A Proof of the Location of a Triangle's Centroid Using Vectors

by: Al Byrnes

What is a centroid? How is it constructed? Where exactly is it located? Why is the centroid so special? The following write-up will attempt to address these questions.

Firstly, what is a centroid? As stated on the EMAT 4680/6680 Explorations 04 Centers of a Triangle webpage, "the centroid (G) of a triangle is the common intersection of three medians. A median of a triangle is the segment from a vertex to the midpoint of the opposite side." (Wilson, 2016)

Follow this link for a construction of a centroid in Geometer's Sketchpad as well as a script tool for constructing your own triangle with centroid. I have also included an illustration of the triangle ABC with its centroid and medians below:

Here we see that the three medians of the triangle ABC are concurrent at the centroid. Is this always the case?

Want to show: the three medians of a triangle are concurrent. Let's start with two vectors

uandv, as shown below. Both vectors originate at the origin and extend to the points (v_{1}, v_{2}) and (u_{1}, u_{2}) respectively; v_{1}, v_{2}, u_{1}, and u_{2}are all inR^{2}:Construct the vector 1/2(

v + u), indicated with the purple highlight:Now, construct the vector

v-u, highlighted below in purple:Aside: Vectors

v - uandv + uare diagonals of the parallelogram formed by the vectorsuandv. See the illustration below:Since we know that diagonals of a parallelogram bisect one another, we know that the vector 1/2(

v + u) will bisect the vectorv - u(the orange vector in the visual above). Stated another way, 1/2(v + u) is a median of our triangle as it represents the segment from a vertex to the midpoint of the opposite side. Let's construct another vector in a similar fashion. The goal will be to construct a vector which will bisect the vectorv. If we use a similar argument as before, we know that the diagonal of the parallelogram with two sides-uand two sidesv-uwill bisectv. See the illustration below:The other diagonal of the parallelogram shown above is

-u + (v - u) orv -2u(indicated by the green vector):Similarly, the diagonal of the parallelogram

vbisects the vectorv- 2u. So 1/2(v- 2u) is a another median of our triangle. Below, see the triangle with the two medians we have constructed:Our final median can be constructed by adding the vectors

-vand-(v - u).The vectoruwill bisect the vector -2v+u. The final median will then have length 1/2(-2v+u). The illustration below shows all three of these medians and the triangle:For any vectors

uandv, is there a point at which the medians will be concurrent? First let's examine the dark blue median and the green median. Where do these two points meet?Starting at the origin point, the dark blue and the green medians are described by the vectors 1/2(

u + v) andu +1/2(v- 2u); note thatuwas added to the length of the median because our original starting point is the origin.Is there scalar

k, such thatk• 1/2(u + v) =k• (u +1/2(v- 2u))? If so, we know that these two vectors meet at a point that is exactlykthe distance along both vectors from their respective verticies in the direction of the opposite side. See algebra below:So, we see that the two medians for any vectors

uandvdo coincide at a point that is exactly 2/3 of the length along each median away from the vertex and towards the opposite side.As a check let's consider the final median and compare it with the original median.

Sure enough! We see that the dark blue and the pink medians also are concurrent at 2/3 of the length of the medians away from the vertex in the direction of the opposite side.

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