在图中， $\overline{OA}=\tfrac{1}{2}\,\overline{AP}$ 且 $\overline{OB}=\tfrac{1}{4}\,\overline{OQ}$ 。M 是 OQ 的中点，且 $\overline{MN}=\tfrac{1}{4}\,\overline{NP}$ 。OAP、OMQ 和 PNM 是直线。 $\overline{OA}=a$ 且 $\overline{OB}=b$ 。

(a)
用 a 和/或 b 尽可能最简地表示：
1. ((i))
  $\overline{PM}$
2. ((ii))
  $\overline{AN}$
3. ((iii))
  $\overline{AQ}$
(b)
证明 A、N 和 Q 共线。

答
(c)
写出比值 $AN : NQ$ 。

答 ___
(d)
求 $\dfrac{\text{Area of triangle }OPM}{\text{Area of triangle }OPQ}.$ 的值。

答 ___

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完整解法

19(a)((i))

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 $M$.

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

答案

$\overline{PM}=-3a+2b$

19(a)((ii))

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
$\overline{AN}=\overline{AP}+\overline{PN}$
$=2a+\left(-\frac{12}{5}a+\frac85b\right)$
$=-\frac25a+\frac85b.$

答案

$\overline{AN}=-\dfrac{2}{5}a+\dfrac{8}{5}b$

19(a)((iii))

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

答案

$\overline{AQ}=-a+4b$

19(b)

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.

答案

Since $\overline{AN}=\dfrac{2}{5}\overline{AQ}$ , $\overline{AN}$ is parallel to $\overline{AQ}$ . As both vectors start from $A$ , the points $A$ , $N$ and $Q$ lie on a straight line.

19(c)

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
$AN:NQ=\frac{2}{5}AQ:\frac{3}{5}AQ=2:3.$

答案

$AN:NQ=2:3$

19(d)

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

答案

$\dfrac{1}{2}$

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