Given a regular hexagon $ABCDEF$, let $A_1, B_1, C_1, D_1, E_1, F_1$ be the midpoints of the sides $AB, BC, CD, DE, EF, FA$ respectively. We need to prove that $\vec{AA_1} + \vec{BB_1} + \vec{CC_1} + \vec{DD_1} + \vec{EE_1} + \vec{FF_1} = \vec{0}$.

1. Problem Description

Given a regular hexagon $ABCDEF$, let $A_1, B_1, C_1, D_1, E_1, F_1$ be the midpoints of the sides $AB, BC, CD, DE, EF, FA$ respectively. We need to prove that $\vec{AA_1} + \vec{BB_1} + \vec{CC_1} + \vec{DD_1} + \vec{EE_1} + \vec{FF_1} = \vec{0}$.

2. Solution Steps

Let $O$ be the center of the hexagon. We can express each vector as the difference of position vectors:

$\vec{AA_1} = \vec{OA_1} - \vec{OA}$

$\vec{BB_1} = \vec{OB_1} - \vec{OB}$

$\vec{CC_1} = \vec{OC_1} - \vec{OC}$

$\vec{DD_1} = \vec{OD_1} - \vec{OD}$

$\vec{EE_1} = \vec{OE_1} - \vec{OE}$

$\vec{FF_1} = \vec{OF_1} - \vec{OF}$

Therefore,

$\vec{AA_1} + \vec{BB_1} + \vec{CC_1} + \vec{DD_1} + \vec{EE_1} + \vec{FF_1} = (\vec{OA_1} - \vec{OA}) + (\vec{OB_1} - \vec{OB}) + (\vec{OC_1} - \vec{OC}) + (\vec{OD_1} - \vec{OD}) + (\vec{OE_1} - \vec{OE}) + (\vec{OF_1} - \vec{OF})$

$= (\vec{OA_1} + \vec{OB_1} + \vec{OC_1} + \vec{OD_1} + \vec{OE_1} + \vec{OF_1}) - (\vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} + \vec{OE} + \vec{OF})$

Since $A_1$ is the midpoint of $AB$, we have $\vec{OA_1} = \frac{\vec{OA} + \vec{OB}}{2}$. Similarly,

$\vec{OB_1} = \frac{\vec{OB} + \vec{OC}}{2}$

$\vec{OC_1} = \frac{\vec{OC} + \vec{OD}}{2}$

$\vec{OD_1} = \frac{\vec{OD} + \vec{OE}}{2}$

$\vec{OE_1} = \frac{\vec{OE} + \vec{OF}}{2}$

$\vec{OF_1} = \frac{\vec{OF} + \vec{OA}}{2}$

Therefore,

$\vec{OA_1} + \vec{OB_1} + \vec{OC_1} + \vec{OD_1} + \vec{OE_1} + \vec{OF_1} = \frac{\vec{OA} + \vec{OB}}{2} + \frac{\vec{OB} + \vec{OC}}{2} + \frac{\vec{OC} + \vec{OD}}{2} + \frac{\vec{OD} + \vec{OE}}{2} + \frac{\vec{OE} + \vec{OF}}{2} + \frac{\vec{OF} + \vec{OA}}{2}$

$= \frac{2(\vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} + \vec{OE} + \vec{OF})}{2} = \vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} + \vec{OE} + \vec{OF}$

Plugging this back into the original equation:

$\vec{AA_1} + \vec{BB_1} + \vec{CC_1} + \vec{DD_1} + \vec{EE_1} + \vec{FF_1} = (\vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} + \vec{OE} + \vec{OF}) - (\vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} + \vec{OE} + \vec{OF}) = \vec{0}$

3. Final Answer

$\vec{AA_1} + \vec{BB_1} + \vec{CC_1} + \vec{DD_1} + \vec{EE_1} + \vec{FF_1} = \vec{0}$