Given a regular hexagon $ABCDEF$, where $\vec{AB} = \vec{p}$ and $\vec{BC} = \vec{q}$, express the vectors $\vec{CD}, \vec{DE}, \vec{EF}, \vec{FA}, \vec{AD}, \vec{EA}, \vec{AC}$ in terms of $\vec{p}$ and $\vec{q}$.

1. Problem Description

Given a regular hexagon $ABCDEF$, where $\vec{AB} = \vec{p}$ and $\vec{BC} = \vec{q}$, express the vectors $\vec{CD}, \vec{DE}, \vec{EF}, \vec{FA}, \vec{AD}, \vec{EA}, \vec{AC}$ in terms of $\vec{p}$ and $\vec{q}$.

2. Solution Steps

In a regular hexagon, all sides have the same length, and all interior angles are $120^\circ$.

Also, note that $\vec{AB} + \vec{BC} + \vec{CD} + \vec{DE} + \vec{EF} + \vec{FA} = \vec{0}$.

$\vec{CD}$: In a regular hexagon, $\vec{CD}$ has the same length as $\vec{AB}$ and $\vec{AB}$ rotated by $120^\circ$ to get $\vec{BC}$, so $\vec{CD}$ is $-\vec{p}$.

$\vec{DE}$: $\vec{DE}$ is parallel to $\vec{CB}$ with the same length, so $\vec{DE} = -\vec{q}$.

$\vec{EF}$: $\vec{EF}$ is parallel to $\vec{BA}$ with the same length, so $\vec{EF} = -\vec{p}$.

$\vec{FA}$: $\vec{FA}$ is parallel to $\vec{CB}$ with the same length, so $\vec{FA} = -\vec{q}$.

$\vec{AD}$: $\vec{AD} = \vec{AB} + \vec{BC} + \vec{CD} = \vec{p} + \vec{q} + \vec{CD} = \vec{p} + \vec{q} + (\vec{AC}-\vec{AD})$. Consider the center $O$ of the hexagon.

Then $\vec{AD} = 2\vec{BC} = 2\vec{q}$.

$\vec{EA}$: $\vec{EA} = -(\vec{FA} + \vec{EF} + \vec{DE}) = \vec{p} + \vec{q} - \vec{p} = -\vec{CD} - \vec{DE} - \vec{EF} = -(-\vec{p} - \vec{q} + \vec{p}) = \vec{q} - \vec{p}$. Since $\vec{CD} = \vec{BA}=-\vec{p}$ and $\vec{DE} = -\vec{BC} = -\vec{q}$, then by symmetry we must have $\vec{EA} = \vec{q} - \vec{p}$.

$\vec{EA} = \vec{DA} = -\vec{AD} = -2\vec{BC}= -\vec{AD}$. $\vec{AD}=\vec{p} + \vec{q} -\vec{p} = \vec{p} + \vec{BC} - p = \vec{BC}$. $\vec{EA}=\vec{BA}= q-p$.

$\vec{EA} = \vec{FA} + \vec{FE} = -\vec{q} + \vec{p}$. $\vec{CD} = \vec{BA}=-\vec{p}$ and $\vec{DE} = -\vec{BC} = -\vec{q}$. Then $\vec{EA} = \vec{BC} - \vec{AB} = \vec{q}-\vec{p}$.

$\vec{AC}$: $\vec{AC} = \vec{AB} + \vec{BC} = \vec{p} + \vec{q}$.

3. Final Answer

$\vec{CD} = -\vec{p}$

$\vec{DE} = -\vec{q}$

$\vec{EF} = -\vec{p}$

$\vec{FA} = -\vec{q}$

$\vec{AD} = 2\vec{q}$

$\vec{EA} = \vec{q}-\vec{p}$

$\vec{AC} = \vec{p}+\vec{q}$