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I got a bit confused about this problem, and its something that my country's academic curriculum does not really delve into or practice as much.

The Problem:

Let $A,B,C$ be three points on a plane and $O$ be the origin point on this plane. Put $\vec{a}=\vec{OA}$, and $\vec{c}=\vec{OC}$, $P$ is a point inside the triangle $ABC$. Suppose that the ratio of the areas of $\triangle PAB,\triangle PBC$ and $\triangle PCA$ is $2:3:5$ respectively.

(1) The straight line $BP$ intersects the side $AC$ at point $Q$. Find $AQ:QC$

(2) Express $OP$ in terms of $\vec {a},\vec{b},\vec{c}$.

What method would be the most time efficient in solving this?

In addition to this, is there a proper naming etiquette in geometry, I noticed in most problems, whithout specifications, that $\angle A$ is always at the 12 o'clock position and the scheme continues counter clockwise.

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Let $AF$ and $CE$ be altitudes of $\Delta PAB$ and $\Delta PBC$ respectively.

Thus, since $\Delta AFQ\sim\Delta CEQ$, we obtain: $$\frac{AQ}{QC}=\frac{AF}{CE}=\frac{\frac{1}{2}BP\cdot AF}{\frac{1}{2}BP\cdot CE}=\frac{S_{\Delta PAB}}{S_{\Delta PBC}}=\frac{2}{3}.$$

Now, you can get $S_{\Delta PQC}$ and $BP:PQ$.

Can you end it now?

I got $$\vec{OP}=\frac{3}{10}\vec{a}+\frac{1}{2}\vec{b}+\frac{1}{5}\vec{c}.$$

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  • $\begingroup$ How would this be represented visually? $\endgroup$ – Alkahest Jul 11 at 5:46
  • $\begingroup$ @Alkahest Draw the picture and ask please a concrete question. $\endgroup$ – Michael Rozenberg Jul 11 at 5:51

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