Let $T$ be the centroid of triangle $ABC$. Prove that the vector sum $\vec{TA} + \vec{TB} + \vec{TC} = \vec{0}$.

1. Problem Description

Let $T$ be the centroid of triangle $ABC$. Prove that the vector sum $\vec{TA} + \vec{TB} + \vec{TC} = \vec{0}$.

2. Solution Steps

Let $A$, $B$, and $C$ be the vertices of a triangle. Let $T$ be the centroid of the triangle. The centroid is the intersection point of the medians of the triangle. A median is a line segment connecting a vertex to the midpoint of the opposite side.

The position vector of the centroid $T$ can be expressed as the average of the position vectors of the vertices:

$$\vec{OT} = \frac{\vec{OA} + \vec{OB} + \vec{OC}}{3}$$

where $O$ is the origin.

Now, let's express the vectors $\vec{TA}$, $\vec{TB}$, and $\vec{TC}$ in terms of the position vectors of the vertices and the centroid:

$$\vec{TA} = \vec{OA} - \vec{OT}$$

$$\vec{TB} = \vec{OB} - \vec{OT}$$

$$\vec{TC} = \vec{OC} - \vec{OT}$$

We want to show that $\vec{TA} + \vec{TB} + \vec{TC} = \vec{0}$. Let's add the expressions above:

$$\vec{TA} + \vec{TB} + \vec{TC} = (\vec{OA} - \vec{OT}) + (\vec{OB} - \vec{OT}) + (\vec{OC} - \vec{OT})$$

$$\vec{TA} + \vec{TB} + \vec{TC} = \vec{OA} + \vec{OB} + \vec{OC} - 3\vec{OT}$$

Substitute $\vec{OT} = \frac{\vec{OA} + \vec{OB} + \vec{OC}}{3}$ into the equation:

$$\vec{TA} + \vec{TB} + \vec{TC} = \vec{OA} + \vec{OB} + \vec{OC} - 3 \left(\frac{\vec{OA} + \vec{OB} + \vec{OC}}{3}\right)$$

$$\vec{TA} + \vec{TB} + \vec{TC} = \vec{OA} + \vec{OB} + \vec{OC} - (\vec{OA} + \vec{OB} + \vec{OC})$$

$$\vec{TA} + \vec{TB} + \vec{TC} = \vec{0}$$

3. Final Answer

$\vec{TA} + \vec{TB} + \vec{TC} = \vec{0}$