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P is the middle of the median line from vertex A, of ABC triangle. Q is the point of intersection between lines AC and BP.

enter image description here

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  • $\begingroup$ I haven't done any 'hard' geometry problems with vectors, this one isn't similar with those that I did. Didn't get anywhere. $\endgroup$ – Desperado Aug 24 '15 at 11:26
  • $\begingroup$ For a vector approach, here is a way to start off. WLOG, let $C$ be the origin and $\mathbb {a, b}$ position vectors for the other vertices. Now express $M, P, Q$ in terms of these. Do some work and you will find others helping you along. $\endgroup$ – Macavity Aug 24 '15 at 11:27
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Starting you off in more detail....

With $\mathbb{a, b}$ denoting the vertices $A, B$ and $C$ being the Origin, one gets $$\mathbb m = \tfrac12\mathbb b, \quad \mathbb p = \tfrac12(\mathbb{m+a})=\tfrac12\mathbb a+\tfrac14 \mathbb b$$

Now $Q$ is located on the intersection of $\vec{BP} = t\mathbb b+(1-t)\mathbb p$ and $\vec {CA} = s\mathbb a$, so we solve to get $t = -\frac13, s = \frac23$, giving $\mathbb q = \frac23 \mathbb a$.

Actually, we have all that is needed to answer the questions by now...

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Let $\vec a=\vec{CA},\vec b=\vec{CB}$. Then, we have $$\vec{CM}=\frac 12\vec b$$$$\vec{CP}=\frac 12\vec{CA}+\frac 12\vec{CM}=\frac 12\vec a+\frac 14\vec b\tag 1$$

Also, setting $QC:AQ=s:1-s,BP:PQ=t:1-t$ gives

$$\vec{CP}=t\vec{CQ}+(1-t)\vec{CB}=ts\vec a+(1-t)\vec b\tag2$$

Now comparing $(2)$ with $(1)$ will give you the answer.

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Hint:

Part 1: Apply Menelaus' theorem on triangle $AMC$ and line $BPQ$.

Part 2: Apply Menelaus' theorem on triangle $BCQ$ and line $APM$ (needs the answer form part 1).

I'm giving only hints but you should be able to figure it out after learning the theorem (you will learn more!). If you need further help, feel free to comment.

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    $\begingroup$ That isn't using vectors... $\endgroup$ – Macavity Aug 24 '15 at 11:27
  • $\begingroup$ I need to know to do this with vectors. Thanks anyway $\endgroup$ – Desperado Aug 24 '15 at 11:28
  • $\begingroup$ Well, I certainly learnt this while on the topic of vectors... It certainly isn't far-fetched (especially given the signed lengths and the $-1$) / a possible proof is by vectors. Won't discount the possibility that OP learnt it in his/her vectors class/book. Guess not >< $\endgroup$ – suncup224 Aug 24 '15 at 11:31

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