Vector in a Triangle

Geometry Level 4

Let $O$ be a point inside $\Delta\text{A}{B}{C}$ such that $AO$ , $BO$ , $CO$ meet opposite sides having points $P$ , $Q$ , $R$ ,then which of the following is correct?

\text{None of These}

\dfrac{OP}{BP}+\dfrac{OQ}{CQ}+\dfrac{OR}{AR}=1

\dfrac{AP}{OP}+\dfrac{BQ}{OQ}+\dfrac{CR}{OR}=1

\dfrac{OP}{AP}+\dfrac{OQ}{BQ}+\dfrac{OR}{CR}=1

1 solution

Parag Zode
Dec 15, 2014

Let $x\vec{a}+y\vec{b}+z\vec{c}=0$ . Then we get $\dfrac{y\vec{b}+z\vec{c}}{y+c}=\dfrac{-x}{y+z}.\vec{a}$ .Now, $\vec{OP}=\dfrac{-x}{y+z}.\vec{a}$ $\implies\vec{AP}=\vec{-a}.\dfrac{-x}{y+z}.\vec{a}$ $\implies\dfrac{-x+y+z}{y+z}.\vec{a}$ ... So we see that $\dfrac{OP}{AP}=\dfrac{x}{x+y+z}$ and so it is similar for other cases $\dfrac{OQ}{BQ}$ and $\dfrac{OR}{CR}$ ..and ultimately we get $\dfrac{x+y+z}{x+y+z}=1$

A slightly easier approach would be to consider the areas of the triangle. In particular, we have $\frac{ OP}{AP} = \frac{ [OBC]}{[ABC]}$ , and hence the sum of these 3 fractions would be $\frac{ [OBC] + [OCA] + [OAB] } { [ABC ] } = 1$ .

Calvin Lin Staff - 6 years, 5 months ago

Well I didn't knew this method but I was keen on sharpening the approach via vectors.

Parag Zode - 6 years, 5 months ago

Vector in a Triangle

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