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In $\triangle ABC$, points $A',B',C'$ are on sides $BC,AC,AB$ respectively. $AA', BB', CC'$ are concurrent at point $O$.
Given $AO:OA' + BO:OB' + CO:OC' = 92$ find the value of $AO:OA' \times BO:OB' \times CO:OC'$.

My work
enter image description here
I can find these relations -
$AO:OA' =[AOB]:[BOA'] = [AOC]:[COA']\\ BO:OB' =[AOB]:[AOB'] = [COB]:[COB']\\ CO:OC' =[AOC]:[AOC'] = [BOC]:[BOC']$
Now how to continue?
Source: BdMO 2015 national secondary.

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    how is this linear algebra? – 2017-01-23
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    @A.Molendijk May not... But this may need algebra at the last... So I tagged it .. – 2017-01-23
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    If you mean high school algebra, the appropriate tag is [tag:algebra-precalculus]. – 2017-01-23
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    @user1551 edited now :) – 2017-01-23

2 Answers 2

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Let $u,v,w$ be the ratios $\frac{BA'}{A'C},\frac{CB'}{B'A},\frac{AC'}{C'B}$. By Van Obel's theorem $$\frac{AO}{OA'}=w+\frac{1}{v}$$ and so on, hence we know that $$ 92=(u+v+w)+\left(\frac{1}{u}+\frac{1}{v}+\frac{1}{w}\right)$$ and $uvw=1$ by Ceva's theorem. It follows that: $$ \left(w+\frac{1}{v}\right)\left(v+\frac{1}{u}\right)\left(u+\frac{1}{w}\right)=(u+v+w)+\left(\frac{1}{u}+\frac{1}{v}+\frac{1}{w}\right)+\frac{1}{uvw}+uvw = \color{red}{94}.$$

In general, if $AA',BB',CC'$ concur at $O$, $$ \prod_{cyc}\frac{AO}{OA'}=2+\sum_{cyc}\frac{AO}{OA'}.$$

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    WoW! didn't know about this theorem.. Thanks .. :) – 2017-01-23
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    @RezwanArefin: you're welcome. – 2017-01-23
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    @JackDAurizio: I have a question... Can I directly refer to this theorem in Olympiads or I need to prove this as a lemma? – 2017-01-23
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    @RezwanArefin: nice question. I learned that theorem during the IMO 2003, and the Chinese team granted me it was a well-known theorem (even if I was not aware of it at that time). You can use it by just mentioning Van Obel, no need to re-prove it. – 2017-01-23
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Another way to solve this problem is to use the method of Mass Points. Assign the masses $a,b,c$ to the points $A, B, C$. This implies that we should assign the masses $(a+b), (b+c), (c+a), (a+b+c)$ to $C', A', B', O$. Then we have:

$$\frac{AO}{A'O} = \frac{b+c}{a} \quad \frac{BO}{B'O} = \frac{c+a}{b} \quad \frac{CO}{C'O} = \frac{a+b}{c}$$

Then using this we have:

$$92 = \frac{AO}{A'O} + \frac{BO}{B'O} + \frac{CO}{C'O} = \frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c} = \frac{bc(b+c) + ac(a+c) + ab(a+b)}{abc} = \frac{(a+b)(b+c)(c+a) - 2abc}{abc} = \frac{AO}{A'O} \times \frac{BO}{B'O} \times \frac{CO}{C'O} - 2$$

Therefore: $\frac{AO}{A'O} \times \frac{BO}{B'O} \times \frac{CO}{C'O} = 94$

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    The van obel's theorem can also be proved from mass points :) This is doing the same :) – 2017-01-23