Argentina_2018

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

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

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

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

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