In the diagram, $\overline{OA}=\tfrac{1}{2}\,\overline{AP}$ and $\overline{OB}=\tfrac{1}{4}\,\overline{OQ}$. M is the midpoint of OQ and $\overline{MN}=\tfrac{1}{4}\,\overline{NP}$. OAP, OMQ and PNM are straight lines. $\overline{OA}=a$ and $\overline{OB}=b$.

1. Express $\overline{OP}$ and $\overline{OM}$ in terms of $a$ and $b$.

Since $\overline{OA}=\frac12\overline{AP}$ and $\overline{OA}=a$, we have
$\overline{AP}=2a.$
Hence
$\overline{OP}=\overline{OA}+\overline{AP}=a+2a=3a.$
Also, $\overline{OB}=\frac14\overline{OQ}$ and $\overline{OB}=b$, so
$\overline{OQ}=4b.$
Since $M$ is the midpoint of $OQ$,
$\overline{OM}=\frac12\overline{OQ}=2b.$

2. Find $\overline{PM}$ by travelling from $P$ to $O$ to $M$.

$\overline{PM}=\overline{PO}+\overline{OM}$
$=-\overline{OP}+\overline{OM}$
$=-3a+2b.$

1. Express $\overline{PN}$ in terms of $\overline{PM}$.

Since $\overline{MN}=\dfrac14\overline{NP}$, the line segment from $P$ to $M$ is divided in the ratio
$PN:NM=4:1.$
Hence
$\overline{PN}=\frac45\overline{PM}.$

2. Find $\overline{AN}$ by travelling from $A$ to $P$ to $N$.

From $\overline{OA}=a$ and $\overline{OA}=\dfrac12\overline{AP}$,
$\overline{AP}=2a.$
Using $\overline{PM}=-3a+2b$ from part (a)(i),
$\overline{PN}=\frac45(-3a+2b)=-\frac{12}{5}a+\frac85b.$
Therefore
$\overline{AN}=\overline{AP}+\overline{PN}$
$=2a+\left(-\frac{12}{5}a+\frac85b\right)$
$=-\frac25a+\frac85b.$

1. Express $\overline{AQ}$ by travelling from $A$ to $O$ to $Q$.

Since $\overline{OB}=\dfrac14\overline{OQ}$ and $\overline{OB}=b$, we have
$\overline{OQ}=4b.$
Also, $\overline{AO}=-\overline{OA}=-a$.

Hence
$\overline{AQ}=\overline{AO}+\overline{OQ}=-a+4b.$

1. Compare $\overline{AN}$ with $\overline{AQ}$ using the earlier vector expressions.

From part (a),
$\overline{AN}=-\frac{2}{5}a+\frac{8}{5}b$
and
$\overline{AQ}=-a+4b.$
Now,
$\frac{2}{5}\overline{AQ}=\frac{2}{5}(-a+4b)=-\frac{2}{5}a+\frac{8}{5}b.$
Hence,
$\overline{AN}=\frac{2}{5}\overline{AQ}.$

2. Conclude that the three points are collinear.

Since $\overline{AN}$ is a scalar multiple of $\overline{AQ}$, the vectors $\overline{AN}$ and $\overline{AQ}$ are parallel.

Both vectors start from $A$, so $N$ lies on the same straight line as $A$ and $Q$.

Therefore, $A$, $N$ and $Q$ lie on a straight line.

1. Use the fraction of $AQ$ represented by $AN$ to find the remaining fraction $NQ$.

From part (b),
$\overline{AN}=\frac{2}{5}\overline{AQ}.$
Hence
$AN=\frac{2}{5}AQ.$
So
$NQ=AQ-AN=\frac{3}{5}AQ.$
Therefore
$AN:NQ=\frac{2}{5}AQ:\frac{3}{5}AQ=2:3.$

1. Compare the bases of the two triangles on the same straight line $OQ$.

Since $M$ is the midpoint of $OQ$,
$OM=\frac{1}{2}OQ.$
Triangles $OPM$ and $OPQ$ have the same height from $P$ to the line $OQ$.

Hence
$\frac{\text{Area of }\triangle OPM}{\text{Area of }\triangle OPQ}=\frac{OM}{OQ}=\frac{1}{2}.$