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:

Rendered by QuickLaTeX.com

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:

Rendered by QuickLaTeX.com

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

  1. PROVE that the slopes of the sides of an equilateral triangle cannot all be negative.
  2. Let m_{1},m_{2},m_{3} 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.
  3. In an equilateral \triangle ABC, assume that the slopes of the sides AB,BC,CA are m_{AB},m_{BC},m_{CA} (all non-zero), and that the slopes of the medians from A,B,C are m_{A},m_{B},m_{C}, 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}.
  4. (Unique quadratic) Suppose that the slopes of the sides of an equilateral triangle form a geometric sequence with common ratio r_{1}. Then, in view of the previous exercise, the slopes of the three medians also form a geometric sequence; let its common ratio be r_{2}. 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.)
  5. (Three nil) For an equilateral triangle ABC with side slopes m_{AB},m_{BC},m_{CA} in geometric progression and median slopes m_{A},m_{B},m_{C} (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.
  6. (“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)
  7. 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.
  8. (“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).
  9. Under what condition(s) can the sum of the slopes of an equilateral triangle be equal to the product of the slopes?
  10. (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.