Equilateral triangles characterized via slopes

Triangles whose side slopes form a geometric progression are very nice objects, and a very special subset of such triangles is our subject for today.

Coordinates for equilateral triangles

Just recently, we paid a visit to the place where equilateral triangles live in the Cartesian plane; consequently, we begin by giving you their full “address”, so you can also locate them without stress, without pain.

Example 1

Let $\triangle ABC$ be a triangle. PROVE that the following are equivalent:

$\triangle ABC$ is equilateral with length $2a$ ;
its vertices are located at $A(x_1,y_1)$ , $B\Big(x_1+2a\cos\theta, y_1+2a\sin\theta\Big)$ , $C\Big(x_1+2a\cos(\theta+60^{\circ}),y_{1}+2a\sin(\theta+60^{\circ})\Big).$

$\theta$ is some parameter that can be thought of as the angle that the “base” of the triangle makes with the positive $x$ -axis.

So a complete description of the coordinates of the vertices of an equilateral triangle requires three things: a starting point $(x_1,y_1)$ , a fixed length $2a$ , and the “base” inclination $\theta$ . In the traditional case, these correspond to $(0,0), 2a,$ and $\theta=0^{\circ}$ , respectively.

Let’s first prove that any triangle having the given coordinates is equilateral. We’ll use the trig identity $\cos^2 A+\sin^2 A=1$ and the distance formula:

$\begin{equation*} \begin{split} AB^2&=(2a\cos\theta)^2+(2a\sin\theta)^2\\ &=4a^2(\cos^2\theta+\sin^2\theta)\\ &=4a^2\\ \therefore AB&=2a\\ AC^2&=(2a\cos(\theta+60^{\circ}))^2+(2a\sin(\theta+60^{\circ}))^2\\ &=4a^2(\cos^2(\theta+60^{\circ})+\sin^2(\theta+60^{\circ}))\\ &=4a^2\\ \therefore AC&=2a.\\ \end{split} \end{equation}$

It now remains side $BC$ ; we’ll compute its length in two steps, starting with the “run” (difference of the $x$ -coordinates):

(1) $\begin{equation*} \begin{split} &=2a\Big(\cos(\theta+60^{\circ})-\cos(\theta)\Big)\\ &=2a\Big(\cos\theta\cos(60^{\circ})-\sin\theta\sin(60^{\circ})-\cos\theta\Big)\\ &=2a\Big(\frac{1}{2}\cos\theta-\frac{\sqrt{3}}{2}\sin\theta-\cos\theta\Big)\\ &=2a\Big(-\frac{1}{2}\cos\theta-\frac{\sqrt{3}}{2}\sin\theta\Big)\\ &=a(-\cos\theta-\sqrt{3}\sin\theta) \end{split} \end{equation*}$

Next, we calculate the “rise” of side $BC$ (difference of the $y$ -coordinates):

(2) $\begin{equation*} \begin{split} &=2a\Big(\sin(\theta+60^{\circ})-\sin\theta\Big)\\ &=2a\Big(\sin\theta\cos(60^{\circ})+\cos\theta\sin(60^{\circ})-\sin\theta\Big)\\ &=2a\Big(\frac{1}{2}\sin\theta+\frac{\sqrt{3}}{2}\cos\theta-\sin\theta\Big)\\ &=2a\Big(\frac{\sqrt{3}}{2}\cos\theta-\frac{1}{2}\sin\theta\Big)\\ &=a(\sqrt{3}\cos\theta-\sin\theta). \end{split} \end{equation*}$

We can now calculate the length of side $BC$ , using the last lines in (1) and (2):

$\begin{equation*} \begin{split} BC^2&=a^2(-\cos\theta-\sqrt{3}\sin\theta)^2+a^2(\sqrt{3}\cos\theta-\sin\theta)^2\\ &=a^2\Big(\cos^2\theta+2\sqrt{3}\cos\theta\sin\theta+3\sin^2\theta+3\cos^2\theta-2\sqrt{3}\cos\theta\sin\theta+\sin^2\theta\Big)\\ &=a^2\Big(4\cos^2\theta+4\sin^2\theta\Big)\\ &=4a^2\\ \therefore BC&=2a. \end{split} \end{equation}$

Since $AB=AC=BC=2a$ , the given coordinates yield an equilateral triangle of length $2a$ . Conversely, to see that every equilateral triangle can be described this way, CLICK HERE.

Example 2

Find coordinates for the vertices of an equilateral triangle $ABC$ of length $2a$ where one side is parallel to the $x$ -axis, and one vertex is at the origin $(0,0)$ .

This is the traditional case in which $\theta=0^{\circ}$ . Since we’re also given $(x_1,y_1)=(0,0)$ , the two other vertices are

$\begin{equation*} \begin{split} (0+2a\cos(0^{\circ}),0+2a\sin(0^{\circ}))&=(2a,0)\\ &\cdots\vdots\cdots\\ (0+2a\cos(0^{\circ}+60^{\circ}),0+2a\sin(0^{\circ}+60^{\circ}))&=(a,a\sqrt{3}) \end{split} \end{equation}$

The vertices are $A(0,0),~B(2a,0),~C(a,a\sqrt{3})$ .

Notice that the slopes of sides $CA,AB,BC$ are $\sqrt{3},0,-\sqrt{3}$ , respectively. They form an arithmetic progression with a common difference of $-\sqrt{3}$ (or a common difference of $\sqrt{3}$ , if we reverse the numbers). So this particular equilateral triangle comes equipped, naturally, with slopes in arithmetic progression.

In an equilateral triangle, SUM OF PRODUCT OF SLOPES, taken two slopes at a time=-3: CLICK HERE for a sample.

We’ll prove this in Example 7, but for now notice that $(-\sqrt{3})\times 0+0\times\sqrt{3}+\sqrt{3}\times(-\sqrt{3})=-3$ .

Example 3

Find coordinates for the vertices of an equilateral triangle $ABC$ of length $4a$ in which one vertex is at $A(0,0)$ , and side $AB$ makes an angle of $45^{\circ}$ with the positive $x$ -axis.

Using $(x_1,y_1)=(0,0)$ and $\theta=45^{\circ}$ in Example 1, we obtain the two other vertices’ coordinates:

$\begin{equation*} \begin{split} \Big(0+4a\cos(45^{\circ}),0+4a\sin(45^{\circ})\Big)&=(2a\sqrt{2},2a\sqrt{2})\\ &\cdots\vdots\cdots\\ \Big(0+4a[\cos(45^{\circ}+60^{\circ})],0+4a[\sin(45^{\circ}+60^{\circ})]\Big)&=(x,y)\\ x&=a\sqrt{2}(1-\sqrt{3})\\ y&=a\sqrt{2}(1+\sqrt{3}). \end{split} \end{equation}$

Thus, the vertices are

$\begin{equation*} \begin{split} A(0,0)&\\ B\Big(2a\sqrt{2},~2a\sqrt{2}\Big)&\\ C\Big(a\sqrt{2}(1-\sqrt{3}),~a\sqrt{2}(1+\sqrt{3})\Big). \end{split} \end{equation}$

We can use these vertices to calculate the slopes of the sides:

$\begin{equation*} \begin{split} \textrm{\textbf{slope of side AB}}&=\frac{2a\sqrt{2}-0}{2a\sqrt{2}-0}\\ &=1\\ &\cdots\vdots\cdots\\ \textrm{\textbf{slope of side BC}}&=\frac{a\sqrt{2}(1+\sqrt{3})-2a\sqrt{2}}{a\sqrt{2}(1-\sqrt{3})-2a\sqrt{2}}\\ &=\frac{\sqrt{3}-1}{-\sqrt{3}-1}\\ &=\Big(\frac{\sqrt{3}-1}{-\sqrt{3}-1}\Big)\times\Big(\frac{-\sqrt{3}+1}{-\sqrt{3}+1}\Big)\\ &=-2+\sqrt{3}\\ &\cdots\vdots\cdots\\ \textrm{\textbf{slope of side CA}}&=\frac{a\sqrt{2}(1+\sqrt{3})-0}{a\sqrt{2}(1-\sqrt{3})-0}\\ &=\frac{1+\sqrt{3}}{1-\sqrt{3}}\\ &=-2-\sqrt{3} \end{split} \end{equation}$

Since $1^2=(-2+\sqrt{3})(-2-\sqrt{3})$ , these slopes form a geometric progression with a common ratio of $-2\pm\sqrt{3}$ , depending on how we arrange the numbers. So, this particular equilateral triangle comes equipped, naturally, with slopes in geometric progression.

It follows that by changing the inclination angle of the base, the sequences formed by the slopes also change. Therefore, the traditional case (having one vertex at $(0,0)$ and one side parallel to the $x$ -axis) is inadequate for what we want to investigate, which then necessitates the alternate — and “ultimate” — coordinates, given in Example 1.

Slopes of equilateral triangles

If we’re given a set of three numbers for the slopes of the sides of a triangle, we can easily tell if the triangle is equilateral.

Example 4

PROVE that any triangle whose sides have slopes $-\sqrt{3},0,\sqrt{3}$ is equilateral.

Imagine a $\triangle ABC$ in which sides $AB,BC,CA$ have slopes $\sqrt{3},0,-\sqrt{3}$ as shown below:

Using the fact that the acute angle $\theta$ between two lines with slopes $m_{1}$ and $m_{2}$ can be given by $\tan \theta=|\frac{m_{1}-m_{2}}{1+m_{1}m_{2}}|$ , we have:

$\begin{equation*} \begin{split} \tan(\angle ABC)&=\Big|\frac{\sqrt{3}-0}{1+(\sqrt{3}\times 0)}\Big|\\ &=\Big|\frac{\sqrt{3}}{1}\Big|\\ &=\sqrt{3}\\ \therefore \angle ABC&=60^{\circ}\\ &\cdots\vdots\cdots\\ \tan(\angle BAC)&=\Big|\frac{-\sqrt{3}-\sqrt{3}}{1+(-\sqrt{3}\times\sqrt{3})}\Big|\\ &=\Big|\frac{-2\sqrt{3}}{1-3}\Big|\\ &=\sqrt{3}\\ \therefore\angle BAC&=60^{\circ}\\ &\cdots\vdots\cdots\\ \therefore\angle ACB&=60^{\circ}. \end{split} \end{equation}$

Since each interior angle of $\triangle ABC$ measures $60^{\circ}$ , we conclude that this triangle is equilateral. (Compare Example 4 with Example 2.)

Example 5

PROVE that any triangle whose sides have slopes $-2-\sqrt{3},1,-2+\sqrt{3}$ is equilateral.

As in the previous example, suppose that a certain $\triangle ABC$ has sides $AB,BC,CA$ with slopes $-2-\sqrt{3},1,-2+\sqrt{3}$ as shown:

Then:

$\begin{equation*} \begin{split} \tan(\angle ABC)&=\Big|\frac{-2-\sqrt{3}-1}{1+(-2-\sqrt{3})(1)} \Big|\\ &=\Big|\frac{-3-\sqrt{3}}{-1-\sqrt{3}}\Big|\\ &=\Big|\frac{-3-\sqrt{3}}{-1-\sqrt{3}}\times\Big(\frac{-1+\sqrt{3}}{-1+\sqrt{3}}\Big)\Big|\\ &=\Big|\frac{3-3\sqrt{3}+\sqrt{3}-3}{1-3}\Big|\\ &=\Big|\frac{-2\sqrt{3}}{-2}\Big|\\ &=\sqrt{3}\\ \therefore \angle ABC&=60^{\circ}\\ &\cdots\vdots\cdots\\ \tan(\angle CAB)&=\Big|\frac{-2+\sqrt{3}-(-2-\sqrt{3})}{1+(-2+\sqrt{3})(-2-\sqrt{3})}\Big|\\ &=\Big|\frac{2\sqrt{3}}{1+(4-3)}\Big|\\ &=\sqrt{3}\\ \therefore \angle CAB&=60^{\circ}\\ &\cdots\vdots\cdots\\ \therefore\angle ACB&=60^{\circ}. \end{split} \end{equation}$

Compare Example 5 with Example 3.

Characterizing slopes of equilateral triangles

Due to the result below, we can always assume that the slopes of the sides of any equilateral triangle are of the form $\tan\theta, ~\frac{\tan\theta-\sqrt{3}}{1+\sqrt{3}\tan\theta},~\frac{\tan\theta+\sqrt{3}}{1-\sqrt{3}\tan\theta}$ .

Example 6

Let $ABC$ be a triangle. PROVE that the following are equivalent:

$\triangle ABC$ is equilateral;
the slopes of its sides are $\tan\theta, ~\frac{\tan\theta-\sqrt{3}}{1+\sqrt{3}\tan\theta},~\frac{\tan\theta+\sqrt{3}}{1-\sqrt{3}\tan\theta}$ .

What happens when the denominators are zero, namely when $\tan \theta=\pm\frac{1}{\sqrt{3}}$ ? Easy. The orientation of the resulting equilateral triangle reduces to the “basic” case where one side is parallel to the $y$ -axis. So there’s no issue with this — simply exclude it.

Now to the proof. First suppose that $\triangle ABC$ is equilateral with length $2a$ . From Example 1 we know that its coordinates can be located at

$\begin{equation*} \begin{split} &A(x_1,y_1)\\ &B\Big(x_1+2a\cos\theta, y_1+2a\sin\theta\Big)\\ &C\Big(x_1+2a\cos(\theta+60^{\circ}),y_{1}+2a\sin(\theta+60^{\circ})\Big). \end{split} \end{equation}$

Using these coordinates, we can easily calculate the slopes of the sides:

$\begin{equation*} \begin{split} m_{AB}&=\frac{y_1+2a\sin\theta-y_1}{x_1+2a\cos\theta-x_1}\\ &=\tan\theta\\ &\cdots\vdots\cdots\\ m_{BC}&=\frac{y_{1}+2a\sin(\theta+60^{\circ})-(y_1+2a\sin\theta)}{x_1+2a\cos(\theta+60^{\circ})-(x_1+2a\cos\theta)}\\ &=\frac{\sin(\theta+60^{\circ})-\sin\theta}{\cos(\theta+60^{\circ})-\cos\theta}\\ &=\frac{\sin\theta\cos(60^{\circ})+\cos\theta\sin(60^{\circ})-\sin\theta}{\cos\theta\cos(60^{\circ})-\sin\theta\sin(60^{\circ})-\cos\theta}\\ &=\frac{-\frac{1}{2}\sin\theta+\frac{\sqrt{3}}{2}\cos\theta}{-\frac{1}{2}\cos\theta-\frac{\sqrt{3}}{2}\sin\theta}\\ &=\frac{\tan\theta-\sqrt{3}}{1+\sqrt{3}\tan\theta}\quad\textrm{dividing by}~-\frac{1}{2}\cos\theta\\ &\cdots\vdots\cdots\\ m_{CA}&=\frac{y_1+2a\sin(\theta+60^{\circ})-y_1}{x_1+2a\cos(\theta+60^{\circ})-x_1}\\ &=\tan(\theta+60^{\circ})\\ &=\frac{\tan\theta+\tan(60^{\circ})}{1-\tan\theta\tan(60^{\circ})}\\ &=\frac{\tan\theta+\sqrt{3}}{1-\sqrt{3}\tan\theta} \end{split} \end{equation}$

Conversely, suppose that the sides $AB,BC,CA$ of $\triangle ABC$ have slopes of the form $\tan\theta, ~\frac{\tan\theta-\sqrt{3}}{1+\sqrt{3}\tan\theta},~\frac{\tan\theta+\sqrt{3}}{1-\sqrt{3}\tan\theta}$ . Then, following similar calculations as in Example 4 and Example 5, we have, for example, that $\angle ABC=60^{\circ}$ :

$\begin{equation*} \begin{split} \tan(\angle ABC)&=\Big|\frac{\frac{m-\sqrt{3}}{1+\sqrt{3}m}-m}{1+\Big(\frac{m-\sqrt{3}}{1+\sqrt{3}m}\Big)m}\Big|\quad\textrm{where}~ m=\tan\theta\\ &=\Big|\frac{m-\sqrt{3}-(1+\sqrt{3}m)m}{1+\sqrt{3}m+(m-\sqrt{3})m}\Big|\\ &=\Big|\frac{m-\sqrt{3}-m-\sqrt{3}m^2}{1+\sqrt{3}m+m^2-\sqrt{3}m}\Big|\\ &=\Big|\frac{-\sqrt{3}-\sqrt{3}m^2}{1+m^2}\Big|\\ &=\Big|\frac{-\sqrt{3}(1+m^2)}{1+m^2}\Big|\\ &=\sqrt{3}. \end{split} \end{equation}$

Example 7

In any equilateral triangle, PROVE that the sum of products of slopes, taken two slopes at a time, is always $-3$ .

In view of Example 6, we can let the slopes of the sides be $m_{1}=m,~m_{2}=\frac{m-\sqrt{3}}{1+\sqrt{3}m}$ , and $m_{3}=\frac{m+\sqrt{3}}{1-\sqrt{3}m}$ . Then, taking two slopes at a time:

$\begin{equation*} \begin{split} m_{1}m_{2}&=m\Big(\frac{m-\sqrt{3}}{1+\sqrt{3}m}\Big)\\ &=\frac{m^2-\sqrt{3}m}{1+\sqrt{3}m}\\ &=\Big(\frac{m^2-\sqrt{3}m}{1+\sqrt{3}m}\Big)\times\Big(\frac{1-\sqrt{3}m}{1-\sqrt{3}m}\Big)\\ &=\frac{-\sqrt{3}m^3+4m^2-\sqrt{3}m}{1-3m^2}\\ m_{2}m_{3}&=\Big(\frac{m-\sqrt{3}}{1+\sqrt{3}m}\Big)\times\Big(\frac{m+\sqrt{3}}{1-\sqrt{3}m}\Big)\\ &=\frac{m^2-3}{1-3m^2}\\ m_{3}m_{1}&=\Big(\frac{m+\sqrt{3}}{1-\sqrt{3}m}\Big)m\\ &=\frac{m^2+\sqrt{3}m}{1-\sqrt{3}m}\\ &=\Big(\frac{m^2+\sqrt{3}m}{1-\sqrt{3}m}\Big)\times\Big(\frac{1+\sqrt{3}m}{1+\sqrt{3}m}\Big)\\ &=\frac{\sqrt{3}m^3+4m^2+\sqrt{3}m}{1-3m^2}\\ &\cdots\vdots\cdots\\ m_{1}m_{2}+m_{2}m_{3}+m_{3}m_{1}&=\frac{9m^2-3}{1-3m^2}\\ &=-3\Big(\frac{1-3m^2}{1-3m^2}\Big)\\ &=-3. \end{split} \end{equation}$

Don’t forget to exclude $m=\pm\frac{1}{\sqrt{3}}$ .

Example 8

Let $m_{1},m_{2},m_{3}$ represent the slopes of an equilateral triangle. PROVE that $m_{1}+m_{2}+m_{3}+3m_{1}m_{2}m_{3}=0$ . In other words, we have: sum of slopes is $-3$ times product of slopes .

We characterized the slopes of equilateral triangles in Example 6, so let’s utilize that characterization. The slopes will be $m,~\frac{m-\sqrt{3}}{1+\sqrt{3}m}$ , and $\frac{m+\sqrt{3}}{1-\sqrt{3}m}$ .

$\begin{equation*} \begin{split} m\Big(\frac{m-\sqrt{3}}{1+\sqrt{3}m}\Big)\Big(\frac{m+\sqrt{3}}{1-\sqrt{3}m}\Big) &=\frac{m^3-3m}{1-3m^2}\\ m+\frac{m-\sqrt{3}}{1+\sqrt{3}m}+\frac{m+\sqrt{3}}{1-\sqrt{3}m}&=\frac{9m-3m^3}{1-3m^2}\\ &=-3\Big(\frac{m^3-3m}{1-3m^2}\Big). \end{split} \end{equation}$

Example 9

PROVE that the slopes of the sides of an equilateral triangle cannot ALL be positive .

Let the slopes be $m_{1},~m_{2},~m_{3}$ . From Example 8 above, we have that $m_{1}+m_{2}+m_{3}+3m_{1}m_{2}m_{3}=0$ . So, the slopes cannot all be positive — otherwise both the sum $m_{1}+m_{2}+m_{3}$ and the product $3m_{1}m_{2}m_{3}$ will be positive, making it impossible to have $m_{1}+m_{2}+m_{3}+3m_{1}m_{2}m_{3}=0$ .

Example 10

Let $\tan\theta=-1$ . PROVE that the slopes $\tan\theta, ~\frac{\tan\theta-\sqrt{3}}{1+\sqrt{3}\tan\theta},~\frac{\tan\theta+\sqrt{3}}{1-\sqrt{3}\tan\theta}$ form a geometric progression.

Let’s use the fact that three numbers $a,b,c$ form a geometric progression precisely when $b^2=ac$ . Then:

$\begin{equation*} \begin{split} (\tan\theta)^2&=(-1)^2\\ &=1\\ \Big(\frac{\tan\theta-\sqrt{3}}{1+\sqrt{3}\tan\theta}\Big)\times\Big(\frac{\tan\theta+\sqrt{3}}{1-\sqrt{3}\tan\theta}\Big)&=\frac{(\tan\theta)^2-3}{1-3(\tan\theta)^2}\\ &=\frac{(-1)^2-3}{1-3(-1)^2}\\ &=1\\ &\cdots\vdots\cdots\\ \Big(\frac{\tan\theta-\sqrt{3}}{1+\sqrt{3}\tan\theta}\Big)\times\Big(\frac{\tan\theta+\sqrt{3}}{1-\sqrt{3}\tan\theta}\Big)&=(\tan\theta)^2. \end{split} \end{equation}$

Takeaway

We’ve seen that equilateral triangles are naturally equipped with slopes in arithmetic and geometric progressions — depending on the “base inclination”. Interestingly, it turns out that every triangle contains a “sub-triangle” whose slopes form a geometric progression or an arithmetic progression. We’ll show you how this works in some subsequent posts.

Tasks

PROVE that the slopes of the sides of an equilateral triangle cannot all be negative.
Let be the slopes of the sides of a triangle. PROVE that the following conditions are equivalent:
- the triangle is equilateral;
- both $m_1}m_{2}+m_{2}m_{3}+m_{3}m_{1}+3=0$ and $m_{1}+m_{2}+m_{3}+3m_{1}m_{2}m_{3}=0$ hold.
In an equilateral , assume that the slopes of the sides are (all non-zero), and that the slopes of the medians from are , respectively. PROVE that:
- $m_{A}\times m_{BC}+m_{B}\times m_{CA}+m_{C}\times m_{AB}=-3$ ;
- if $m_{AB},m_{BC},m_{CA}$ form a geometric progression, then so do the median slopes $m_{A},m_{B},m_{C}$ .
(Unique quadratic) Suppose that the slopes of the sides of an equilateral triangle form a geometric sequence with common ratio . Then, in view of the previous exercise, the slopes of the three medians also form a geometric sequence; let its common ratio be . PROVE that:
- $r_{1}r_{2}=1$
- $r_{1}+r_{2}=-4$
- Hence, deduce that both $r_{1}$ and $r_{2}$ satisfy the quadratic equation $r^2+4r+1=0$ . (The two solutions of this quadratic equation are the only admissible values for the common ratio, should the slopes of the sides — and medians — of an equilateral triangle form geometric progressions.)
(Three nil) For an equilateral triangle with side slopes in geometric progression and median slopes (also in geometric progression, in view of a previous exercise), PROVE that:
- $m_{A}+m_{CA}=0$ ;
- $m_{B}+m_{BC}=0$ ;
- $m_{C}+m_{AB}=0$ .
(“Cancellation” property) If the slopes of the sides of an equilateral triangle $ABC$ form a geometric progression and $m_{BC}\times m_{CA}=1$ , PROVE that $m_{B}\times m_{A}=1$ also (thus appearing to “cancel” out $C$ ).
(We might be able to do some group theory here. $\cdots\vdots\cdots$ )
PROVE that the slopes of the sides of an equilateral triangle form an arithmetic progression if, and only if, one of the slopes is $0$ .
(“Cancellation” property) If the slopes of the sides of an equilateral triangle $ABC$ form an arithmetic progression and $m_{BC}+ m_{CA}=0$ , PROVE that $m_{B}+m_{A}=0$ also (thus appearing to “cancel” out $C$ ).
Under what condition(s) can the sum of the slopes of an equilateral triangle be equal to the product of the slopes?
(Unique sum) Let $m_{1},m_{2},m_{3}$ be the slopes of the sides of an equilateral triangle. PROVE that $m_{1},m_{2},m_{3}$ form a geometric progression if, and only if, $m_{1}+m_{2}+m_{3}=\pm 3.$