Argentina competition problems

Let $AM$ be the median through $A$, $CJ$ the bisector through $C$, and let the parallel to $AB$ through $I$ intersect $AM$ at $P$. Set $AP:PM = \lambda$; in our problem, $\lambda = \frac{7}{3}$.

If $N$ is the midpoint of $CL$ then $MN \parallel AB$ as $M$ is the midpoint of $BC$. Hence $MN \parallel IP$. Now Thales' theorem yields $\frac{IL}{IN} = \frac{AP}{PM} = \lambda$. Indeed, let $AM$ and $CL$ meet at $Q$. Then

$$\frac{IL}{AP} = \frac{QI}{QP} = \frac{QN}{QM} = \frac{IN}{PM}, \quad \text{implying} \quad \frac{IL}{IN} = \frac{AP}{PM} = \lambda.$$

Because $N$ is the midpoint of $CL$, the equality
$$\frac{IL}{IN} = \lambda \text{ gives } NL = CN = (\lambda + 1)IN,$$
$$CI = (\lambda + 2)IN. \text{ Hence, } \frac{CI}{LI} = \frac{\lambda + 2}{\lambda}.$$

On the other hand $\frac{CI}{LI} = \frac{AC}{AL}$ by the bisector theorem in triangle $ACL$. Under standard notation $BC = a$, $CA = b$, $AB = c$ we have $AL = \frac{bc}{a+b}$, so

$$\frac{CI}{LI} = \frac{a+b}{c}. \text{ (The last equality is generally known as a fact). It follows that } \frac{\lambda+2}{\lambda} = \frac{a+b}{c}, \text{ implying}$$
$$c = \frac{\lambda}{\lambda + 2}(a + b), \quad c = \frac{\lambda}{2\lambda + 2}(a + b + c).$$

For $\lambda = \frac{7}{3}$ and $a+b+c = 100$ the outcome is $c = 35$.