\overline{BA} = 2\vec{v}$,
$\overline{CB} = 4\vec{u}$
$\overline{CB}+\overline{BA}=\overline{CA}=4\vec{u}+2\vec{v}$
$\overline{NC}+\overline{CA}+\overline{NA}=4\vec{u}+2\vec{v}-\vec{u}=3\vec{u}+2\vec{v}$
$\overline{CB}+\overline{BM}=\overline{CM}=4\vec{u}+\vec{v}$
Am I missing some relation between $\overline{CX}$ and $\overline{CM}$, or $\overline{NX}$ and $\overline{NA}$?

REPLY [0 votes]: Note that $\dfrac{\operatorname{Area}\triangle ACX}{\operatorname{Area}\triangle ACM}=\dfrac{CX}{CM}$, so it suffices to find $\dfrac{CX}{CM}=:s$.
Since $X$ lies on the segment $AN$, from $\overrightarrow{CA}=4\vec{u}+2\vec{v}$ and $\overrightarrow{CN}=\vec{u}$, we have
$\overrightarrow{CX}=t\overrightarrow{CA}+(1-t)\overrightarrow{CN}=(1+3t)\vec{u}+2t\vec{v}$ for some $t\in[0,1]$.
On the other hand, from $\overrightarrow{CM}=4\vec{u}+\vec{v}$ and $\dfrac{CX}{CM}=s$, we have $\overrightarrow{CX}=s\overrightarrow{CM}=4s\vec{u}+s\vec{v}$.
So equating components in the two expressions, we want
$$
\begin{cases}
4s=1+3t\\
s=2t
\end{cases}
$$
which gives $s=2/5$
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