Given a triangle ABC, M is the midpoint of [AB] and I is the midpoint of [MC]. We need to construct a point K such that $\vec{CK} = \frac{1}{3}\vec{CB}$. Then, we need to prove that points A, I, and K are collinear.

1. Problem Description

Given a triangle ABC, M is the midpoint of [AB] and I is the midpoint of [MC]. We need to construct a point K such that $\vec{CK} = \frac{1}{3}\vec{CB}$. Then, we need to prove that points A, I, and K are collinear.

2. Solution Steps

First, express $\vec{AI}$ in terms of $\vec{AC}$ and $\vec{AB}$.

Since I is the midpoint of MC, we have $\vec{AI} = \frac{1}{2}(\vec{AM} + \vec{AC})$.

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

Therefore,

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

Next, express $\vec{AK}$ in terms of $\vec{AC}$ and $\vec{AB}$.

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

Now, we need to check if $\vec{AI}$ and $\vec{AK}$ are collinear. In other words, we need to find a scalar $\lambda$ such that $\vec{AK} = \lambda \vec{AI}$.

If such a $\lambda$ exists, then A, I, and K are collinear.

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

$\frac{1}{3}\vec{AB} + \frac{2}{3}\vec{AC} = \frac{\lambda}{4}\vec{AB} + \frac{\lambda}{2}\vec{AC}$

Equating the coefficients of $\vec{AB}$ and $\vec{AC}$, we get:

$\frac{1}{3} = \frac{\lambda}{4}$ and $\frac{2}{3} = \frac{\lambda}{2}$

From the first equation, $\lambda = \frac{4}{3}$.

From the second equation, $\lambda = \frac{4}{3}$.

Since both equations yield the same value for $\lambda$, we have $\lambda = \frac{4}{3}$.

Therefore, $\vec{AK} = \frac{4}{3}\vec{AI}$.

Since $\vec{AK}$ is a scalar multiple of $\vec{AI}$, the points A, I, and K are collinear.

3. Final Answer

$\vec{AK} = \frac{4}{3}\vec{AI}$