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:

Rendered by QuickLaTeX.com

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:

    \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}

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

  1. 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.]
  2. 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.
  3. 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.]
  4. [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.]

  5. 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.
  6. 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.
  7. 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.]
  8. 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.
  9. 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.]
  10. 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:

    Rendered by QuickLaTeX.com

    You’re welcome.

One thought on “Geometric properties involving medians”

Comments are closed.