# Centroid of a triangle proof

1. Sep 17, 2012

### phospho

We got given the formula for finding the centroid of a triangle which was to $(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3})$ but we the teacher said he didn't have time to show the proof, however I'm still curious. Could anyone possible hint on how I could go about to prove this? My maths experience is of A level Maths & Further maths, or high school senior (AP Calc I think).

2. Sep 17, 2012

### voko

Take a triangle and consider the median from one of vertices. This median divides the triangle into two triangles. Can you see they have the same area? That means the centroid must be on the median. By the same token, we can see this must hold for the other medians. All the medians intersect in one point. This, then, must be the centroid.

3. Sep 17, 2012

### phospho

I've done it the way you mentioned but I don't see how that proves the mean of the 3 vertices is equal to the center of mass.

thanks

4. Sep 17, 2012

### voko

Let's denote the vertices by A, B, C. The equation of the median from vertex A is given by $z = A + (\vec{AB} + \vec{AC})s/2$, where s is the parameter of the line (its length). Can you see why this is true? Likewise, for the median from B, we have $z = B + (\vec{BA} + \vec{BC})t/2$. Where the medians intersect, these must be equal, so we have $A + (\vec{AB} + \vec{AC})s/2 = B + (\vec{BA} + \vec{BC})t/2$. If we express this in terms of the coordinates, we have: $$\begin{cases} x_A + (x_B - x_A + x_C - x_A)s/2 = x_B + (x_A - x_B + x_C - x_B)t/2 \\y_A + (y_B - y_A + y_C - y_A)s/2 = y_B + (y_A - y_B + y_C - y_B)t/2 \end{cases}$$And this is solved with $$\begin{cases}s = 2/3 \\ t = 2/3\end{cases}$$(Show it.)

This means that the centroid lies at 2/3 along any median. Taking one of the equations of the medians, and substituting s or t in it, we obtain the coordinates of the centroid: $$(\frac {x_A + x_B + x_C} 3, \frac { y_A + y_B + y_C } 3)$$