Let $ABC$ be a triangle. Let $M$ be the midpoint of segment $AB$, and let $I$ be the midpoint of segment $MC$. Construct the point $K$ such that $\vec{CK} = \frac{1}{3} \vec{CB}$. Prove that the points $A$, $I$, and $K$ are collinear.

1. Problem Description

Let $ABC$ be a triangle. Let $M$ be the midpoint of segment $AB$, and let $I$ be the midpoint of segment $MC$. Construct the point $K$ such that $\vec{CK} = \frac{1}{3} \vec{CB}$. Prove that the points $A$, $I$, and $K$ are collinear.

2. Solution Steps

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

Since $I$ is the midpoint of $MC$, we have $\vec{MI} = \frac{1}{2} \vec{MC}$.

We want to show that $A$, $I$, and $K$ are collinear, which means that $\vec{AI} = \lambda \vec{AK}$ for some scalar $\lambda$.

We can write

$\vec{AI} = \vec{AM} + \vec{MI} = \frac{1}{2} \vec{AB} + \frac{1}{2} \vec{MC}$

Since $\vec{MC} = \vec{AC} - \vec{AM} = \vec{AC} - \frac{1}{2} \vec{AB}$,

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

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

We want to find $\lambda$ such that $\vec{AI} = \lambda \vec{AK}$. Thus,

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

Equating the coefficients, we have

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

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

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

Thus, $\vec{AI} = \frac{3}{4} \vec{AK}$, which means $A$, $I$, and $K$ are collinear.

3. Final Answer

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