Given a triangle $ABC$, $M$ is the midpoint of segment $AB$, and $I$ is the midpoint of segment $MC$. Point $K$ is defined such that $\vec{CK} = \frac{1}{3} \vec{CB}$. We need to show that points $A, I, K$ are collinear.

1. Problem Description

Given a triangle $ABC$, $M$ is the midpoint of segment $AB$, and $I$ is the midpoint of segment $MC$. Point $K$ is defined such that $\vec{CK} = \frac{1}{3} \vec{CB}$. We need to show that points $A, I, K$ are collinear.

2. Solution Steps

We want to show that $A, I, K$ are collinear. This is equivalent to showing that the vectors $\vec{AI}$ and $\vec{AK}$ are collinear, meaning that $\vec{AI} = \lambda \vec{AK}$ for some scalar $\lambda$.

Since $I$ is the midpoint of $MC$, we have

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

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

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

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

Now, we need to find a scalar $\lambda$ such that $\vec{AI} = \lambda \vec{AK}$.

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

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

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

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

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

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

Since both equations give the same value for $\lambda$, the vectors $\vec{AI}$ and $\vec{AK}$ are collinear.

Therefore, the points $A, I, K$ are collinear.

3. Final Answer

The points A, I, and K are collinear.