How Is the Centroid of a Triangle Determined Mathematically?

[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).

 

[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.

 

[phospho]

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.

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

 

[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: $$(\frac {x_A + x_B + x_C} 3, \frac { y_A + y_B + y_C } 3)$$

 

[Nickie]

The centroid of a triangle is the point where the three medians intersect. A median is a line segment that connects a vertex of the triangle to the midpoint of the opposite side.

To prove the formula for the centroid, we can use the concept of vectors. Let's consider a triangle with vertices A(x1,y1), B(x2,y2), and C(x3,y3). We can represent these points as vectors:

A = (x1, y1)
B = (x2, y2)
C = (x3, y3)

Now, let's find the midpoint of each side of the triangle. The midpoint of AB is M, which can be represented as:

M = (x1+x2, y1+y2)/2

Similarly, the midpoint of BC and CA can be represented as:

N = (x2+x3, y2+y3)/2
P = (x3+x1, y3+y1)/2

Now, let's consider the vector from point A to the centroid G. This vector can be represented as:

AG = (x-x1, y-y1)

Since G lies on the median from A to BC, it must also lie on the line segment MN. Therefore, we can write the following equation:

AG = t(MN)

where t is a scalar.

Using the vector representations of A, M, and N, we can write:

(x-x1, y-y1) = t[(x1+x2, y1+y2)/2 - (x2+x3, y2+y3)/2]

Expanding this equation and simplifying, we get:

(x-x1, y-y1) = t[(x1-x3, y1-y3)]

Equating the x and y components, we get:

x-x1 = t(x1-x3)
y-y1 = t(y1-y3)

Solving for t, we get:

t = (x-x1)/(x1-x3) = (y-y1)/(y1-y3)

Similarly, we can show that t = (x-x2)/(x2-x3) = (y-y2)/(y2-y3)

Since t is the same for both equations, we can equate them and solve for x and y.

(x-x1)/(x1-x