Geometric properties involving medians

Let’s start with a small piece of art that is quite pleasing to look at; afterward we move on to the math:

You feel that? No doubt.

Apart from being a fine art, the above diagram will play a main part in the path we’re going to chart in today’s task. Got all of that?

Example 1

In the diagram above, $AM, ~BN,~CL$ are the three medians of the original triangle $ABC$ ; $R,~S,~T$ are the respective midpoints of these medians. Find the coordinates of the points $L,M,N,R,S,T,$ given $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3).$

By definition, $L,M,N$ are the midpoints of sides $AB,BC,CA$ , respectively; so:

$\begin{equation*} \begin{split} L:=&\Big(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\Big)\\ M:=&\Big(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\Big)\\ N:=&\Big(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2}\Big) \end{split} \end{equation}$

Similarly, because $R,S,T$ are the midpoints of medians $AM, BN, CL$ , we have:

$\begin{equation*} \begin{split} R:=&\Big(\frac{x_1+\frac{x_2+x_3}{2}}{2},\frac{y_1+\frac{y_2+y_3}{2}}{2}\Big)\\ =&\Big(\frac{2x_1+x_2+x_3}{4},\frac{2y_1+y_2+y_3}{4}\Big)\\ &\cdots\vdots\cdots\\ S:=&\Big(\frac{x_2+\frac{x_1+x_3}{2}}{2},\frac{y_2+\frac{y_1+y_3}{2}}{2}\Big)\\ =&\Big(\frac{x_1+2x_2+x_3}{4},\frac{y_1+2y_2+y_3}{4}\Big)\\ &\cdots\vdots\cdots\\ T:=&\Big(\frac{x_3+\frac{x_1+x_2}{2}}{2},\frac{y_3+\frac{y_1+y_2}{2}}{2}\Big)\\ =&\Big(\frac{x_1+x_2+2x_3}{4},\frac{y_1+y_2+2y_3}{4}\Big)\\ \end{split} \end{equation}$

Let those six click. Let them stick.

Parallelism, collinearity, and similarity

Example 2

The midpoint of a median which originates from a given vertex and the midpoints of the two other sides that originate from the same vertex are collinear. (With reference to our diagram, points $L,R,N$ are collinear, so are $L,S,M$ and $M,T,N$ .)

Let’s prove that $L,R,N$ are collinear; the others are similar. From Example 1 above we already have the coordinates of these points, so we now use them to find slopes.

$\begin{equation*} \begin{split} \textrm{\textbf{slope of LR}}&=\frac{\frac{2y_1+y_2+y_3}{4}-\frac{y_1+y_2}{2}}{\frac{2x_1+x_2+x_3}{4}-\frac{x_1+x_2}{2}}\\ &=\frac{\frac{2y_1+y_2+y_3}{4}-\frac{2y_1+2y_2}{4}}{\frac{2x_1+x_2+x_3}{4}-\frac{2x_1+2x_2}{2}}\\ &=\frac{\frac{2y_1-2y_1+y_2-2y_2+y_3}{4}}{\frac{2x_1-2x_1+x_2-2x_2+x_3}{4}}\\ &=\frac{y_3-y_2}{x_3-x_2}\\ &\cdots\vdots\cdots\\ \textrm{\textbf{slope of RN}}&=\frac{\frac{y_1+y_3}{2}-\frac{2y_1+y_2+y_3}{4}}{\frac{x_1+x_3}{2}-\frac{2x_1+x_2+x_3}{4}}\\ &=\frac{\frac{2y_1+2y_3}{4}-\frac{2y_1+y_2+y_3}{4}}{\frac{2x_1+2x_3}{4}-\frac{2x_1+x_2+x_3}{4}}\\ &=\frac{\frac{2y_1-2y_1-y_2-y_3+2y_3}{4}}{\frac{2x_1-2x_1-x_2-x_3+2x_3}{4}}\\ &=\frac{y_3-y_2}{x_3-x_2} \end{split} \end{equation}$

Since $LR$ and $RN$ have the same slope, it follows that $L,R,N$ are on the same line.

Example 3

The midpoint of a median which originates from a given vertex, and the midpoint of the midpoints of the two other sides that originate from the same vertex, coincide. (With reference to our diagram, the midpoint of median $AM$ and the midpoint of line segment $LN$ coincide. Similarly, the midpoint of median $BN$ and the midpoint of line segment $LM$ coincide; the midpoint of median $CL$ and the midpoint of line segment $MN$ also coincide.)

We’ve encountered this before; see Example 10 in our post on median of a triangle.

Example 4

The triangle formed by joining the midpoints of medians is similar to the original triangle. (With reference to our diagram, $\triangle RST$ is similar to $\triangle ABC$ .)

Using the coordinates obtained in Example 1 above, we simply calculate the lengths of the sides of $\triangle RST$ :

$\begin{equation*} \begin{split} RS&=\frac{1}{4}\sqrt{(x_1-x_2)^2+(y_1-y_2))^2}\\ &=\frac{1}{4}AB\\ &\cdots\vdots\cdots\\ ST&=\frac{1}{4}\sqrt{(x_3-x_2)^2+(y_3-y_2))^2}\\ &=\frac{1}{4}BC\\ &\cdots\vdots\cdots\\ TR&=\frac{1}{4}\sqrt{(x_1-x_3)^2+(y_1-y_3))^2}\\ &=\frac{1}{4}CA \end{split} \end{equation}$

Since the lengths are proportional, the two triangles are similar.

Example 5

The triangle formed by joining the “feet” of the three medians is similar to the original triangle. (With reference to our diagram, $\triangle LMN$ is similar to $\triangle ABC$ .)

Like in the previous example, we simply calculate the lengths of the sides of $\triangle LMN$ and compare them with those of $\triangle ABC$ :

$\begin{equation*} \begin{split} LM&=\frac{1}{2}\sqrt{(x_1-x_3)^2+(y_1-y_3)^2}\\ &=\frac{1}{2}AC\\ &\cdots\vdots\cdots\\ MN&=\frac{1}{2}\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\ &=\frac{1}{2}AB\\ &\cdots\vdots\cdots\\ NL&=\frac{1}{2}\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}\\ &=\frac{1}{2}BC \end{split} \end{equation}$

Since the side lengths are proportional, the two triangles are similar.

Example 6

The triangle formed by joining the “feet” of the three medians is similar to the triangle formed by joining the midpoints of the three medians. (With reference to our diagram, $\triangle LMN$ is similar to $\triangle RST$ .)

Based on the calculations in Example 4 and Example 5, we saw that:

$\begin{equation*} \begin{split} RS&=\frac{1}{4}AB\\ MN&=\frac{1}{2}AB\\ \implies RS&=\frac{1}{2}MN\\ \cdots\vdots\cdots\\ ST&=\frac{1}{4}BC\\ NL&=\frac{1}{2}BC\\ \implies ST&=\frac{1}{2}NL\\ \cdots\vdots\cdots\\ TR&=\frac{1}{4}CA\\ LM&=\frac{1}{2}CA\\ \implies TR&=\frac{1}{2}LM \end{split} \end{equation}$

Since the side lengths of $\triangle LMN$ and $\triangle RST$ are proportional, the two triangles are similar.

Example 7

The line joining the midpoints of two medians is parallel to the third side and equal to one-fourth of its length.

Consider medians $AM$ and $BN$ in our “fine-tistic” diagram. The midpoints of these medians are $R$ and $S$ , respectively. We show that line $RS$ is parallel to side $AB$ . This immediately follows by calculating slopes:

$\begin{equation*} \begin{split} \textrm{\textbf{slope of RS}}&=\frac{\frac{y_1+2y_2+y_3}{4}-\frac{2y_1+y_2+y_3}{4}}{\frac{x_1+2x_2+x_3}{4}-\frac{2x_1+2x_2+x_3}{4}}\\ &=\frac{y_2-y_1}{x_2-x_1}\\ &\cdots\vdots\cdots\\ \textrm{\textbf{slope of AB}}&=\frac{y_2-y_1}{x_2-x_1} \end{split} \end{equation}$

Since $RS$ and $AB$ have the same slopes, they are parallel. Furthermore, in Example 4 we saw that the length of $RS$ is a quarter of the length of $AB$ , so this concludes the proof.

Example 8

The line joining the midpoints of two sides of a triangle is parallel to the third side and equal to one-half of its length.

This is more familiar.

Preservation of the centroid

Example 9

The triangle formed by joining the midpoints of the three medians has the same centroid as the original triangle. (With reference to our diagram, $\triangle RST$ has the same centroid as $\triangle ABC$ .)

In view of Example 3, note that this result is expected. But let’s prove it using the centroid formula. Since the coordinates of $\triangle ABC$ are $(x_1,y_1), (x_2,y_2),(x_3,y_3)$ , we know that the centroid of $\triangle ABC$ is located at the point

$\Big(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\Big).$

Next, the coordinates of points $R,S,T$ were calculated in Example 1; they are:

$\begin{equation*} \begin{split} R=&\Big(\frac{2x_1+x_2+x_3}{4},\frac{2y_1+y_2+y_3}{4}\Big)\\ &\cdots\vdots\cdots\\ S=&\Big(\frac{x_1+2x_2+x_3}{4},\frac{y_1+2y_2+y_3}{4}\Big)\\ &\cdots\vdots\cdots\\ T=&\Big(\frac{x_1+x_2+2x_3}{4},\frac{y_1+y_2+2y_3}{4}\Big)\\ \end{split} \end{equation}$

To find the centroid of $\triangle RST$ , we add up the $x$ -coordinates and divide by $3$ , then repeat the same thing for the $y$ -coordinates. Adding the $x$ -coordinates gives

$\begin{equation*} \begin{split} &=\frac{2x_1+x_2+x_3}{4}+\frac{x_1+2x_2+x_3}{4}+\frac{x_1+x_2+2x_3}{4}\\ &=\frac{4x_1+4x_2+4x_3}{4}\\ &=x_1+x_2+x_3 \end{split} \end{equation}$

Similarly, adding the $y$ -coordinates gives $y_1+y_2+y_3$ . Finally we divide each of these sums by $3$ and then it follows that the centroid of $\triangle RST$ is the same as that of $\triangle ABC$ .

Example 10

The triangle formed by joining the “feet” of the three medians has the same centroid as the original triangle. (With reference to our diagram, $\triangle LMN$ has the same centroid as $\triangle ABC$ .)

Just like we did in Example 9 above, we use the coordinates of points $L,M,N$ to find the centroid. Recall, from Example 1, that the coordinates of $L,M,N$ are:

Adding the $x$ -coordinates, we obtain

$\begin{equation*} \begin{split} &=\frac{x_1+x_2}{2}+\frac{x_2+x_3}{2}+\frac{x_1+x_3}{2}\\ &=\frac{2x_1+2x_2+2x_3}{2}\\ &=x_1+x_2+x_3 \end{split} \end{equation}$

Adding the $y$ -coordinates also gives $y_1+y_2+y_3$ . Therefore, the centroid of $\triangle LMN$ is located at $\Big(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\Big)$ , same point as that of $\triangle ABC.$

Takeaway

The triangle obtained by joining the midpoints of the sides of a triangle is usually called the medial triangle. The process of constructing the medial triangle can be repeated until a point that approximates the centroid (of the original triangle) is reached.

In passing, by appearing somewhat playful — especially by “touting” our “artistic” diagram — we’ve exhibited a trait that can be traced to our trade.

Tasks

PROVE that the triangle formed by joining the midpoints of the sides of an isosceles triangle is also an isosceles triangle.
[Same for an equilateral triangle.]
Let $\triangle ABC$ be such that $\angle C=90^{\circ}$ . Prove that the length of the median from vertex $C$ is half of the length of the hypotenuse $AB$ .
Let $\triangle ABC$ be such that $\angle C=90^{\circ}$ . Prove that the distance from the centroid to vertex $C$ is one-third of the length of the hypotenuse $AB$ .
[You’ll love right triangles. They can be so nice $\cdots$ . See Exercise 10 below.]
[A little “calculus”] Let $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ be the vertices of $\triangle ABC$ . Define $M_1$ as the midpoint of the median from $A$ , namely $M_1=\Big(\frac{2x_1+x_2+x_3}{4},\frac{2y_1+y_2+y_3}{4}\Big)$ . Define $M_2$ as the midpoint of $A$ and $M_1$ . Define $M_3$ to be the midpoint of $A$ and $M_2$ , and so on. PROVE that
$M_n=\Big(\frac{(2^{n+1}-2)x_1+x_2+x_3}{2^{n+1}},\frac{(2^{n+1}-2)y_1+y_2+y_3}{2^{n+1}}\Big),\quad n\geq 0.$

[Observe that if $n=0$ , then we obtain the midpoint of side $BC$ , as expected. In the limit $n\rightarrow\infty$ , we obtain $(x_1,y_1)$ , the coordinates of $A$ . Beautiful, simple calculus.]
Let $\triangle LMN$ be the triangle obtained by joining the midpoints of the sides of $\triangle ABC$ . PROVE that the area of $\triangle LMN$ is a quarter of the area of $\triangle ABC$ .
Let $\triangle RST$ be the triangle obtained by joining the midpoints of the three medians of $\triangle ABC$ . PROVE that the area of $\triangle RST$ is $\frac{1}{16}$ the area of $\triangle ABC$ .
Find a triangle $ABC$ that satisfies the condition that the product of the slopes of the three sides is equal to $1$ , and the product of the slopes of the three medians is also equal to $1$ .
[An example is $\triangle ABC$ with coordinates at $A(3,5),B(1,1),C(7,-1)$ . Find a different example.]
Find a triangle in which the product of the slopes of the three sides is $1$ , while the product of the slopes of the three medians is $-1$ .
PROVE that a triangle is isosceles if, and only if, two of its medians are equal in length. [Similarly, a triangle being equilateral is equivalent to having three medians that are equal in length.]
PROVE that the triangle formed by joining the midpoints of the sides of a right triangle is also a right triangle. Moreover, the $90^{\circ}$ angle of the new triangle is on the hypotenuse of the original triangle. [Also, PROVE that the triangle formed by joining the midpoints of the three medians of a right triangle is also a right triangle, and that its $90^{\circ}$ angle is on the hypotenuse of the right triangle obtained by joining the midpoints of the original, parent right triangle.] A picture is worth a thousand words. And then a good picture??? In order to provide a visual aid to this exercise, below is a “mouth-watering” diagram, simultaneously serving as a befitting addition — and conclusion — to the “eye-catching” one we had at the beginning of the post:

You’re welcome.