Slopefessor Fr1day – Page 36 – High School Geometry

Arithmetic and geometric slopes

Two posts ago, we promised to discuss the general theory behind triangles whose sides have slopes of $1,2,3$ — and here we go!!! We also considered, in the previous post, triangles whose sides have slopes of the form $k,-2k,4k$ . If you loved those two, and you read this one through, you’ll love this one too. Or even “befriend” it as we do.

That’s true.

Slopes in arithmetic progression

Example 1

Find coordinates for the vertices of a triangle whose sides have slopes of $a,~a+d,~a+2d$ .

Let $\triangle ABC$ have vertices as shown below:

Since the slopes of sides $AB,~BC,~CA$ are $a,~a+d,~a+2d$ , respectively, we have the following linear equations:

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

which we can recast into a “homogeneous” system:

(1) $\begin{equation*} -ax_1+y_1+ax_2-y_2=0 \end{equation*}$

(2) $\begin{equation*} (-a-2d)x_1+y_1+(a+2d)x_3-y_3=0 \end{equation*}$

(3) $\begin{equation*} (-a-d)x_2+y_2+(a+d)x_3-y_3=0 \end{equation*}$

Notice that there are six unknowns in three equations, so the system is underdetermined. Being also homogeneous, it has an infinite number of solutions, as our friend from L.A. (a.k.a Linear Algebra, not Los Angeles), just informed us. By row reduction, our system is found to have a rank of $3$ as well as three degrees of freedom — the latter meaning that the solution can be expressed in terms of three parameters (or “free” variables). In fact, we have:

$\begin{equation*} \begin{split} x_1&=\frac{1}{2(a+d)}y_2+x_3-\frac{1}{2(a+d)}y_3\\ y_1&=\frac{a+2d}{2(a+d)}y_2+\frac{a}{2(a+d)}y_3\\ x_2&=\frac{1}{a+d}y_2+x_3-\frac{1}{a+d}y_3 \end{split} \end{equation}$

where $y_2,x_3,y_3$ are our three “free” variables. Observe that the coefficients of these variables add up to one in the above equations, so in a sense they’re weighted.

To sum up, the required coordinates are:

$\begin{equation*} \begin{split} (x_1,y_1)&\mapsto\Big(\frac{1}{2(a+d)}y_2+x_3-\frac{1}{2(a+d)}y_3,~\frac{a+2d}{2(a+d)}y_2+\frac{a}{2(a+d)}y_3\Big)\\ (x_2,y_2)&\mapsto \Big(\frac{1}{a+d}y_2+x_3-\frac{1}{a+d}y_3,\quad y_2\Big)\\ (x_3,y_3)&\mapsto(x_3,y_3) \end{split} \end{equation}$

As you may have thought, the formula breaks down when $a+d=0$ , but this leads to an important special case that deserves a separate treatment.

Example 2

In the notation of our first diagram, PROVE that $a+d=0$ if, and only if, $y_3-y_2=0$ .

This is immediate; just another way of saying that the slope of side $BC$ is zero. However, if we want to be fancy, we can use equation (3). First assume that $a+d=0$ . Then (3) reduces to $y_3-y_2=0$ . Conversely, if $y_3-y_2=0$ , then we’re left with

$(-a-d)x_2+(a+d)x_3=0\implies (a+d)(x_3-x_2)=0,$

which in turn leaves us with two choices: $a+d=0$ or $x_3-x_2=0$ . However, $x_3-x_2=0$ is impossible because otherwise it would mean that $x_3=x_2$ . Together with the fact that $y_3=y_2$ (by the assumption in the converse), the points $(x_2,y_2)$ and $(x_3,y_3)$ will coincide, meaning that we don’t have a triangle. So we choose $a+d=0$ and this completes the proof.

As a by-product, we also obtain that $a+d\neq 0$ is equivalent to $y_3-y_2\neq 0$ . We’ll have more to say about this case later.

Example 3

Find the $x$ -coordinate of the centroid of a triangle whose sides have slopes of $a,~a+d,~a+2d$ .

The $x$ -coordinate of the centroid is given by $\frac{x_1+x_2+x_3}{3}$ . Using the coordinates we found in Example 1, we have:

$\begin{equation*} \begin{split} &=\frac{\Big(\frac{1}{2(a+d)}y_2+x_3-\frac{1}{2(a+d)}y_3\Big)+\Big(\frac{1}{a+d}y_2+x_3-\frac{1}{a+d}y_3\Big)+x_3}{3}\\ &=\frac{\Big(\frac{3}{2(a+d)}y_2+3x_3-\frac{3}{2(a+d)}y_3\Big)}{3}\\ &=\frac{1}{2(a+d)}y_2+x_3-\frac{1}{2(a+d)}y_3\\ &=x_1 \end{split} \end{equation}$

Example 4

PROVE that if the slopes of the sides of a triangle follow an arithmetic progression, then the triangle contains one vertical median.

Let $\triangle ABC$ have vertices $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$ and let sides $AB,BC,CA$ have slopes of $a,~a+d,~a+2d$ , respectively. In Example 3, we saw that the $x$ -coordinate of the centroid is precisely $x_1$ . Since the centroid shares an $x$ -coordinate with a triangle’s vertex (in this case $A(x_1,y_1)$ ), we know — from our previous post — that the median through that vertex is vertical.

Example 5

PROVE that if the slopes of the sides of a triangle follow an arithmetic progression $a,~a+d,~a+2d$ , then so do the x-coordinates.

Note that the converse is not true.

To prove the above statement, we can use a result from our previous post, but let’s calculate it directly here. Enumerate the $x$ -coordinates as follows:

$\begin{equation*} \begin{split} x_3&=x_3\\ &\cdots\vdots\cdots\\ x_1&=\frac{1}{2(a+d)}y_2+x_3-\frac{1}{2(a+d)}y_3\\ &=x_3+\frac{1}{2(a+d)}\Big(y_2-y_3\Big)\\ &\cdots\vdots\cdots\\ x_2&=\frac{1}{a+d}y_2+x_3-\frac{1}{a+d}y_3\\ &=x_3+\frac{1}{a+d}\Big(y_2-y_3\Big)\\ &=x_3+\frac{1}{2(a+d)}\Big(y_2-y_3\Big)+\frac{1}{2(a+d)}\Big(y_2-y_3\Big)\\ &=x_1+\frac{1}{2(a+d)}\Big(y_2-y_3\Big). \end{split} \end{equation}$

So the $x$ -coordinates differ consecutively by $\frac{1}{2(a+d)}\Big(y_2-y_3\Big)$ .

The fact that the $x$ -coordinates follow an arithmetic progression whereas the $y$ -coordinates do not (necessarily) follow an arithmetic progression makes the resulting triangle diagram quite weird, which was what we called “ugly-looking” two posts ago.

Slopes in geometric progression

Example 6

Find coordinates for the vertices of a triangle whose sides have slopes of $k,~kr,~kr^2$ .

Of course, we want $k\neq 0$ and $r\neq\pm 1$ . If there be any other restriction, it will become obvious when we’re done with the computations.

To find the coordinates, suppose we work with $\triangle ABC$ whose vertices are as shown below:

As before, it boils down to solving a linear system:

$\begin{equation*} \begin{split} y_2-y_1&=k(x_2-x_1)\\ y_3-y_2&=kr(x_3-x_2)\\ y_3-y_1&=kr^2(x_3-x_1) \end{split} \end{equation}$

where, after row reduction and back-substitution, the solution is:

$\begin{equation*} \begin{split} x_1&=\frac{1}{kr(1+r)}y_2+x_3-\frac{1}{kr(1+r)}y_3\\ y_1&=\frac{r}{1+r}y_2+\frac{1}{1+r}y_3\\ x_2&=\frac{1}{kr}y_2+x_3-\frac{1}{kr}y_3 \end{split} \end{equation}$

In each of the above expressions, the coefficients of the “free variables” $y_2,x_3,y_3$ add up to one, so in a sense they’re weighted.

Example 7

Find the centroid of the triangle with vertices given in Example 6 above.

The $x$ -coordinate of the centroid is $\frac{x_1+x_2+x_3}{3}$ :

$\begin{equation*} \begin{split} &=\frac{\Big(\frac{1}{kr(1+r)}y_2+x_3-\frac{1}{kr(1+r)}y_3\Big)+\Big(\frac{1}{kr}y_2+x_3-\frac{1}{kr}y_3\Big)+x_3}{3}\\ &=\frac{\frac{2+r}{kr(1+r)}y_2-\frac{2+r}{kr(1+r)}y_3+3x_3}{3}\\ &=\frac{(2+r)(y_2-y_3)}{3kr(1+r)}+x_3 \end{split} \end{equation}$

If $r=-2$ , then the $x$ -coordinate coincides with $x_3$ . (Note that we can’t have $y_2=y_3$ since otherwise one of the slopes will be zero, and we’re working with slopes in geometric progression.)

For the $y$ -coordinate of the centroid, we use $\frac{y_1+y_2+y_3}{3}$ :

$\begin{equation*} \begin{split} &=\frac{\Big(\frac{r}{1+r}y_2+\frac{1}{1+r}y_3\Big)+y_2+y_3}{3}\\ &=\frac{\frac{1+2r}{1+r}y_2+\frac{2+r}{1+r}y_3}{3}\\ &=\frac{1+2r}{3(1+r)}y_2+\frac{2+r}{3(1+r)}y_3 \end{split} \end{equation}$

If $r=-2$ , the $y$ -coordinate of the centroid coincides with $y_2$ .

COOL.

Example 8

Let $\triangle ABC$ be such that sides $AB,BC,CA$ have slopes $k,kr,kr^2$ , respectively. PROVE that the slope of the median from vertex $A$ is $-kr$ .

The fact that the slope of the median from vertex $A$ to side $BC$ and the slope of side $BC$ are negatives of each other has an important implication and application (see Example 9 below).

To prove the present statement, let’s refer to an earlier diagram:

We found that

$\begin{equation*} \begin{split} x_1&=\frac{1}{kr(1+r)}y_2+x_3-\frac{1}{kr(1+r)}y_3\\ y_1&=\frac{r}{1+r}y_2+\frac{1}{1+r}y_3\\ x_2&=\frac{1}{kr}y_2+x_3-\frac{1}{kr}y_3 \end{split} \end{equation}$

The midpoint of $BC$ is $\Big(x_3+\frac{1}{2kr}(y_2-y_3),\frac{y_2+y_3}{2}\Big)$ and so the slope of the median from vertex $A$ is:

$\begin{equation*} \begin{split} &=\frac{y_1-\Big(\frac{y_2+y_3}{2}\Big)}{x_1-\Big(x_3+\frac{1}{2kr}(y_2-y_3)\Big)}\\ &=\frac{\frac{r}{1+r}y_2+\frac{1}{1+r}y_3-\Big(\frac{y_2+y_3}{2}\Big)}{\frac{1}{kr(1+r)}y_2+x_3-\frac{1}{kr(1+r)}y_3-\Big(x_3+\frac{1}{2kr}(y_2-y_3)\Big)}\\ &=\frac{\frac{r-1}{2(1+r)}(y_2-y_3)}{\frac{1-r}{2kr(1+r)}(y_2-y_3)}\\ &=\frac{r-1}{2(1+r)}(y_2-y_3)\times \frac{2kr(1+r)}{(1-r)(y_2-y_3)}\\ &=-kr. \end{split} \end{equation}$

Example 9

Let $\triangle ABC$ be such that sides $AB,BC,CA$ have slopes $k,kr,kr^2$ , respectively. PROVE that $AB=CA$ if, and only if, $kr=\pm 1$ .

Suppose that $kr=1$ (a similar argument holds for $kr=-1$ ). By Example 8 above, the slope of the median from vertex $A$ is $-1$ . But then this means that the median from vertex $A$ is perpendicular to side $BC$ . Therefore, $AB=CA$ and $\triangle ABC$ is isosceles.

Conversely, suppose that $AB=CA$ . Then the median from vertex $A$ is perpendicular to side $BC$ , which in turn means that the product of their slopes is $-1$ . Since their slopes are $kr$ and $-kr$ , it follows that

$\begin{equation*} \begin{split} kr\times(-kr)&=-1\\ kr\times kr&=1\\ (kr)^2&=1\\ kr&=\pm 1. \end{split} \end{equation}$

So we obtain a characterization of isosceles triangles, using slopes in geometric progression.

Example 10

Let $\triangle ABC$ be a right-triangle in which sides $AB,BC,CA$ have slopes $k,kr,kr^2$ , respectively. PROVE that $r$ is necessarily negative, and that side $BC$ cannot be the hypotenuse.

Since the location of the right-angle was not specified, we have to consider all possibilities. We use the fact that the product of the slopes of two perpendicular lines is $-1$ , and take the product of two slopes at a time:

$\begin{equation*} \begin{split} k\times kr&=-1\\ k\times kr^2&=-1\\ kr\times kr^2&=-1 \end{split} \end{equation}$

From the first equation, $r=-\frac{1}{k^2}$ , which is always negative because the denominator $k^2$ is always positive whatever (the real number) $k$ is. From the third equation, we have $r^3=-\frac{1}{k^2}$ , and again $r$ is negative since it’s only the cube of a negative (real) number that can be negative.

Now from the second equation, we have $(kr)^2=-1$ . Since the left side of this equation is always positive whereas the right side is always negative, the equation doesn’t have real solution. In turn, this means that side $AB$ (with slope $k$ ) and side $CA$ (with slope $kr^2$ ) do not form a right-angle. Therefore, side $BC$ cannot be the hypotenuse.

Numerical examples

We now move from the abstract angle to actual examples.

Let and . Then the arithmetic sequence is and the corresponding coordinates for the vertices of a triangle with slopes are:
$x_1=x_3+\frac{1}{4}(y_2-y_3),~y_1=\frac{3}{4}y_2+\frac{1}{4}y_3,~x_2=x_3+\frac{1}{2}(y_2-y_3).$

$x_3,y_2,y_3$ are “free” variables. We obtain an assemblage of triangles by specifying values for $x_3,y_2,y_3$ , so long as $y_2\neq y_3$ .
- Put $x_3=0,~y_3=0$ and $y_2=4$ . Then $x_1=1,~y_1=3,~x_2=2$ . The resulting triangle has coordinates at $(1,3),(2,4),(0,0)$ :
  
  Notice the presence of a vertical median $PB$ .
- Put $x_3=-1,~y_3=-2$ and $y_2=6$ . Then $x_1=1,~y_1=4,~x_2=3$ . We obtain a triangle with coordinates at $(1,4),(3,6),(-1,-2)$ :
  
  Notice the presence of a vertical median $PB$ .
Let $a=1$ and $d=-2$ . Then the corresponding arithmetic sequence of slopes is $1,~-1,~-3$ ; moreover:
$x_1=x_3-\frac{1}{2}(y_2-y_3),~y_1=\frac{3}{2}y_2-\frac{1}{2}y_3,~x_2=x_3-(y_2-y_3).$

Depending on what we choose for the “free variables”, we obtain different sets of triangles. Let $x_3=1,~y_3=2,~y_2=3$ . Then $x_1=\frac{1}{2},~y_1=\frac{7}{2},~x_2=0$ . This results in a triangle with coordinates at $(\frac{1}{2},\frac{7}{2}),~(1,2),~(0,3)$ :

Notice the presence of a vertical median $PB$ . Also this is a right triangle.
An example of a triangle with slopes in geometric progression. Let $k=\frac{1}{2}$ and $r=-4$ ; the corresponding geometric sequence of slopes is $\frac{1}{2},~-2,~8$ . If we put $k=\frac{1}{2}$ and $r=-4$ in our formula, we obtain
$x_1=x_3+\frac{1}{6}(y_2-y_3),~y_1=\frac{4}{3}y_2-\frac{1}{3}y_3,~x_2=x_3-\frac{1}{2}(y_2-y_3).$

Next stop: choose values for the “free variables” $x_3,y_3,y_2$ , bearing in mind that $y_2\neq y_3$ . Let $y_2=6$ and $x_3=y_3=0$ . We obtain $x_1=1,~y_1=8,~x_2=-3$ and the resulting triangle has vertices located at $(1,8),(-3,6),(0,0)$ :

Notice that this is a right triangle.
Let $k=1$ and $r=2$ ; we obtain the geometric sequence of slopes $1,~2,~4$ . In turn:
$x_1=x_3+\frac{1}{6}(y_2-y_3),~y_1=\frac{2}{3}y_2+\frac{1}{3}y_3,~x_2=x_3+\frac{1}{2}(y_2-y_3).$

For convenience, let’s choose our “free variables” $x_3=0=y_3$ and $y_2=6$ . We then obtain $x_1=1,~y_1=4,~x_2=3$ and a triangle with coordinates at $(1,4),(3,6),(0,0)$ results:
Here’s an example in which some of the coordinates are irrational numbers. Let $k=\sqrt{5}$ and $r=\frac{1}{2}$ . Let’s also choose our “free variables” right away: $x_3=0=y_3$ and $y_2=3\sqrt{5}$ . We obtain $x_1=4,~y_1=\sqrt{5},~x_2=6$ and a triangle with coordinates at $(4,\sqrt{5}),(6,3\sqrt{5}),(0,0)$ results:

There’s something fascinating about this example that is worth mentioning. Observe that the given triangle doesn’t contain $90^{\circ}$ . Now the midpoints of sides $PQ,~QR,~RP$ are

$\Big(3,\frac{3}{2}\sqrt{5}\Big),~\Big(2,\frac{\sqrt{5}}{2}\Big),~\Big(5,2\sqrt{5}\Big).$

Consequently, the lengths of the medians from $R,~P,~Q$ are $\frac{3}{2},~\frac{3\sqrt{21}}{2},~3\sqrt{5}$ , respectively. How are these related?

$\begin{equation*} \begin{split} \Big(\frac{3\sqrt{21}}{2}\Big)^2&=\frac{189}{4}\\ \Big(\frac{3}{2}\Big)^2+(3\sqrt{5})^2&=\frac{9}{4}+45\\ &=\frac{9}{4}+\frac{45\times 4}{1\times 4}\\ &=\frac{9}{4}+\frac{180}{4}\\ &=\frac{189}{4}\\ &=\Big(\frac{3\sqrt{21}}{2}\Big)^2. \end{split} \end{equation}$

COOL.

Takeaway

If the slopes of the sides of a triangle are $k,kr,kr^2$ (a geometric progression), and the slopes of the sides of another triangle are $a,a+d,a+2d$ (an arithmetic progression), then properties that correspond to $kr=\pm 1$ have analogues in $a+d=0$ . These properties are mainly lateral properties — isosceles or equilateral.

We encourage you to investigate different cases of slopes in arithmetic progression and slopes in geometric progression. You’ll encounter exciting surprises.

Tasks

PROVE that if the slopes of the sides of a triangle form a geometric progression $k, kr, kr^2$ , then the slopes of its three medians also form a geometric progression (with a different common ratio), provided $r\neq -2,-\frac{1}{2}$ .
PROVE that if the slopes of the sides of a triangle form an arithmetic progression, then the slopes of its three medians do not form an arithmetic progression.
Let $k,kr,kr^2$ ( $r\neq -2,-\frac{1}{2}$ ) be the slopes of the sides of a $\triangle ABC$ . PROVE that the product of the slopes of the three medians is $-(kr)^3$ .
PROVE that if the slopes of the sides of a triangle are $1,r,r^2$ , then the triangle can never be a right triangle.
$\Big($ Of course $r\neq \pm 1$ . $\Big)$
PROVE that if the slopes of the sides of a triangle are $k,kr,kr^2$ , where $k$ and $r$ are rational numbers, then the triangle can never be equilateral.
$\Big($ Actually if either $k$ or $r$ is a rational number, same conclusion. $\Big)$
PROVE that if the slopes of the sides of a triangle are $a,a+d,a+2d$ , where $a$ and $d$ are rational numbers, then the triangle can never be equilateral.
$\Big($ It appears that equilateral triangles don’t like rational numbers? $\Big)$
Let $\triangle ABC$ be such that sides $AB,BC,CA$ have slopes $k,kr,kr^2$ , respectively. PROVE that the median from $B$ and the median from $C$ can never intersect at $90^{\circ}$ , except when $r=-2$ .
$\Big($ Any triangle whose side slopes form a geometric progression with common ratio $r=-2$ is a peach. $\Big)$
PROVE that if the $x$ -coordinates of the vertices of a triangle form an arithmetic progression, then the slopes of the sides do not necessarily form an arithmetic progression.
$\Big($ A counter-example will do. $\Big)$
If the slopes of the sides of a triangle form a geometric sequence with a common ratio of $-2$ , PROVE that both the $x$ -coordinates and the $y$ -coordinates form arithmetic progressions.
$\Big($ Cool. $\Big)$
Let $\triangle ABC$ be such that sides $AB,BC,CA$ have slopes of $k,kr,kr^2$ , respectively. PROVE that $AB^2+BC^2=CA^2$ if, and only if $k^2r=-1$ .
$\Big($ Pretend that you’re unaware of the Pythagorean theorem and perpendicular slopes. Solve the problem without recourse to these two. $\Big)$

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:

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

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:

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:

$PQ=\sqrt{41},~QR=\sqrt{29},~RP=\sqrt{104}=2\sqrt{26},$ 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.

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:

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

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. $\Big)$
PROVE that an equilateral triangle can never contain a vertical median and a horizontal median at the same time.
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.)
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.
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. $\Big)$
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.
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.
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. $\Big)$
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. $\Big)$
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 $\Big(\frac{a-\beta}{b-\alpha}\Big)\Big(\frac{2(a-\beta)^2-(b-\alpha)^2}{2(b-\alpha)^2-(a-\beta)^2}\Big)$