Two propositions on geometric slopes

Basically, geometric slopes involve three numbers $m_{1},m_{2},m_{3}$ that are in geometric progression — like $1,2,4$ — but at the same time are slopes of the sides of a triangle. Note that this is not a standard terminology; we’re using it here to keep the title short.

We present two “propositions” that describe precisely when the slopes of the sides of a triangle form a geometric progression.

Notation and characterization

In $\triangle ABC$ , let’s denote the slopes of the three medians from vertices $A,B,C$ by $m_{A},~m_{B},~m_{C}$ , respectively. Similarly, the slopes of sides $AB,BC,CA$ will be denoted by $m_{AB},~m_{BC},~m_{CA}$ .

If $m_{A}+m_{BC}=0$ , PROVE that $x_{1}=\frac{x_{2}(y_1-y_2)+x_{3}(y_3-y_1)}{y_3-y_2}$ , where $\triangle ABC$ has vertices at $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ .

$m_{A}$ is the slope of the line segment joining $A(x_1,y_1)$ to the midpoint of $BC=\Big(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\Big)$ ; $m_{BC}$ is the slope of side $BC$ . So we have:

$m_{A}=\frac{\frac{y_2+y_3}{2}-y_1}{\frac{x_2+x_3}{2}-x_1}=\frac{y_2+y_3-2y_1}{x_2+x_3-2x_1};\quad m_{BC}=\frac{y_3-y_2}{x_3-x_2}.$

Now, if $m_{A}+m_{BC}=0$ , write $m_{A}=-m_{BC}$ . Clear fractions:

$(x_3-x_2)(y_2+y_3-2y_1)=-(y_3-y_2)(x_2+x_3-2x_1).$

Expand and combine like terms:

$x_3y_2+x_3y_3-2x_3y_1-x_2y_2-x_2y_3+2x_2y_1=-(x_2y_3-x_2y_2+x_3y_3-x_3y_2-2x_1y_3+2x_1y_2).$

Obtain $2x_3y_3-2x_3y_1+2x_2y_1-2x_2y_2=2x_1y_3-2x_1y_2$ , from which $x_1$ can be isolated as:

$x_1=\frac{x_2(y_1-y_2)+x_3(y_3-y_1)}{y_3-y_2}.$

Example 2 (First proposition)

For $\triangle ABC$ with vertices at $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ and non-zero side slopes, PROVE that the slopes of sides $AB,BC,CA$ form a geometric progression if, and only if, $m_{A}+m_{BC}=0$ .

In other words, once one of the medians has a slope that is the negative of the slope of the side that contains its “foot”, then the slopes of the sides of the triangle form a geometric progression.

The non-zero requirement for the slopes of the sides is crucial. See Example 3 below. Also, click here for a pdf version of some of the equations in this post.

Suppose that $m_{A}+m_{BC}=0$ ; we’ll show that the slopes of sides $AB,BC,CA$ (namely $\frac{y_1-y_2}{x_1-x_2}, \frac{y_2-y_3}{x_2-x_3}, \frac{y_3-y_1}{x_3-x_1}$ ) form a geometric progression. It suffices to show that $\frac{y_1-y_2}{x_1-x_2}\times \frac{y_3-y_1}{x_3-x_1}=\Big(\frac{y_2-y_3}{x_2-x_3}\Big)^2$ .

Since $m_{A}+m_{BC}=0$ , we can use $x_1=\frac{x_2(y_1-y_2)+x_3(y_3-y_1)}{y_3-y_2}$ from Example 1 and obtain $\frac{y_1-y_2}{x_1-x_2}\times \frac{y_3-y_1}{x_3-x_1}$ :

$\begin{equation*} \boldmath\begin{split} &=\scriptstyle\frac{(y_1-y_2)(y_3-y_1)}{\Big[\frac{x_2(y_1-y_2)+x_3(y_3-y_1)}{y_3-y_2}-x_2\Big]\Big[x_3-\Big(\frac{x_2(y_1-y_2)+x_3(y_3-y_1)}{y_3-y_2}\Big)\Big]}\\ &=\scriptscriptstyle\frac{(y_1-y_2)(y_3-y_1)(y_3-y_2)^2}{[x_2(y_1-y_2)+x_3(y_3-y_1)-x_2(y_3-y_2)][x_3(y_3-y_2)-x_2(y_1-y_2)-x_3(y_3-y_2)]}\\ &=\scriptstyle\frac{(y_1-y_2)(y_3-y_1)(y_3-y_2)^2}{[x_2(y_1-y_2-y_3+y_2)+x_3(y_3-y_1)][x_3(y_3-y_2-y_3+y_1)-x_2(y_1-y_2)]}\\ &=\frac{(y_1-y_2)(y_3-y_1)(y_3-y_2)^2}{(x_3-x_2)(y_3-y_1)(x_3-x_2)(y_1-y_2)}\\ &=\Big(\frac{y_3-y_2}{x_3-x_2}\Big)^2. \end{split} \end{equation*}$

$\therefore \frac{(y_1-y_2)(y_3-y_1)}{(x_1-x_2)(x_3-x_1)}&=\Big(\frac{y_3-y_2}{x_3-x_2}\Big)^2$ , and so the slopes $\frac{y_1-y_2}{x_1-x_2}, \frac{y_2-y_3}{x_2-x_3}, \frac{y_3-y_1}{x_3-x_1}$ form a geometric progression. The converse was proved in Example 8 of our post on January 28, 2020.

Consider the right triangle $ABC$ with vertices at $A(0,0),~B(k,0),~C(0,l)$ . PROVE that the slope of the median from vertex $A$ is the negative of the slope of the hypotenuse $BC$ .

Using the given coordinates and the mid-point formula, the mid-point of the hypotenuse $BC$ is

$\Big(\frac{0+k}{2},\frac{l+0}{2}\Big)=\Big(\frac{k}{2},\frac{l}{2}\Big),$

and so the slope of the median from vertex $A$ is

$\frac{\frac{l}{2}-0}{\frac{k}{2}-0}=\frac{l}{k}.$

Now, the slope of the hypotenuse $BC$ is

$\frac{l-0}{0-k}=-\frac{l}{k},$

which is the negative of the slope of the median from vertex $A$ .

However, the slopes of the sides of $\triangle ABC$

, namely $(0,~-\frac{l}{k},~\infty)$

do not form a geometric progression. This doesn’t contradict the equivalent statements in our first proposition; it simply illustrates the fact that a hypothesis cannot be dismissed, else things can go amiss.

Example 4 (Second proposition)

In $\triangle ABC$ , PROVE that $m_{A}+m_{BC}=0$ if, and only if, $m_{a}=\frac{x_2+x_3-2x_1}{2(x_3-x_2)}\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}$ , where $m_{a}$ is the length of the median from vertex $A$ and the triangle’s vertices are located at $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ .

Consequently, we obtain a second characterization of slopes in geometric progression.

First suppose that $m_{A}+m_{BC}=0$ . Since

$m_{A}=\frac{\frac{y_2+y_3}{2}-y_1}{\frac{x_2+x_3}{2}-x_1}=\frac{y_2+y_3-2y_1}{x_2+x_3-2x_1};\quad m_{BC}=\frac{y_3-y_2}{x_3-x_2}$

$m_{A}+m_{BC}=0\implies y_2+y_3-2y_1=-\Big(\frac{y_3-y_2}{x_3-x_2}\Big)(x_2+x_3-2x_1)$ .

Now, the length $m_{a}$ of the median from vertex $A$ is the distance from $(x_1,y_1)$ to the midpoint of $BC=\Big(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\Big)$ :

$\begin{equation*} \begin{split} m_{a}^2&=\Big(\frac{x_2+x_3}{2}-x_1\Big)^2+\Big(\frac{y_2+y_3}{2}-y_1\Big)^2\\ &=\Big(\frac{x_2+x_3-2x_1}{2}\Big)^2+\Big(\frac{y_2+y_3-2y_1}{2}\Big)^2\\ &=\frac{1}{4}\Big((x_2+x_3-2x_1)^2+(y_2+y_3-2y_1)^2\Big)\\ &=\frac{1}{4}\Big[(x_2+x_3-2x_1)^2+\Big(-(\frac{y_3-y_2}{x_3-x_2})(x_2+x_3-2x_1)\Big)^2\Big]\\ &=\frac{1}{4}\frac{(x_2+x_3-2x_1)^2}{(x_3-x_2)^2}\Big((x_3-x_2)^2+(y_3-y_2)^2\Big)\\ \therefore m_{a}&=\frac{x_2+x_3-2x_1}{2(x_3-x_2)}\sqrt{(x_3-x_2)^2+(y_3-y_2)^2} \end{split} \end{equation}$

A little care should be taken pertaining the sign of $\frac{x_2+x_3-2x_1}{x_3-x_2}$ , so its absolute value should be used instead.

Conversely, suppose that $m_{a}=\frac{x_2+x_3-2x_1}{2(x_3-x_2)}\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}$ . Usually, $m_{a}^2=\Big(\frac{x_2+x_3-2x_1}{2}\Big)^2+\Big(\frac{y_2+y_3-2y_1}{2}\Big)^2$ , in view of the given coordinates. Together with the assumption of the converse, this amounts to:

$\begin{equation*} \begin{split} \Big(\frac{x_2+x_3-2x_1}{2}\Big)^2+\Big(\frac{y_2+y_3-2y_1}{2}\Big)^2&=\Big(\frac{x_2+x_3-2x_1}{2(x_3-x_2)}\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}\Big)^2\\ \Big(\frac{y_2+y_3-2y_1}{2}\Big)^2&=\Big(\frac{x_2+x_3-2x_1}{2(x_3-x_2)}\Big)^2[(x_3-x_2)^2+(y_3-y_2)^2]-\Big(\frac{x_2+x_3-2x_1}{2}\Big)^2\\ \Big(\frac{y_2+y_3-2y_1}{2}\Big)^2&=(x_2+x_3-2x_1)^2\Big(\frac{(x_3-x_2)^2+(y_3-y_2)^2}{4(x_3-x_2)^2}-\frac{1}{4}\Big)\\ \therefore \frac{(y_2+y_3-2y_1)^2}{4}&=\frac{1}{4}(x_2+x_3-2x_1)^2\Big(\frac{y_3-y_2}{x_3-x_2}\Big)^2\\ \Big(\frac{y_2+y_3-2y_1}{x_2+x_3-2x_1}\Big)^2&=\Big(\frac{y_3-y_2}{x_3-x_2}\Big)^2 \end{split} \end{equation}$

Extracting square roots, we get $\frac{y_2+y_3-2y_1}{x_2+x_3-2x_1}=\pm(\frac{y_3-y_2}{x_3-x_2})$ . Since the slope of a median cannot be equal to the slope of the side it meets, we must then take the negative square root, and so $\frac{y_2+y_3-2y_1}{x_2+x_3-2x_1}=-(\frac{y_3-y_2}{x_3-x_2})$ . In terms of our notation, this translates to $m_{A}=-m_{BC}$ , or $m_{A}+m_{BC}=0$ .

Combining our first and second propositions, we obtain the following three equivalent statements for $\triangle ABC$

with vertices at $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$

$m_{A}+m_{BC}=0$ (that is, the slope of the median from vertex $A$ is the negative of the slope of side $BC$ );

the (non-zero) slopes of sides $AB,BC,CA$ form a geometric progression;

the length of the median from vertex $A$ is $m_{a}=\frac{x_2+x_3-2x_1}{2(x_3-x_2)}\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}$ .

Applying the first proposition

Our first application goes to the right triangle considered in Example 3; it will be the basis of a future post — where we aim to unveil a hidden characteristic in every triangle.

PROVE that the right triangle $ABC$ with vertices at $A(0,0),~B(k,0),~C(0,l)$ contains a “sub-triangle” whose side slopes form a geometric progression with a common ratio of $3$ .

In view of Example 3, we know that the median from vertex $A$ has a slope that’s the negative of the slope of the hypotenuse $BC$ . Consider the midpoint of median from $A$ to $BC$ , located at $G\Big(\frac{k}{4},\frac{l}{4}\Big)$ , say.

The triangle with vertices $C(0,l),~G\Big(\frac{k}{4},\frac{l}{4}\Big),~B(k,0)$ is the desired “sub-triangle”. Indeed, the slopes of its sides are:

$\begin{equation*} \begin{split} m_{BC}&=-\frac{l}{k}\\ &\cdots\vdots\cdots\\ m_{CG}&=\frac{l-\frac{l}{4}}{0-\frac{k}{4}}\\ &=\frac{\frac{3l}{4}}{-\frac{k}{4}}\\ &=3\Big(-\frac{l}{k}\Big)\\ &\cdots\vdots\cdots\\ m_{GB}&=\frac{\frac{l}{4}-0}{\frac{k}{4}-k}\\ &=\frac{\frac{l}{4}}{-\frac{3k}{4}}\\ &=\frac{1}{3}\Big(-\frac{l}{k}\Big) \end{split} \end{equation}$

This verifies that the slopes of the “sub-triangle” $BCG$ form a geometric progression with a common ratio of $3$ .

PROVE that if the slopes of the sides of a triangle form a geometric progression (with common ratio $r$ ), then so do the slopes of the three medians, provided $r\neq -2$ (or $-\frac{1}{2}$ ).

In Example 7 and Example 8 below, we explain what underlies the restriction $r\neq -2$ .

If $\triangle ABC$ has sides $AB,BC,CA$ with slopes $m_{AB},m_{BC},m_{CA}$ in geometric progression, we’ll prove that the median slopes $m_{C},m_{A},m_{B}$ form a new geometric progression.

Let the coordinates of $\triangle ABC$ be $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ . We have

$m_{AB}=\frac{y_1-y_2}{x_1-x_2},~m_{BC}=\frac{y_2-y_3}{x_2-x_3},~m_{CA}=\frac{y_3-y_1}{x_3-x_1}$

and

$m_{A}=\frac{y_2+y_3-2y_1}{x_2+x_3-2x_1},~m_{B}=\frac{y_1+y_3-2y_2}{x_1+x_3-2x_2},~m_{C}=\frac{y_1+y_2-2y_3}{x_1+x_2-2x_3}.$

From our first proposition, $m_{AB},~m_{BC},~m_{CA}$ being in geometric progression means that $m_{A}+m_{BC}=0$ , and so $m_{A}^2=m_{BC}^2$ . That is, $\Big(\frac{y_2+y_3-2y_1}{x_2+x_3-2x_1}\Big)^2=\Big(\frac{y_3-y_2}{x_3-x_2}\Big)^2$ . From Example 1, $x_1=\frac{x_3(y_3-y_1)+x_2(y_1-y_2)}{y_3-y_2}$ .

$\begin{equation*} \begin{split} m_{B}\times m_{C}&=\Big(\frac{y_1+y_3-2y_2}{x_1+x_3-2x_2}\Big)\Big(\frac{y_1+y_2-2y_3}{x_1+x_2-2x_3}\Big)\\ &=\frac{y_1+y_2-2y_3}{\Big[\frac{x_3(y_3-y_1)+x_2(y_1-y_2)}{y_3-y_2}+x_2-2x_3\Big]}\times\frac{y_1+y_3-2y_2}{\Big[\frac{x_3(y_3-y_1)+x_2(y_1-y_2)}{y_3-y_2}+x_3-2x_2\Big]}\\ &=\frac{y_1+y_2-2y_3}{\frac{x_3(y_3-y_1)+x_2(y_1-y_2)+x_2(y_3-y_2)-2x_3(y_3-y_2)}{y_3-y_2}}\times\frac{y_1+y_3-2y_2}{\frac{x_3(y_3-y_1)+x_2(y_1-y_2)+x_3(y_3-y_2)-2x_2(y_3-y_2)}{y_3-y_2}}\\ &=\scriptscriptstyle\frac{(y_1+y_2-2y_3)(y_3-y_2)}{\Big[x_3(y_3-y_1-2y_3+2y_2)+x_2(y_1-y_2+y_3-y_2)\Big]}\times\frac{(y_1+y_3-2y_2)(y_3-y_2)}{\Big[x_3(y_3-y_1+y_3-y_2)+x_2(y_1-y_2-2y_3+2y_2)\Big]}\\ &=\scriptscriptstyle\frac{(y_1+y_2-2y_3)(y_3-y_2)}{[x_3(-y_1-y_3+2y_2)+x_2(y_1+y_3-2y_2)]}\times\frac{(y_1+y_3-2y_2)(y_3-y_2)}{[x_3(-y_1-y_2+2y_3)+x_2(y_1+y_2-2y_3)]}\\ &=\frac{(y_1+y_2-2y_3)(y_3-y_2)}{(x_2-x_3)(y_1+y_3-2y_2)}\times\frac{(y_1+y_3-2y_2)(y_3-y_2)}{(x_2-x_3)(y_1+y_2-2y_3)}\\ &=\Big(\frac{y_3-y_2}{x_3-x_2}\Big)^2\\ &=m_{A}^2 \end{split} \end{equation*}$

So, the median slopes $m_{C},~m_{A},~m_{B}$ form a geometric progression, with $m_{A}$ as the geometric mean.

Suppose that the slopes of the sides of a triangle form a geometric progression with a common ratio of $-2$ . PROVE that the triangle contains a horizontal median.

We’ve given a proof of this before by explicitly determining the coordinates of the vertices in terms of the common ratio $r=-2$ ; here we present a direct approach that avoids row reduction (linear algebra).

Assume the geometric progression of the slopes is $m_{AB},~m_{BC},~m_{CA}$ , where $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ are the triangle’s vertices. Since the common ratio is $-2$ , we have:

$\frac{y_2-y_3}{x_2-x_3}=-2\Big(\frac{y_1-y_2}{x_1-x_2}\Big),~\frac{y_1-y_3}{x_1-x_3}=-2\Big(\frac{y_2-y_3}{x_2-x_3}\Big).$

Isolate $x_1$ from the first equation:

$x_1=x_2-\frac{2(y_1-y_2)(x_2-x_3)}{y_2-y_3}.$

Also, from the second equation, isolating $x_1$ gives

$x_1=x_3-\frac{(y_1-y_3)(x_2-x_3)}{2(y_2-y_3)}.$

Equate the two expressions obtained for $x_1$ :

$\begin{equation*} \begin{split} x_2-\frac{2(y_1-y_2)(x_2-x_3)}{y_2-y_3}&=x_3-\frac{(y_1-y_3)(x_2-x_3)}{2(y_2-y_3)}\\ x_2-x_3&=\frac{2(y_1-y_2)(x_2-x_3)}{y_2-y_3}-\frac{(y_1-y_3)(x_2-x_3)}{2(y_2-y_3)}\\ 2(x_2-x_3)(y_2-y_3)&=4(y_1-y_2)(x_2-x_3)-(y_1-y_3)(x_2-x_3)\\ \therefore 2(y_2-y_3)&=4(y_1-y_2)-(y_1-y_3),\quad\textrm{since}~x_2\neq x_3\\ 2y_2-2y_3&=4y_1-4y_2-y_1+y_3\\ 3y_1+3y_3-6y_2&=0\\ y_1+y_3-2y_2&=0 \end{split} \end{equation}$

Thus, the slope of the median from vertex $B$ is zero, that is, $m_{B}=\frac{y_1+y_3-2y_2}{x_1+x_3-2x_2}=\frac{0}{x_1+x_3-2x_2}=0$ , giving a horizontal median.

Suppose that the slopes of the sides of a triangle form a geometric progression with a common ratio of $-2$ . PROVE that the triangle contains a vertical median.

As in Example 7, the fact the geometric progression has a common ratio of $-2$ means we can write:

$\frac{y_2-y_3}{x_2-x_3}=-2\Big(\frac{y_1-y_2}{x_1-x_2}\Big),~\frac{y_1-y_3}{x_1-x_3}=-2\Big(\frac{y_2-y_3}{x_2-x_3}\Big).$

We obtain $\frac{y_1-y_3}{x_1-x_3}=4\Big(\frac{y_1-y_2}{x_1-x_2}\Big)$ , from which

$y_3=y_1-\frac{4(y_1-y_2)(x_1-x_3)}{x_1-x_2}.$

Also, from $\frac{y_2-y_3}{x_2-x_3}=-2\Big(\frac{y_1-y_2}{x_1-x_2}\Big)$ , isolating $y_3$ gives

$y_3=y_2+\frac{2(y_1-y_2)(x_2-x_3)}{x_1-x_2}.$

Equate the two expressions obtained for $y_3$ :

$\begin{equation*} \begin{split} y_1-\frac{4(y_1-y_2)(x_1-x_3)}{x_1-x_2}&=y_2+\frac{2(y_1-y_2)(x_2-x_3)}{x_1-x_2}\\ y_1-y_2&=\frac{2(y_1-y_2)(x_2-x_3)}{x_1-x_2}+\frac{4(y_1-y_2)(x_1-x_3)}{x_1-x_2}\\ y_1-y_2&=\frac{2(y_1-y_2)(x_2-x_3+2x_1-2x_2)}{x_1-x_2}\\ (y_1-y_2)(x_1-x_2)&=2(y_1-y_2)(2x_1+x_2-3x_3)\\ \therefore x_1-x_2&=2(2x_1+x_2-3x_3)\quad\textrm{since}~y_1\neq y_2\\ 0&=3x_1+3x_2-6x_3\\ 0&=x_1+x_2-2x_3 \end{split} \end{equation}$

Therefore, the slope of the median from vertex $C$ , given by $m_{C}=\frac{y_1+y_2-2y_3}{x_1+x_2-2x_3}=\frac{y_1+y_2-2y_3}{0}$ , is undefined, so this median is vertical.

Suppose that $\triangle ABC$ is isosceles and that the slope of the median from the “apex” $A$ is $m_{A}=\pm 1$ . PROVE that the slopes of its sides form a geometric progression.

This is based on the fact that the slope of the median from the “apex” of an isosceles triangle is the negative reciprocal of the slope of the base. In particular, if $m_{A}=1$ , then $m_{BC}=\frac{-1}{1}=-1$ , giving $m_{A}+m_{BC}=0$ . By our first proposition, this means the slopes of sides $AB,BC,CA$ form a geometric progression. Similarly, if $m_{A}=-1$ , then $m_{BC}=\frac{-1}{-1}=1$ , also giving $m_{A}+m_{BC}=0$ , and we again obtain a geometric progression for the slopes of sides $AB,BC,CA$ .

A numerical problem

$\triangle ABC$ has vertices at $A(r,s),~B(-2,1),~C(4,-3)$ . Determine suitable values of $r$ and $s$ for which the slopes of sides $AB,BC,CA$ form a geometric progression.

Using our first proposition, we can tackle this problem by somehow “forcing” a median from side $BC$ to have a slope that’s the negative of side $BC$ ‘s slope. (Note that we can also use Exercise 1 below, which may be easier.)

Let’s find the equation of this “forced” median. The midpoint of $BC$ is $(1,-1)$ , and its slope is $-\frac{2}{3}$ . So we want a median through $(1,-1)$ with slope $\frac{2}{3}$ . Its equation is:

$\begin{equation*} \begin{split} y-(-1)&=\frac{2}{3}\Big(x-1\Big)\\ y&=\frac{2}{3}x-\frac{5}{3} \end{split} \end{equation}$

Thus, any point on the line $y=\frac{2}{3}x-\frac{5}{3}$ — except the trio $(4,1),~(1,-1),~(-2,-3)$ — will do for the coordinates of vertex $A$ . Why exclude these three? First, the point $(1,-1)$ is the midpoint of $BC$ , so we can’t form a triangle with it. Next, the point $(4,1)$ shares the same $y$ -coordinate with vertex $B$ , which will make the slope of $AB$ zero if $A$ is chosen as the point $(4,1)$ . Similarly, we can’t take $(-2,-3)$ for $A$ because then $A$ will have the same $x$ -coordinate with vertex $C$ , resulting in an undefined slope.

With the exclusions out of the way, we can take any other point, say $(7,3)$ , as $A$ . Then $\triangle ABC$ has vertices at $A(7,3),~B(-2,1),~C(4,-3)$ , and the slopes of sides $AB,BC,CA$ become

$\frac{2}{9},-\frac{2}{3},2$

respectively. They form a geometric progression with a common ratio of $-3$ (or $-\frac{1}{3}$ if we reverse the numbers).

A few other points on the line $y=\frac{2}{3}x-\frac{5}{3}$ that can be chosen in place of $A(7,3)$ — and the corresponding geometric progressions for the slopes of the sides — are listed below:

$\begin{equation*} \begin{split} \textbf{(10,5)},(-2,1),(4,-3)&\mapsto\Big\{\frac{1}{3},-\frac{2}{3},\frac{4}{3}\Big\},~r=-2;\\ \textbf{(13,7)},(-2,1),(4,-3)&\mapsto\Big\{\frac{2}{5},-\frac{2}{3},\frac{10}{9}\Big\},~r=-\frac{5}{3};\\ \textbf{(16,9)},(-2,1),(4,-3)&\mapsto\Big\{\frac{4}{9},-\frac{2}{3},1\Big\},~r=-\frac{3}{2};\\ \textbf{(19,11)},(-2,1),(4,-3)&\mapsto\Big\{\frac{10}{21},-\frac{2}{3},\frac{14}{15}\Big\},~r=-\frac{7}{5};\\ \textbf{(-5,-5)},(-2,1),(4,-3)&\mapsto\Big\{\frac{2}{9},-\frac{2}{3},2\Big\},~r=-3 \end{split} \end{equation}$

Observe that the choice $(-5,-5)$ yields the same set of geometric progression as our initial choice $(7,3)$ .

As in, this whole thing is COOL!

Takeaway

In the domain of triangles, the GOAT (Greatest Of All Triangles) is the triangle whose slopes form a geometric progression. No jokes.

That told, our first proposition is the basis for our future application. In particular, what we did for the right triangle in Example 5 will be done for any triangle. As in, A-N-Y!

Tasks

(Generating formula) Let $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ be the coordinates of the vertices of $\triangle ABC$ . Require that $x_1\neq x_2\neq x_3$ and that $y_1\neq y_2\neq y_3$ . PROVE that the slopes of sides $AB,BC,CA$ form a geometric progression if, and only if, $x_1(y_3-y_2)+x_2(y_2-y_1)+x_3(y_1-y_3)=0.$
$\Big($ Thus, in order to obtain coordinates for the vertices of a triangle whose slopes form a geometric progression, begin by selecting “any” five numbers — ensuring that no two $x'$ s are equal and no two $y'$ s are equal — then use the above relation to determine the value of the remaining coordinate. $\Big)$
(Special ratio) Let be the coordinates of the vertices of . PROVE that:
- $m_{a}=\frac{3}{2}a$ if, and only if, $\frac{y_1+y_3-2y_2}{x_1+x_3-2x_2}\times\frac{y_1+y_2-2y_3}{x_1+x_2-2x_3}=-1$ (that is, the medians through $B$ and $C$ are perpendicular)
- if the slopes $m_{AB},~m_{BC},~m_{CA}$ form a geometric progression, then $m_{a}=\frac{3}{2}a$ if, and only if, $x_1+x_3-2x_2=0$ (that is, the median through $B$ is vertical)
- if the slopes $m_{AB},~m_{BC},~m_{CA}$ form a geometric progression and $m_{a}=\frac{3}{2}a$ , then the median through $C$ is horizontal
- if the slopes $m_{AB},~m_{BC},~m_{CA}$ form a geometric progression and $m_{a}=\frac{3}{2}a$ , then the common ratio of the geometric progression is $r=-2.$
  (If the slopes of the sides of a triangle form a geometric progression, no other value for the common ratio yields the relation $m_{a}=\frac{3}{2}a$ ; only $r=-2$ does.)
(Half right) Let $ABC$ be a triangle in which the slopes $m_{AB},m_{BC},m_{CA}$ form a geometric progression. PROVE that the acute angle between the median from $A$ and side $BC$ is $45^{\circ}$ if, and only if, $m_{BC}=-1\pm\sqrt{2}$ or $m_{BC}=1\pm\sqrt{2}$ .
(Pseudo perpendicularity) Suppose that the slopes $m_{AB},m_{BC},m_{CA}$ of the sides $AB,BC,CA$ of $\triangle ABC$ form a geometric progression with common ratio $r$ . Let the lengths of sides $AB,BC,CA$ be $l_1,l_2,l_3$ , respectively. PROVE that the following pseudo-pythagorean property holds: $l_1^2+l_3^2=\frac{1+r^2}{(1+r)^2}l_2^2$ .
(If we let $r=0$ , then we obtain a full pythagorean relationship. However, $r=0$ is unacceptable for a geometric progression. Note that this doesn’t imply the impossibility of having right triangles with slopes in geometric progression; there are actually many of them — with an appropriately chosen first term and a negative common ratio. Now, if we let $r=-2$ , we get $l_1^2+l_3^2=5l_2^2$ , a peach.)
Suppose that the slopes of a right triangle form a geometric progression. PROVE that the slope of the hypotenuse cannot be the geometric mean of the progression.
(For a change, attempt this exercise without recourse to perpendicular slopes. Use our second proposition (Example 4) and the fact that the length of the median to the hypotenuse is half of the length of the hypotenuse.)
Let $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ be the coordinates of the vertices of an equilateral $\triangle ABC$ . If the slopes of sides $AB,BC,CA$ form a geometric progression, prove that $x_{1}=\frac{(1+\sqrt{3})x_2+(1-\sqrt{3})x_{3}}{2}$ , or $x_{1}=\frac{(1-\sqrt{3})x_{2}+(1+\sqrt{3})x_{3}}{2}.$
(Note the order of the slopes.)
$\triangle ABC$ has vertices at $A(-2,1),~B(4,-3),~C(r,s)$ . Find three different pairs $(r,s)$ for which the slopes of the sides form a geometric progression with a common ratio of $-2$ .
(Three possibilities) Let be the coordinates of the vertices of a triangle. If the slopes of the sides form a geometric progression, PROVE that at least one of these three is true:
- $r+s=5$ ;
- $rs-2r-2s+3=0$ ;
- $rs-3r-3s+8=0$ .
  (Use the generating formula.)
PROVE that every right triangle $ABC$ with legs $AB$ and $AC$ on — or parallel to — the coordinate axes contains a “sub-triangle” $GBC$ that has the following property: the length of the median from vertex $G$ is one-fourth the length of the hypotenuse $BC$ ; that is, $m_{g}=\frac{1}{4}\times BC$ .
(For instance, consider the “sub-triangle” we constructed in Example 5.)
Find suitable choice of coordinates for the vertices of $\triangle ABC$ that has the property that $m_{a}=\frac{1}{3}a$ , where $m_{a}$ is the length of the median from vertex $A$ and $a$ is the length of side $BC$ .
(Consider the case where the slopes of sides $AB,BC,CA$ form a geometric progression with a common ratio of $5$ .)