Given a triangle $ABC$, $M$ is the midpoint of $[AB]$ and $N$ is the midpoint of $[MC]$. 1) a) Place the point $L$ such that $\vec{CL} = \frac{1}{3} \vec{CB}$. b) Show that the points $A$, $N$ and $L$ are collinear. 2) a) Place the points $I$, $J$ and $K$ such that $\vec{BI} = \frac{3}{2} \vec{BC}$, $\vec{CJ} = \frac{1}{3} \vec{CA}$ and $\vec{AK} = \frac{2}{5} \vec{AB}$. b) Show that the points $I$, $J$ and $K$ are collinear.

1. Problem Description

Given a triangle $ABC$, $M$ is the midpoint of $[AB]$ and $N$ is the midpoint of $[MC]$.

1) a) Place the point $L$ such that $\vec{CL} = \frac{1}{3} \vec{CB}$.

b) Show that the points $A$, $N$ and $L$ are collinear.

2) a) Place the points $I$, $J$ and $K$ such that $\vec{BI} = \frac{3}{2} \vec{BC}$, $\vec{CJ} = \frac{1}{3} \vec{CA}$ and $\vec{AK} = \frac{2}{5} \vec{AB}$.

b) Show that the points $I$, $J$ and $K$ are collinear.

2. Solution Steps

1) a) We simply place the point $L$ such that $\vec{CL} = \frac{1}{3} \vec{CB}$. This means that $L$ lies on the segment $[CB]$, and $CL = \frac{1}{3}CB$.

b) To show that $A$, $N$ and $L$ are collinear, we need to show that $\vec{AN}$ and $\vec{AL}$ are collinear.

Since $M$ is the midpoint of $[AB]$, we have $\vec{AM} = \frac{1}{2} \vec{AB}$.

Since $N$ is the midpoint of $[MC]$, we have $\vec{AN} = \frac{1}{2} (\vec{AM} + \vec{AC}) = \frac{1}{2}(\frac{1}{2} \vec{AB} + \vec{AC}) = \frac{1}{4} \vec{AB} + \frac{1}{2} \vec{AC}$.

$\vec{AL} = \vec{AC} + \vec{CL} = \vec{AC} + \frac{1}{3} \vec{CB} = \vec{AC} + \frac{1}{3} (\vec{AB} - \vec{AC}) = \frac{1}{3} \vec{AB} + \frac{2}{3} \vec{AC}$.

Let's express $\vec{AL}$ as a multiple of $\vec{AN}$, i.e., $\vec{AL} = k \vec{AN}$ for some scalar $k$.

$\frac{1}{3} \vec{AB} + \frac{2}{3} \vec{AC} = k (\frac{1}{4} \vec{AB} + \frac{1}{2} \vec{AC})$.

Equating coefficients:

$\frac{1}{3} = \frac{k}{4} \implies k = \frac{4}{3}$

$\frac{2}{3} = \frac{k}{2} \implies k = \frac{4}{3}$

Since the value of $k$ is consistent, $\vec{AL} = \frac{4}{3} \vec{AN}$, which means that $A$, $N$ and $L$ are collinear.

2) a) We simply place the points according to the given conditions.

$\vec{BI} = \frac{3}{2} \vec{BC}$ means that $I$ lies on the line $(BC)$, and $BI = \frac{3}{2} BC$.

$\vec{CJ} = \frac{1}{3} \vec{CA}$ means that $J$ lies on the segment $[CA]$, and $CJ = \frac{1}{3} CA$.

$\vec{AK} = \frac{2}{5} \vec{AB}$ means that $K$ lies on the segment $[AB]$, and $AK = \frac{2}{5} AB$.

b) To show that $I, J, K$ are collinear, we need to show that $\vec{IJ}$ and $\vec{IK}$ are collinear.

$\vec{IJ} = \vec{IC} + \vec{CJ} = \vec{BC} - \vec{BI} + \frac{1}{3} \vec{CA} = \vec{BC} - \frac{3}{2} \vec{BC} + \frac{1}{3} \vec{CA} = -\frac{1}{2} \vec{BC} + \frac{1}{3} \vec{CA}$.

$\vec{IK} = \vec{IA} + \vec{AK} = \vec{BA} - \vec{BI} + \frac{2}{5} \vec{AB} = \vec{BA} - \frac{3}{2} \vec{BC} + \frac{2}{5} \vec{AB} = -\vec{AB} - \frac{3}{2} \vec{BC} + \frac{2}{5} \vec{AB} = -\frac{3}{5} \vec{AB} - \frac{3}{2} \vec{BC}$.

Since $\vec{AB} = \vec{CB} - \vec{CA} = -\vec{BC} - \vec{CA}$,

$\vec{IK} = -\frac{3}{5} (-\vec{BC} - \vec{CA}) - \frac{3}{2} \vec{BC} = \frac{3}{5} \vec{BC} + \frac{3}{5} \vec{CA} - \frac{3}{2} \vec{BC} = (\frac{3}{5} - \frac{3}{2}) \vec{BC} + \frac{3}{5} \vec{CA} = (\frac{6 - 15}{10}) \vec{BC} + \frac{3}{5} \vec{CA} = -\frac{9}{10} \vec{BC} + \frac{3}{5} \vec{CA}$.

If $\vec{IK} = k \vec{IJ}$, then $-\frac{9}{10} \vec{BC} + \frac{3}{5} \vec{CA} = k(-\frac{1}{2} \vec{BC} + \frac{1}{3} \vec{CA})$.

Equating coefficients:

$-\frac{9}{10} = -\frac{k}{2} \implies k = \frac{18}{10} = \frac{9}{5}$

$\frac{3}{5} = \frac{k}{3} \implies k = \frac{9}{5}$

Since the value of $k$ is consistent, $I$, $J$ and $K$ are collinear.

3. Final Answer

1) b) The points $A$, $N$ and $L$ are collinear.

2) b) The points $I$, $J$ and $K$ are collinear.