Vertical and horizontal medians

In conjunction with our newly found companion, whose name is Analytic Geometry, we wish you a happy new year!

Right now as we write, we’re situated right at the coordinates of the point where our interest intersects with this fine, familiar, friendly, fascinating companion.

And, today’s post stems from a task we asked you to do in the past. Our own solution sample yielded this post’s title, together with what we’ve assembled in the accompanying examples.

You’ll love this!!!

Example 1

Let x_1,~x_2,~x_3 be the x-coordinates of the vertices of a triangle. PROVE that the x-coordinate of the centroid is x_1 if, and only if, the sequence x_2,~x_1,~x_3 is arithmetic.

Trivial fact presented in a technical way. Note that the result can be easily adapted for the y-coordinates.

To prove this, recall that the x-coordinate of the centroid is the “average” of the x-coordinates of the vertices of a triangle. In the present case we have

    \begin{equation*} \begin{split} \frac{x_1+x_2+x_3}{3}&=x_1\\ x_1+x_2+x_3&=3x_1\\ x_2+x_3&=2x_1 \end{split} \end{equation}

and so the sequence x_2,~x_1,~x_3 is arithmetic. Conversely, if the sequence x_2,~x_1,~x_3 is arithmetic, then x_1 is the arithmetic mean of x_2 and x_3; that is, x_1=\frac{x_2+x_3}{2}, which in turn gives 2x_1=x_2+x_3. Add x_1 to both sides: 3x_1=x_1+x_2+x_3, then divide both sides by 3:

    \[\frac{3x_1}{3}=\frac{x_1+x_2+x_3}{3}\]

and so the x-coordinate of the centroid is x_1.

The next result can be combined with the above one to form three equivalent statements, but we’ve separated it for emphasis.

Example 2

Let x_1,~x_2,~x_3 be the x-coordinates of the vertices of a triangle. PROVE that the x-coordinate of the centroid is x_1 if, and only if, the median through the vertex containing x_1 is vertical.

First suppose that the x-coordinate of the centroid is x_1. From Example 1 above we saw that this means the sequence x_2,x_1,x_3 is arithmetic, from which x_1=\frac{x_2+x_3}{2}. Consider the diagram below:

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from which it can be seen that the median through A(x_1,y_1) goes through (\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}). Since x_1=\frac{x_2+x_3}{2}, the median actually goes through A(x_1,y_1) and (x_1,\frac{y_2+y_3}{2}) , and so its equation is x=x_1, a vertical line.

Conversely, if the equation of the median through vertex A(x_1,y_1) is x=x_1, then the coordinates of the midpoint of the opposite side BC must satisfy this equation, giving \frac{x_2+x_3}{2}=x_1, which can be manipulated to obtain

    \[\frac{x_1+x_2+x_3}{3}=x_1,\]

and so the x-coordinate of the centroid is x_1.

Example 3

Let y_1,~y_2,~y_3 be the y-coordinates of the vertices of a triangle. PROVE that the y-coordinate of the centroid is y_2 if, and only if, the median through the vertex containing y_2 is horizontal.

Since this is similar to Example 2, we omit the proof. Note that we used y_2 instead of y_1 here because of the next result.

Example 4

Let A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3) be the vertices of \triangle ABC. If the centroid is (x_1,y_2), PROVE that the slopes of the sides follow a geometric progression whose common ratio is -2.

Beautiful, colourful, simpleful, \cdots, whateverful that’s useful.

The slopes of the sides (AB,BC,CA), as seen from the diagram below

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are:

    \[\frac{y_1-y_2}{x_1-x_2},\quad \frac{y_2-y_3}{x_2-x_3},\quad \frac{y_3-y_1}{x_3-x_1}\]

respectively, and we assume that these expressions are well-behaved. We prove that the enumeration

(1)   \begin{equation*} \frac{y_2-y_3}{x_2-x_3},\quad \frac{y_1-y_2}{x_1-x_2},\quad \frac{y_3-y_1}{x_3-x_1} \end{equation*}

is a geometric sequence and that its common ratio is -2.

Since the x-coordinate of the centroid is x_1, we have, by Example 1, that

(2)   \begin{equation*} x_1=\frac{x_2+x_3}{2}\quad\quad\Longrightarrow x_3=2x_1-x_2 \end{equation*}

Similarly, since the y-coordinate of the centroid is y_2, we have that

(3)   \begin{equation*} y_2=\frac{y_1+y_3}{2}\quad\quad\Longrightarrow y_1=2y_2-y_3 \end{equation*}

Consider the second term of the enumeration in (1), namely \frac{y_1-y_2}{x_1-x_2}. We have (by (2) and (3)) that:

    \begin{equation*} \begin{split} \frac{y_1-y_2}{x_1-x_2}&=\frac{(2y_2-y_3)-y_2}{\Big(\frac{x_2+x_3}{2}\Big)-x_2}\\ &=\frac{y_2-y_3}{\frac{x_2+x_3-2x_2}{2}}\\ &=\frac{y_2-y_3}{\frac{-x_2+x_3}{2}}\\ &=\frac{y_2-y_3}{\frac{-(x_2-x_3)}{2}}\\ &=-2\Big(\frac{y_2-y_3}{x_2-x_3}\Big) \end{split} \end{equation}

and so the second term in (1) is -2 times the first. Now consider the third term, namely \frac{y_3-y_1}{x_3-x_1}. Again, by (2) and (3), we have that:

    \begin{equation*} \begin{split} \frac{y_3-y_1}{x_3-x_1}&=\frac{(2y_2-y_1)-y_1}{(2x_1-x_2)-x_1}\\ &=\frac{2y_2-2y_1}{x_1-x_2}\\ &=-2\Big(\frac{y_1-y_2}{x_1-x_2}\Big) \end{split} \end{equation}

and so the third term is -2 times the second. Therefore, the enumeration given by (1) is indeed a geometric sequence with a common ratio of -2.

The VHM property in triangles

By the VHM property, we mean that a triangle possesses a vertical median as well as a horizontal median (view the VHM property as a special case of two perpendicular medians).

Example 5

Give an example of a triangle that contains a vertical median and a horizontal median.

Easy. Very easy — as 1,2,3.

Indeed, the vertices of our triangle will be constructed from the numbers 1,2,3. Take \triangle PQR with vertices (1,2),~(2,1),~(3,3) shown below:

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Notice the horizontal median PB and the vertical median QA. Also, the slope of PQ is -1; the slope of QR is 2; the slope of RP is \frac{1}{2}. Re-arranging, we have the geometric sequence of slopes:

    \[\frac{1}{2},~-1,~2;\]

The common ratio is -2 as you’ve seen. Also worth noting is that the above triangle is isosceles; in the next example, we show that not all VHM triangles are isosceles.

Example 6

Give an example of a scalene triangle that contains a vertical median and a horizontal median.

Consider \triangle PQR shown below:

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PQ=\sqrt{41},~QR=\sqrt{29},~RP=\sqrt{104}=2\sqrt{26}, so this triangle is scalene. Its centroid is located at (3,1), the median RB is vertical, while the median QA is horizontal. Moreover, the slopes of sides PQ,QR,RP are \frac{5}{4},~-\frac{5}{2},~5, respectively. They form a geometric sequence whose common ratio is -2, as you know.

Example 7

Find a general set of coordinates for the vertices of a triangle that satisfies the VHM property.

Let a,~b,~\alpha, and \beta be real numbers. Set

    \[A=(a,\alpha),\quad B=(\beta, b),\quad C=(2a-\beta, 2b-\alpha).\]

Then \triangle ABC satisfies the VHM property. Indeed, its centroid is located at

    \[\Big(\frac{a+\beta+2a-\beta}{3},\frac{\alpha+b+2b-\alpha}{3} \Big)=(a,b),\]

and so it shares an x-coordinate with vertex A and a y-coordinate with vertex B. By Example 2 and Example 3, it contains a vertical median and a horizontal median, which is the VHM property.

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Notice that the slopes of sides BC,~AB,~CA are

    \[\frac{b-\alpha}{2(a-\beta)}, \quad-\frac{b-\alpha}{a-\beta},\quad\frac{2(b-\alpha)}{a-\beta} ,\]

respectively. These slopes form a geometric progression with a common ratio of -2, as expected. Note that it is important to impose the restrictions a\neq \beta and b\neq \alpha.

Example 8

PROVE that if a triangle contains a vertical median and a horizontal median, then the product of the slopes of the sides is a “perfect cube”.

The slopes of the sides of a VHM triangle were calculated in Example 7; their product is

    \[\frac{b-\alpha}{2(a-\beta)}~\times -\frac{b-\alpha}{a-\beta}~\times\frac{2(b-\alpha)}{a-\beta} =\Big(-\frac{b-\alpha}{a-\beta}\Big)^{3},\]

which is a “perfect cube”. (We’ve used the term “perfect cube” in a loose way here.)

A VHM triangle contains a vertical median (undefined slope) and a horizontal median (zero slope). The remaining “non-degenerate” median has a slope that’s related to the slope of the side it meets.

Example 9

PROVE that the slope of the “non-degenerate” median in a VHM triangle is the negative of the slope of the side that contains its foot.

Consider the “non-degenerate” median (dashed blue line) in the diagram below:

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The slope of side AB is \frac{b-\alpha}{\beta-a}. Let’s calculate the slope of median CL:

    \begin{equation*} \begin{split} \textrm{\textbf{slope of CL}}&=\frac{2b-\alpha-\Big(\frac{b+\alpha}{2}\Big)}{2a-\beta-\Big(\frac{a+\beta}{2}\Big)}\\ &=\frac{4b-2\alpha-\alpha-b}{4a-2\beta-a-\beta}\\ &=\frac{3b-3\alpha}{3a-3\beta}\\ &=\frac{b-\alpha}{a-\beta}\\ &=-\Big(\frac{b-\alpha}{\beta-a}\Big)\\ &=\textrm{\textbf{negative of the slope of side AB.}} \end{split} \end{equation}

We’ve saved the last for last (you read that correctly), which has to do with the converse of what we saw in Example 4.

Example 10

PROVE that if the slopes of the sides of a triangle form a geometric progression with a common ratio of -2, then the triangle contains a vertical median and a horizontal median.

Beautiful, colourful, \cdots, whateverful.

Let A(a,b), ~B(c,d),~C(e,f) be the vertices of a triangle whose sides slopes form a geometric progression with a common ratio of -2. Enumerate this geometric progression as

    \[r,~-2r,~4r\]

and let the terms correspond, respectively, to the slopes of sides AB,BC,CA. Then we have the following linear system:

(4)   \begin{equation*} b-d=r(a-c),\quad \implies ra-b-rc+d=0 \end{equation*}

(5)   \begin{equation*} f-d=-2r(e-c),\quad\implies (2r)c+d-(2r)e-f=0 \end{equation*}

(6)   \begin{equation*} f-b=4r(e-a),\quad\implies -(4r)a+b+(4r)e-f=0 \end{equation*}

This is a homogeneous linear system that is underdetermined (it contains fewer equations than unknowns). By a result in linear algebra, such a system always has non-trivial solutions. In fact, our solution is:

    \[a=\Big(\frac{1}{2r}\Big)d+e-\Big(\frac{1}{2r}\Big)f,~ b=2d-f, ~c=-\Big(\frac{1}{2r})d+e+(\frac{1}{2r}\Big)f.\]

Note that a,b,c have been expressed in terms of d,e,f because the system given by (4), (5),(6) has three degrees of freedom. Don’t worry about these technical terms. The centroid of \triangle ABC is

    \[\Big(\frac{a+c+e}{3},\frac{b+d+f}{3}\Big)=(e,d),\]

after simplification. Since the centroid shares an x-coordinate with vertex C and a y-coordinate with vertex B, it contains a vertical median and a horizontal median.

It follows that a triangle satisfies the VHM property if, and only if, the slopes of its sides form a geometric progression whose common ratio is -2.

Takeaway

The centroid of a triangle is properly located within the triangle (unlike the circumcenter or orthocenter that are sometimes situated outside). The centroid hardly “shares” its coordinates with any of the triangle’s vertices, but when it does decide to share, then it “straightens” one median and “flattens” the other.

Tasks

  1. Let x_1,x_2,x_3 be an arithmetic sequence, and let y_1,y_2,y_3 be another arithmetic sequence. PROVE that the points (x_1,y_1),(x_2,y_2),(x_3,y_3) lie on a straight line.
    \Big(This shows that, in order to form a triangle, the order in which we select coordinates from the sequences matter.\Big)
  2. PROVE that an equilateral triangle can never contain a vertical median and a horizontal median at the same time.
  3. PROVE that if a right triangle is to contain both a vertical median and a horizontal median simultaneously, then the slopes of its sides have to be \frac{\sqrt{2}}{2},~-\sqrt{2},~2\sqrt{2}, or -\frac{\sqrt{2}}{2},~\sqrt{2},~-2\sqrt{2}.
    (Further, the medians to the two legs cannot be vertical and horizontal at the same time.)
  4. Suppose that \triangle ABC contains a vertical median and a horizontal median. PROVE that if one of its sides has a slope of \pm 1, then \triangle ABC is isosceles.
  5. Suppose that \triangle ABC contains a vertical and a horizontal median. PROVE that the sum of the slopes of its sides is \frac{3}{2} times the slope which is the geometric mean of the other two slopes.
    \Big( For such a triangle, the slopes of the sides form a geometric sequence with a common ratio of -2, so it makes sense to talk about the “geometric mean” of the slopes. \Big)
  6. PROVE that if a triangle contains a vertical and a horizontal median, then the sum of product of the slopes of the sides, taken two at a time, is always negative.
  7. The three medians of a triangle divide the triangle into six smaller triangles of equal areas. PROVE that if the original triangle satisfies the VHM property, then two out of the six smaller triangles are right triangles. If, in addition, the original triangle is isosceles and satisfies the VHM property, PROVE that four out of the six smaller triangles are right triangles.
  8. Let A(a,\alpha),~ B(\beta, b), ~C(2a-\beta, 2b-\alpha) be the coordinates of \triangle ABC. PROVE AC^2+CB^2=5AB^2.
    \Big( This triangle contains a vertical median and a horizontal median, so the above equation is an instance of a well-known result; see this article. \Big)
  9. Let A(a,\alpha),~ B(\beta, b), ~C(2a-\beta, 2b-\alpha) be the coordinates of \triangle ABC. PROVE that its area is \frac{3(\alpha-b)(\beta-a)}{2}.
    \Big( Note that this triangle contains a vertical median and a horizontal median, and so its area takes a simple form. \Big)
  10. Let A(a,\alpha),~ B(\beta, b), ~C(2a-\beta, 2b-\alpha) be the coordinates of \triangle ABC. PROVE that the slope of its Euler line can be given by \Big(\frac{a-\beta}{b-\alpha}\Big)\Big(\frac{2(a-\beta)^2-(b-\alpha)^2}{2(b-\alpha)^2-(a-\beta)^2}\Big)

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