Problem 9.2. A triangle $A B C$ with centroid $G$ and incenter $I$ is given. If $A B=42, G I=2$ and $A B \| G I$, find $A C$ and $B C$.

Ivailo Kortezov
9.2. Let $C M$ be the median and $C L$ be the bisector of $\triangle A B C(I \in C L)$. Using the standard notation for $\triangle A B C$ we have $\frac{A L}{B L}=\frac{A C}{B C}=\frac{b}{a}$, whence $A L=\frac{b c}{a+b}$.

Since $A I$ is the bisector of $\triangle A L C$ through $A$ we get $\frac{C I}{I L}=\frac{C A}{A L}=\frac{a+b}{c}$.

By the Thales theorem we have $\frac{C I}{I L}=\frac{C G}{G M}$, i.e. $a+b=2 c$, whence $a+b=84$. For $a \leq b$ we have $L M=\frac{3}{2} I G=3, A M=21, A L=$ $A M+L M=24$ and $L B=B M-L M=18$. Therefore $\frac{b}{a}=\frac{24}{18}=\frac{4}{3}$. Hence $3 x+4 x=84$, i.e. $x=12, A C=48$ and $B C=36$. For $b \geq a$ we

![](https://cdn.mathpix.com/cropped/2024_06_03_f8ccd71b4db367e7c2a9g-138.jpg?height=510&width=507&top_left_y=1021&top_left_x=1159)
get $B C=48$ and $A C=36$.