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In triangle ABC, points D and E are on sides BC and CA respectively, and points F and G are on side AB with G between F and B. BE intersects CF at point O_1 and BE intersects DG at point O_2. If FG = 1, AE = AF = DB = DC = 2, and BG = CE = 3, compute $\tfrac{O_1O_2}{BE}$.

I couldn't think of other method to solve this(besides, iI don't know mass point geometry) will choose the most reasonable answer!

Answer is in problem 3 in this article:

Thanks in advance.

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Please don't put things like "Thanks in advance?" in the title -- it clutters the main page. The rest of the title should also be more concise and descriptive, usually not a complete sentence but a noun phrase or a short question. – joriki Oct 27 '11 at 5:06
up vote 2 down vote accepted

You can do it with vectors in the plane. Choose $B$ as origin and put $BD=:d$, $BG=:g$. Then from the given data we see that $BF={4g\over3}$, $BA=2g$, $BC=2d$ and $BE={4d+6g\over5}$. Obviously $BO_2={1\over2}BE={2d+3g\over5}$.

It remains to compute the point $O_1$. Now $$BO_1=BC+s\ CF\quad{\rm and}\quad BO_1=t\ BE$$ for some $s>0$ and some $t>{1\over2}$. It follows that $$2d + s\Bigl({4g\over3}-2d\Bigr)= t{4d+6g\over 5}\ .$$ Since the vectors $d$ and $g$ are linearly independent this equation determines $s$ and $t$ uniquely.

The rest is easy. Actually we only need the value of $t$, and the final result is $t-{1\over2}$.

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