The first three explorations, on logarithms and parabolas, have focused primarily on algebra. We've plotted Cartesian geometry through points and curves, but our main focus has been symbolic math. This exploration begins our transition to figure based geometry, even though we'll keep some algebraic steps. It is designed for a student in early high school, such as Georgia Math 1 or beginning Geometry.

Though a more serious mathematician, like those at Wolfram Mathworld, might cringe at this definition, we might label geometry the study of shapes. Geometry has a long history. Plato and Socrates spoke about geometry. A little later, in the 3rd century BC, Euclid's Elements were written. The books are still available and useful today. For comparison, our earlier topics (logarithms and Cartesian coordinates) were first published around 1615 AD. Most statistical topics are relative babies, from the last 100 years!

The Elements begin with some definitions. In this write up, we'll look at one of the basic figures, defined by three straight lines. Book I, Definition 19 calls this a trilateral, based on the number of sides. In modern language, we call this a triangle, because the figure has three interior angles. The figure below illustrates some common types of triangles. In particular, we need to remember that some triangles are obtuse, with one angle larger than a right angle. They sometimes don't look as neat as non-obtuse acute triangles (not just because I colored that triangle brown).

We can draw many things given a triangle. In this assignment, we'll focus on line segments called medians. A triangle median connects one of the corner points of a triangle, called a vertex, to the midpoint of the opposite side. There are three medians inside each triangle, one from each vertex.

Let's construct the medians for a triangle. I'll demonstrate with pictures from the free Java-based package GeoGebra. We'll start with a triangle in a coordinate system, labeling the vertices A, B, and C. To make things easier, I'll fix point A at the origin (0, 0). We can make any kind of triangle by moving B and C.

The Greeks, including Euclid, had rules about what could be constructed. Their constructions were only allowed to use a compass and an unmarked straightedge. If you want to see how they constructed a median, you might examine this page at Math Open Reference. Instead of that approach, we'll use properties of the coordinate system.

Let's label the coordinates of point B as (b_x, b_y). The midpoint of side AB is halfway between A and B. In the horizontal x direction, that's \frac{0 + b_x}{2} = \frac{b_x}{2}. In the vertical y direction, the midpoint will be at \frac{0 + b_y}{2} = \frac{b_y}{2}. Similarly, if the coordinates of point C are (c_x, c_y). the midpoint of side AC is \left( \frac{c_x}{2}, \frac{c_y}{2} \right) . The midpoint of side BC is \left( \frac{b_x + c_x}{2} , \frac{b_y + c_y}{2} \right) . On GeoGebra, we'll call these points midAB, midAC, and midBC respectively. The medians connect these midpoints to the opposite vertex, such as from midAB to C. The picture shows each median in a different color - purple, green, and black.

Interestingly, it appears that the purple, green, and black median lines all intersect at the same point. Is this true, or are they just close together? We'll go through the math and show our eyes do not deceive us, that all three medians are joined.

In a plane, any two non-parallel lines have an intersection point. We need to find equations for the median lines, then search for an intersection. To find line equations, we will find the slope of the line. Based on two points (x_1, y_1) , (x_2, y_2) , slope m equals change in y divided by change in x, m = \frac{y_2 - y_1}{x_2 - x_1} . (There's a special case we can't ignore. If the line is vertical, the two x coordinates would be the same and we'd be attempting to divide by zero. Since that's no good, we'll remember to write vertical lines properly as x = x_1 = x_2 if we must.) Given the slope, we can specify the line using a point and slope.

y - y_1 = m (x - x_1) ~~ = \frac{y_2 - y_1}{x_2 - x_1} (x - x_1)

Since we know the coordinates of the vertices and midpoints, we substitute to find the line equations. A little algebra finds the following equations:

\mbox{Purple median from C}: y = \frac{b_y - 2 c_y}{b_x - 2 c_x} (x - c_x) + c_y

\mbox{Green median from B}: y = \frac{c_y - 2 b_y}{c_x - 2 b_x} (x - b_x) + b_y

\mbox{Black median from A}: y = \frac{b_y + c_y}{b_x + c_x} ~ x

To find the intersection of the purple and green medians, since we have them in the form ( y = something ) , we set y = y and solve for x.

\frac{b_y - 2 c_y}{b_x - 2 c_x} (x - c_x) + c_y ~~ = ~~ \frac{c_y - 2 b_y}{c_x - 2 b_x} (x - b_x) + b_y

The goal is to isolate the two terms involving x on one side, the terms not involving x on the other side, then perform the necessary division. There are a few algebraic steps involved, but eventually we reach the result

x = \frac{b_x + c_x}{3}

and then substituting that x value back into either line gives us y coordinate y = \frac{b_y + c_y}{3} . The point \left( \frac{b_x + c_x}{3} , \frac{b_y + c_y}{3} \right) lies on the purple and green segments. Does it also lie on the third median? To check, let's try substituting the point into the equation for the black median from A.

\left( \frac{b_y + c_y}{3} \right) = \frac{b_y + c_y}{b_x + c_x} ~ \left( \frac{b_x + c_x}{3} \right)

\left( \frac{b_y + c_y}{3} \right) = \frac{b_y + c_y}{3}

The two sides match, meaning the point does lie on the median from A. Since we found this point as the intersection of the medians from B and C, all three medians do share a common intersection. This point has a special name, the centroid. In this picture, I added numeric labels to the points.

In the prior part, we saw that the fractions included the numbers 2 and 3. It turns out that the centroid has a special location. Not only does it intersect all three medians, it lies two-thirds of the way from the vertex to the midpoint. For instance, the distance from A to the Centroid is twice the distance from the Centroid to midBC. There are several ways to demonstrate this. The fabulous Khan Academy provides Youtube videos of two approaches. One might use similar triangles, or apply 3-D distance. To show something different, I'm going to apply the distance formula in two dimensions.

When GeoGebra provides a numeric distance between two points (x_1, y_1) , (x_2, y_2) , it has calculated using the distance formula.

d = \sqrt{ (x_2 - x_1)^{2} + (y_2 - y_1)^{2} }

Since we know the coordinates of the centroid are \left( \frac{b_x + c_x}{3} , \frac{b_y + c_y}{3} \right) , the distance from the origin point A to the centroid is

d = \sqrt{ \left( \frac{b_x + c_x}{3} - 0 \right)^{2} + \left( \frac{b_y + c_y}{3} - 0 \right)^{2} } ~ = \sqrt{ \left( \frac{b_x + c_x}{3} \right)^{2} + \left( \frac{b_y + c_y}{3} \right)^{2} } ~ = \frac{1}{3} ~ \sqrt{ \left( b_x + c_x \right)^{2} + \left( b_y + c_y \right)^{2} }

The coordinates of midBC, the midpoint of side BC, are \left( \frac{b_x + c_x}{2} , \frac{b_y + c_y}{2} \right) . Therefore, the distance from the origin point A to midBC is

d_m = \sqrt{ \left( \frac{b_x + c_x}{2} - 0 \right)^{2} + \left( \frac{b_y + c_y}{2} - 0 \right)^{2} } ~ = \sqrt{ \left( \frac{b_x + c_x}{2} \right)^{2} + \left( \frac{b_y + c_y}{2} \right)^{2} } ~ = \frac{1}{2} ~ \sqrt{ \left( b_x + c_x \right)^{2} + \left( b_y + c_y \right)^{2} }

The distance to the centroid is one-third times a square root, while the longer second distance equals one-half times the same square root. The ratio is \frac{1/3}{1/2} = \frac{2}{3} , as was claimed.

I've just shown the calculations for the black median from A. The procedure is the same for the other two medians - it's just that the computations are a little longer. For experimentation through manipulation, I suggest clicking through to the centroid.html page for a GeoGebra applet. You can click and move all three vertices of the triangle - A, B and C. The applet automatically updates the midpoints, medians, and distances.