Assignment 4

Jonathan Lawson

This exploration will look at the centroid of a triangle and prove that the three medians are concurrent at the centroid.

The median of a triangle is the segment that connects one of the vertices of a triangle with the opposite side.  The triangle below shows all three medians of a triangle and the centroid (in red).

Looking at the median of a triangle it is easy to see that it would always be contained within the interior of the triangle.  So if all three medians are contained within the interior of the triangle, then the medians will intersect with each other.  We will now prove that these intersections are at the same point.

So we will assume without loss of generality that one of the vertices of the triangle are located at the origin of the Cartesian plane.  One of the vertices will be located on the x-axis at the point (a, 0) and the third vertex will be located in the first quadrant at the point (b, c).

The midpoint of each leg can be found by taking the average of the x- and y-values of each endpoint of the leg.

The equation for median 1 is

y = x,

and the equation for median 2 is

y = (x - )

and the equation for median 3 is

y = (x - a).

So we will first find the point of intersection for median 1 and median 2 by setting each equation equal to the other,

x = (x - )

cx(2b - a) = c(2x - a)(b + a)

2bcx - acx = 2bcx + 2acx - abc - a2c

3acx = abc + a2c

x = , so y =

The point of intersection for median 1 and median 2 is the point (, ).

Now we will find the point of intersection for median 2 and median 3 by setting each equation equal to the other,

(x - ) = (x - a)

c(2x - a)(b - 2a) = c(x - a)(2b - a)

2bcx - 4acx - abc + 2a2c = 2bcx - acx - 2abc + a2c

3acx = abc + a2c

x = , so y =

The point of intersection for median 2 and median 3 is the point (, ).  If median one and median 3 share the same point and only intersect at the same (they do not have the same slopes) then each median must intersect at one point, and thus must be concurrent.

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