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Here are a parallelogram $PQRS$.

Let the internal dividing point of the $\overline{PQ}$ by $2:1$ is $A$, middle point of the $\overline{PS}$ is $B$ and intersection point between those is $C$

Express $\overrightarrow {PC} $as a linear combination of the $\overrightarrow {PQ}$ and $\overrightarrow {PS}$. enter image description here


Here is my attempt

enter image description here

All I have to do find the ratio $\alpha$.

What should I do next? Thanks.

P.s.) Lately checking the answer sheet, it said $\alpha$ is ${1 \over 4}$

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  • $\begingroup$ That cannot be right: $PQ$ and $RS$ are parallel so any linear combination will also be parallel to them, and $PC$ isn't. Please proof read your question carefully and edit it. $\endgroup$ – David Aug 14 at 1:29
  • $\begingroup$ Thanks for your point out. I will edit it. $\endgroup$ – se-hyuck yang Aug 14 at 1:30
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$\alpha$ can be obtained from the area ratios as follows,

$$\frac{\alpha}{1-\alpha} =\frac{\triangle BAS}{\triangle RAS}=\frac{\frac{2}{3} \cdot \frac{1}{2}\cdot\frac{1}{2}}{\frac{1}{2}}=\frac{1}{3} $$

where we observe that △BAS and △RAS share the same base AS. This allows their area ratio to be expressed as $\alpha/(1-\alpha)$, which is proportional to their heights. Furthermore, from the side partitions given, △RAS is $\frac{1}{2}$ of the area PQRS and, similarly, △BAS is $\frac{1}{6}$ of the area PQRS.

Thus,

$$\alpha=\frac{1}{4}$$

and $$\vec{PC}=\frac{1}{4}\vec{PQ}+\frac{5}{8}\vec{PS}$$

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  • $\begingroup$ Why does $$\frac{\alpha}{1-\alpha} =\frac{△BAS}{△RAS}=\frac{\frac{2}{3} \cdot \frac{1}{2}\cdot\frac{1}{2}}{\frac{1}{2}}=\frac{1}{3} $$ holds? $\endgroup$ – se-hyuck yang Aug 14 at 1:43
  • $\begingroup$ Please see the added explanation in the answer. $\endgroup$ – Quanto Aug 14 at 1:49
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So your work shows that $$\overrightarrow{PC}=\alpha(\overrightarrow{PQ}+\overrightarrow{PS})+\frac{1}{2}(1-\alpha)\overrightarrow{PS}$$ $$=\alpha\overrightarrow{PQ}+\frac{1}{2}(1+\alpha)\overrightarrow{PS}$$ for some $\alpha\in(0,1)$ since C is on $\overline{BR}$. By the same reasoning, since C is also on $\overline{AS}$ it is also $$\overrightarrow{PC}=\frac{2}{3}\beta\overrightarrow{PQ}+(1-\beta)\overrightarrow{PS}$$ for some $\beta\in(0,1)$. So, $$\alpha=\frac {2}{3}\beta$$ $$\frac{1}{2}(1+\alpha)=1-\beta$$ which is solved by $\beta=\frac{6}{11}$. So $$\overrightarrow{PC}=\frac{4}{11}\overrightarrow{PQ}+\frac{5}{11}\overrightarrow{PS}$$.

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