<s>[INST] Given $AO:OA' + BO:OB' + CO:OC' = 92$ find the value of $AO:OA' \times BO:OB' \times CO:OC'$. 
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

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. 
 [/INST] 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: $\frac{AO}{A'O} \times \frac{BO}{B'O} \times \frac{CO}{C'O} = 94$
</s>