Meaning of center of mass : triangle Consider a triangle ABC, where $A=(x_1,y_1), B=(x_2,y_2)$ and $C=(x_3,y_3).$ Then it is well known that the centroid is $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}).$
Why is the following sentence true? 

Centroid of a triangle is the point at which the triangle could be perfectly balanced on the tip of a pin.

I know the definition and properties of the centroid of triangle, but I couldn’t understand the sentence exactly. 
For the definition and properties of it, see https://www.mathwarehouse.com/geometry/triangles/triangle-concurrency-points/centroid-of-triangle.php#ixzz66i8NwJ3q 
My another question is

In general, is the centroid of polygon ( for example, rectangle, pentagon, etc) unique?

Would you give me a hint or reference for it? 
Thanks in advance!
 A: Is this a definition of centroid or has the centroid been defined and this is being stated as a property?  If you try to balance something on top of a pin the pin cannot exert any torque on the object.  The torque due to gravity around the pin must be zero.  Try cutting a shape (it doesn't have to be a triangle) out of cardboard and balancing it on your finger.  
We can consider the torque around the horizontal $x$ axis.  If the density per area is constant over the shape, the torque around the $x$ axis is $\int y \ dA$ because a small area $dA$ will have a downward force proportional to $dA$ and will exert a torque of $y\ dA$.  The $x$ axis will pass through the centroid when this integral is zero.
A: 
Centroid of a triangle is the point at which the triangle could be perfectly balanced on the tip of a pin.


Here you are! 
In mathematics and physics, the centroid (or geometric center) of a plane figure is the arithmetic mean ("average") position of all the points in the shape.  The geometric centroid of a triangle is therefore also its center of mass, meaning that we could imagine all the mass of the plane object concentrated in this single point. For each point outside the centroid that has a certain little weight ($dW$) there is another point or more that balance this weight. Thus, the triangle is in equilibrium: there are no torques (you can think of the torque as the tendency of a force, here the weight, to rotate the body onto which it acts). If there wasn't such a balance around the centroid, the triangle would incline from one side and fall down.
A: Let $\vec{r_0} = (x_0, y_0, z_0)$ be the point on which the polygon could be balanced. The torque $d\vec{M}$ of an infinitesimal part of a polygon at coordinates $\vec{r} = (x,y,z)$ of mass $dm$ w.r.t the point $\vec{r_0}$ is given by
$$d\vec{M} = (\vec{r} - \vec{r_0})\times \vec{g}\,dm$$
The total torque $\vec{M}$ w.r.t. the point $\vec{r_0}$ is $0$ by definition of $\vec{r_0}$ so
\begin{align}
0 &= \vec{M} \\
&= \int_{\text{polygon}} d\vec{M} \\
&= \int_{\text{polygon}}(\vec{r} - \vec{r_0})\times \vec{g}\,dm \\
&= \left(\int_{\text{polygon}}(\vec{r} - \vec{r_0})\,dm\right) \times \vec{g}
\end{align}
And hence $\int_{\text{polygon}}(\vec{r} - \vec{r_0})\,dm$ is parallel to $\vec{g}$. The integral $\int_{\text{polygon}}(\vec{r} - \vec{r_0})\,dm$ is a vector always orthogonal to $\vec{g}$ so
\begin{align}
0 &= \int_{\text{polygon}}(\vec{r} - \vec{r_0})\,dm \\
&= \int_{\text{polygon}}\vec{r}\,dm - \int_{\text{polygon}}\vec{r_0}\,dm\\
&= \int_{\text{polygon}}\vec{r}\,dm - \vec{r_0}\int_{\text{polygon}}dm\\
&= \int_{\text{polygon}}\vec{r}\,dm - m\vec{r_0}
\end{align}
where $m$ is the mass of the triangle.
We conclude
$$\vec{r_0} = \frac1m \int_{\text{polygon}} \vec{r}\,dm$$
which is precisely the definition of the centroid.
Coordinate-wise, this means
$$(x_0, y_0, z_0) = \left(\int_{\text{polygon}} x\,dm, \int_{\text{polygon}} y\,dm, \int_{\text{polygon}} z\,dm\right)$$
