Slopefessor Fr1day – Page 35 – High School Geometry

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, $m_{1}+m_{2}+m_{3}=\pm 3.$

Easy condition for perpendicular medians

With us in the crew today are two perpendicular medians, who have something new to say. It’s possible you already knew what they’ll say, but let’s continue all the same. Or, you can jump straight to Example 8.

Join us to welcome perpendicular medians, chief host of today’s post!!!

Median lengths in terms of side lengths

Let $\triangle ABC$ be a triangle whose side lengths are $a,b,c$ . Throughout we’ll adopt the usual convention of denoting the lengths of the medians from vertices $A,B,C$ by $m_{a},m_{b},m_{c}$ , respectively.

In the above diagram, the median $AM$ has length $m_{a}$ . Also, $\angle AMB=\alpha$ , and so $\angle AMC=180^{\circ}-\alpha.$ Notice the side lengths of the triangle as indicated.

Example 1

PROVE that $m^2_{a}=\frac{2b^2+2c^2-a^2}{4},~m^2_{b}=\frac{2a^2+2c^2-b^2}{4},$ and $m^2_{c}=\frac{2a^2+2b^2-c^2}{4}.$

We only prove the first one, as the other two follow similarly (or by “symmetry”). Let’s use the cosine law to this end.

Applied to $\triangle ABM$ , the cosine law gives, for side $AB$ :

$\begin{equation*} \begin{split} AB^2&=AM^2+MB^2-2\times AM\times MB\times \cos(\angle AMB)\\ c^2&=m_{a}^2+\Big(\frac{a}{2}\Big)^2-2\times m_{a}\times \Big(\frac{a}{2}\Big)\times\cos(\alpha)\\ c^2&=m_{a}^2+\frac{a^2}{4}-am_{a}\cos(\alpha) \end{split} \end{equation}$

Similarly, the cosine law applied to $\triangle AMC$ gives, for side $AC$ :

$\begin{equation*} \begin{split} AC^2&=AM^2+MC^2-2\times AM\times MC\times\cos(\angle AMC)\\ b^2&=m_{a}^2+\Big(\frac{a}{2}\Big)^2-2\times m_{a}\times\Big(\frac{a}{2}\Big)\times\cos(180^{\circ}-\alpha)\\ b^2&=m_{a}^2+\frac{a^2}{4}-am_{a}\cos(180^{\circ}-\alpha)\\ b^2&=m_{a}^2+\frac{a^2}{4}+am_{a}\cos(\alpha)\\ \end{split} \end{equation}$

In the last step we used the trigonometric identity $\cos(180^{\circ}-\alpha)=-\cos(\alpha)$ . So we’ve got two equations:

(1) $\begin{equation*} c^2&=m_{a}^2+\frac{a^2}{4}-am_{a}\cos(\alpha). \end{equation*}$

(2) $\begin{equation*} b^2&=m_{a}^2+\frac{a^2}{4}+am_{a}\cos(\alpha). \end{equation*}$

Adding (1) and (2), the term involving $am_{a}\cos(\alpha)$ gets eliminated, leaving behind

$b^2+c^2=2m_{a}^2+\frac{2a^2}{4},$

which can be re-arranged to give $m^2_{a}=\frac{2b^2+2c^2-a^2}{4},$ as desired.

Example 2

PROVE that $m^2_{a}+m^2_{b}+m^2_{c}=\frac{3}{4}(a^2+b^2+c^2).$

We simply use the expressions from Example 1: $m^2_{a}+m^2_{b}+m^2_{c}$

$\begin{equation*} \begin{split} &=\Big(\frac{2b^2+2c^2-a^2}{4}\Big)+\Big(\frac{2a^2+2c^2-b^2}{4}\Big)+\Big(\frac{2a^2+2b^2-c^2}{4}\Big)\\ &=\frac{3}{4}\Big(a^2+b^2+c^2\Big). \end{split} \end{equation}$

For our purposes, the above relationship is very vital.

Example 3

PROVE that $m_{a}^2-m_{b}^2=\frac{3}{4}(b^2-a^2).$

Again, we use the expressions from Example 1:

$\begin{equation*} \begin{split} m_{a}^2-m_{b}^2&=\Big(\frac{2b^2+2c^2-a^2}{4}\Big)-\Big(\frac{2a^2+2c^2-b^2}{4}\Big)\\ &=\frac{3}{4}\Big(b^2-a^2\Big). \end{split} \end{equation}$

Example 4

PROVE that in any triangle, the longer sides have shorter medians.

We use the result of Example 3 above. Suppose that in $\triangle ABC$ we have $b>a$ . Since

$m_{a}^2-m_{b}^2=\frac{3}{4}(b^2-a^2),$

the right side is positive. The left side must be positive as well, so $m_{a}^2>m_{b}^2$ . Since median lengths are positive quantities, we can extract square roots to obtain $m_{a}>m_{b}$ .

If then there’s an ordering $c>b>a$ , it will follow that $m_{a}>m_{b}>m_{c}$ , using the transitivity property of inequalities.

Example 5

Find coordinates for the vertices of a $\triangle ABC$ that has the following property: $m_{a}=a.$

If all we needed were the side lengths, then we could use the equation $m_{a}=a$ and the fact that $m^2_{a}=\frac{2b^2+2c^2-a^2}{4}$ to obtain the relation $5a^2=2b^2+2c^2$ and thereby find suitable side lengths.

We’ll instead appeal to triangles whose slopes are in geometric progression, as these provide us with a ready-made solution. Consider $\triangle ABC$ below, with vertices at $A(1,9),~B(0,0),~C(-2,6)$ :

Using the given coordinates, we find, by the distance formula, that:

$\begin{equation*} \begin{split} AM^2&=(1--1)^2+(9-3)^2\\ &=40\\ AM&=\sqrt{40}\\ &=2\sqrt{10}\\ \therefore m_{a}&=2\sqrt{10}\quad\textrm{(\textbf{in view of the diagram})}\\ BC^2&=(-2-0)^2+(6-0)^2\\ &=40\\ BC&=\sqrt{40}\\ &=2\sqrt{10}\\ \therefore a&=2\sqrt{10}\quad\textrm{(\textbf{since BC=a, by usual notation})}. \end{split} \end{equation}$

So we obtain for the above triangle,whose vertices are $A(1,9),~B(0,0),~C(-2,6)$ , that $m_{a}=a$ .

Notice that the slopes of the sides of $\triangle ABC$ given above are $1,-3,9$ ; so they form a geometric progression with a common ratio of $-3$ . There’s something quite interesting that’s hidden there: in any triangle whose sides slopes form a geometric progression with a common ratio of $-3$ , there is a vertex $A$ such that $m_{a}=a$ .

Conditions for perpendicular medians

Example 6

PROVE that medians from vertex $A$ and vertex $B$ are perpendicular if, and only if, $a^2+b^2=5c^2$ .

First suppose that the median $AM$ through vertex $A$ and the median $BN$ through vertex $B$ are perpendicular. Consider the diagram below, where $G$ is the centroid of $\triangle ABC$ :

Since the centroid divides the median in the ratio $2:1$ , we have, in $\triangle AGB$ , that $AG=\frac{2}{3}m_{a}$ and $GB=\frac{2}{3}m_{b}$ . By assumption, $\triangle AGB$ is a right triangle, so the Pythagorean theorem applied to it gives:

$\begin{equation*} \begin{split} AB^2&=AG^2+GB^2\\ c^2&=\Big(\frac{2}{3}m_{a}\Big)^2+\Big(\frac{2}{3}m_{b}\Big)^2\\ c^2&=\frac{4}{9}(m_{a}^2+m_{b}^2)\\ c^2&=\frac{4}{9}\Big(\frac{2b^2+2c^2-a^2}{4}+\frac{2a^2+2c^2-b^2}{4}\Big)\\ c^2&=\frac{4}{9}\Big(\frac{a^2+b^2+4c^2}{4}\Big)\\ 9c^2&=a^2+b^2+4c^2\\ 5c^2&=a^2+b^2. \end{split} \end{equation}$

Conversely, if we suppose that the relation $a^2+b^2=5c^2$ holds, then we need to show that $\angle AGB=90^{\circ}$ . We’ll use the cosine rule to this end.

$\begin{equation*} \begin{split} \cos(\angle AGB)&=\frac{AG^2+GB^2-AB^2}{2\times AG\times GB}\\ &=\frac{\Big(\frac{2}{3}m_{a}\Big)^2+\Big(\frac{2}{3}m_{b}\Big)^2-c^2}{2\times \frac{2}{3}m_{a}\times\frac{2}{3}m_{b} }\\ &=\frac{\frac{4}{9}(m_{a}^2+m_{b}^2)-c^2}{\frac{8}{9}m_{a}m_{b}}\\ &=\frac{\frac{4}{9}\Big(\frac{2b^2+2c^2-a^2}{4}+\frac{2a^2+2c^2-b^2}{4}\Big)-c^2}{\frac{8}{9}m_{a}m_{b}}\\ &=\frac{\frac{4}{9}\Big(\frac{a^2+b^2+4c^2}{4}\Big)-c^2}{\frac{8}{9}m_{a}m_{b}}\\ &=\frac{\frac{4}{9}\Big(\frac{5c^2+4c^2}{4}\Big)-c^2}{\frac{8}{9}m_{a}m_{b}}\\ &=0\\ \angle AGB&=90^{\circ}. \end{split} \end{equation}$

This shows that the medians are perpendicular.

Example 7

PROVE that $a^2+b^2=5c^2$ if, and only if, $m^2_{a}+m^2_{b}=m^2_{c}$ .

We first prove: $a^2+b^2=5c^2\implies m^2_{a}+m^2_{b}=m^2_{c}$ .

$\begin{equation*} \begin{split} m^2_{a}+m^2_{b}&=\Big(\frac{2b^2+2c^2-a^2}{4}\Big)+\Big(\frac{2a^2+2c^2-b^2}{4}\Big)\\ &=\frac{a^2+b^2+4c^2}{4}\\ &=\frac{5c^2+4c^2}{4}\\ &=\frac{9c^2}{4}\\ m_{c}^2&=\frac{2a^2+2b^2-c^2}{4}\\ &=\frac{2(a^2+b^2)-c^2}{4}\\ &=\frac{2(5c^2)-c^2}{4}\\ &=\frac{9c^2}{4}\\ &=m^2_{a}+m^2_{b}. \end{split} \end{equation}$

Next, let’s prove $m^2_{a}+m^2_{b}=m^2_{c}\implies a^2+b^2=5c^2$ .

$\begin{equation*} \begin{split} \frac{2b^2+2c^2-a^2}{4}+\frac{2a^2+2c^2-b^2}{4}&=\frac{2a^2+2b^2-c^2}{4}\\ a^2+b^2+4c^2&=2a^2+2b^2-c^2\\ 5c^2&=a^2+b^2. \end{split} \end{equation}$

If all math proofs were this simple!

An easier, equivalent condition

What we’ve featured in our next example is somewhat absent in literature because it’s so miniature. Put differently, it is not found because it is not profound. But it still counts!

Example 8

PROVE that $a^2+b^2=5c^2$ if, and only if, $m_{c}=\frac{3}{2}c$ .

As you can see, this is extremely easy. To obtain the implication $a^2+b^2=5c^2\implies m_{c}=\frac{3}{2}c$ , begin with the expression for $m_{c}^2$ :

$\begin{equation*} \begin{split} m_{c}^2&=\frac{2a^2+2b^2-c^2}{4}\\ &=\frac{2(a^2+b^2)-c^2}{4}\\ &=\frac{2(5c^2)-c^2}{4}\\ &=\frac{9c^2}{4}\\ \therefore m_{c}&=\frac{3}{2}c. \end{split} \end{equation}$

Conversely, if $m_{c}=\frac{3}{2}c$ , then $m_{c}^2=\frac{9}{4}c^2$ . So using the expression for $m_{c}^2$ again, we have:

$\begin{equation*} \begin{split} \frac{2a^2+2b^2-c^2}{4}&=\frac{9}{4}c^2\\ 2a^2+2b^2-c^2&=9c^2\\ 2a^2+2b^2&=10c^2\\ a^2+b^2&=5c^2. \end{split} \end{equation}$

Example 9

For any $\triangle ABC$ with the usual notation, PROVE that each of the following statements implies the others:

median through vertex $A$ and median through vertex $B$ are perpendicular;
$a^2+b^2=5c^2$ ;
$m^2_{a}+m^2_{b}=m^2_{c}$ ;
$m_c=\frac{3}{2}c.$

Actually, this has been done in Example 6, Example 7, and Example 8; we just collect and connect them together here.

Notice that $(1)\iff(2)$ by virtue of Example 6. Also, $(2)\iff(3)$ is due to Example 7. Then, Example 8 verifies that $(2)\iff(4)$ .

(As a result, we obtain $(1)\iff(3)$ , $(1)\iff(4)$ , and $(3)\iff(4)$ . Each statement therefore implies the others.)

Example 10

Find coordinates for the vertices of a triangle that contains two perpendicular medians.

The easiest way to do this is to start with any three “consecutive” numbers, let’s say $1,2,3$ . Then, form three pairs as follows: $(1,2),(2,1),(3,3)$ . That’s it!!! Though with a little caveat: the order is important. You won’t get a triangle with something like $(1,1),(2,2),(3,3)$ .

To get a “bigger” triangle, we can space out the numbers, like so: $1,4,7$ , then use the three pairs $(1,4),(4,1),(7,7)$ as our triangle’s vertices, as shown below:

Notice the horizontal median $AM$ and the vertical median $BN$ . So these medians are perpendicular. With the given coordinates and the usual notation, let’s write out the three other equivalent conditions:

$\begin{equation*} \begin{split} a^2&=(7-4)^2+(7-1)^2\\ \therefore a^2&=45\\ b^2&=(7-1)^2+(7-4)^2\\ \therefore b^2&=45\\ c^2&=(1-4)^2+(4-1)^2\\ \therefore c^2&=18\\ &\cdots\vdots\cdots\\ a^2+b^2&=45+45\\ \therefore a^2+b^2&=90\\ 5c^2&=5\times 18=90\\ \therefore 5c^2&=a^2+b^2\\ &\cdots\vdots\cdots\\ m_{a}^2&=(1-5.5)^2+(4-4)^2\\ \therefore m_{a}^2&=20.25\\ m_{b}^2&=(4-4)^2+(1-5.5)^2\\ \therefore m_{b}^2&=20.25\\ m_{c}^2&=(7-2.5)^2+(7-2.5)^2\\ \therefore m_{}^2&=40.50\\ &\cdots\vdots\cdots\\ m_{a}^2+m_{b}^2&=20.25+20.25\\ &=40.50\\ \therefore m_{a}^2+m_{b}^2&=m_{c}^2\\ &\cdots\vdots\cdots\\ \textrm{Using}~m_{c}^2&=40.50\\ \textrm{We obtain}~m_{c}&=\sqrt{40.50}\\ &=4.5\sqrt{2}\\ \textrm{Using}~c^2&=18\\ \textrm{We obtain}~c&=\sqrt{18}\\ &=3\sqrt{2}\\ \therefore m_{c}&=\frac{3}{2}c. \end{split} \end{equation}$

In providing examples of perpendicular medians using coordinates, it is always convenient to use the special case in which one median is vertical and the other horizontal. But there are other cases in which the (two perpendicular) medians are neither vertical nor horizontal — for example, the triangle with vertices at $A(-2,\frac{1}{3}),B(-3,-4),C(4,-\frac{1}{3})$ is such. It may happen that we devote another post to those.

Takeaway

Of all the three other equivalent conditions for perpendicular medians, it is clear that $m_c=\frac{3}{2}c$ is the simplest in “appearance”, but maybe not in application — particularly because it uses the length of one of the medians, so one cannot tell immediately based on the side lengths.

If all the side lengths $a,b,c$ are provided, always use $a^2+b^2=5c^2$ in checking for perpendicular medians (assuming you’re concerned with the medians through vertices $A$ and $B$ ), but be aware of its equivalent formulations.

Lastly, to every triangle is a median triangle, whose side lengths are the lengths of the medians of the original triangle. Thus, the median triangle is a right triangle if, and only if, the original triangle contains two perpendicular medians.

Tasks

Suppose that $\triangle ABC$ is such that $m_{a}=\frac{\sqrt{3}}{2}a$ and $m_{b}=\frac{\sqrt{3}}{2}b$ . PROVE that the triangle is equilateral.
$\Big($ As shown in the next exercise, having $m_{a}=\frac{\sqrt{3}}{2}a$ alone is not enough to make a triangle equilateral. Also, observe that both $m_{a}=\frac{\sqrt{3}}{2}a$ and $m_{b}=\frac{\sqrt{3}}{2}b$ together imply that $m_{c}=\frac{\sqrt{3}}{2}c$ , so requiring all three conditions is asking for too much. $\Big)$
(This exercise shows that it is possible to have a median whose length is of the form $m_{a}=\frac{\sqrt{3}}{2}a$ , even when the triangle is not equilateral.) To this end, let , , and be the vertices of . PROVE that:
- $m_{a}^2=\frac{3}{4}a^2$
- $m_{b}^2=\frac{21}{4}b^2$
- $m_{c}^2=\frac{3}{4}b^2$
- $m_{c}^2=\frac{3}{28}c^2$
- $b^2+c^2=2a^2$
- $m_{b}^2+m_{c}^2=2m_{a}^2$
- the slopes of sides $AC,CB,BA$ form a geometric progression whose common ratio is $-2-\sqrt{3}.$
(Nice Nine)For any triangle with the usual notation, PROVE that the following NINE statements are equivalent:
- $m^2_{a}=\frac{3}{4}a^2;$
- $m^2_{b}=\frac{3}{4}c^2;$
- $m^2_{c}=\frac{3}{4}b^2;$
- $b^2+c^2=2a^2$ ;
- $m_{b}^2+m_{c}^2=2m_{a}^2$ ;
- $m^2_{a}+m^2_{b}=\frac{3}{4}(a^2+c^2)$ ;
- $m^2_{a}+m^2_{c}=\frac{3}{4}(a^2+b^2)$ ;
- $m^2_{b}+m^2_{c}=\frac{3}{2}a^2$ ;
- $m^2_{a}+m^2_{c}+m^2_{c}=\frac{9}{4}a^2$ .
Suppose that the side lengths $a,b,c$ of a $\triangle ABC$ are related via $c^2=k(a^2+b^2)$ . PROVE that $0< k < 2$ .
$\Big($ More generally, we can find constants $k_{1},k_{2},k_{3}$ such that $c^2=k_{1}(a^2+b^2),~b^2=k_{2}(a^2+c^2),~a^2=k_{3}(b^2+c^2)$ , and $0< k_{1}< 2,~0< k_{2}< 2, ~0< k_{3}< 2$ . $\Big)$
PROVE that there is no $\triangle ABC$ in which the side lengths $a,b,c$ satisfy $c^2=2(a^2+b^2)$ .
$\Big($ Just to buttress the point that the strict inequality in the preceding exercise is crucial. Simply calculate the length $m_{c}$ of the median from vertex $C$ to see why the claim is valid. $\Big)$
Let be such that the side lengths are related via . PROVE that:
- $\triangle ABC$ is either scalene or else equilateral;
- $m_{c}>m_{a}>m_{b}$ or $m_{b}>m_{a}>m_{c}$ (both for the scalene case);
- $\frac{1}{2}<\cos A<1$ (and so $0^{\circ}< A<60^{\circ}$ , for the scalene case).
Let $k$ be a positive constant. PROVE that $m_{c}=kc$ if, and only if, $(4k^2+1)c^2=2a^2+2b^2$ .
$\Big($ In particular, if $k=\frac{1}{2}$ , we obtain $m_{c}=\frac{1}{2}c\iff c^2=a^2+b^2$ , which expresses the popular fact that the length of the median to the hypotenuse of a right triangle is half the length of the hypotenuse. And, only right triangles have this property. $\Big)$
Suppose that $m_{c}=kc$ for some constant $k$ . What are the admissible values of $k$ ?
PROVE that $m_{a}^2+2m_{b}^2+3m_{c}^2=\frac{3}{4}\Big(3a^2+2b^2+c^2\Big)$ and that $m_a}^2+2m_{b}^2-3m_{c}^2=\frac{3}{4}\Big(3c^2-2b^2-a^2\Big)$ .
PROVE that the three medians in a right triangle satisfy $m_{a}^2+m_{b}^2=5m_{c}^2$ , where $c$ is the length of the hypotenuse.
PROVE that $m_{a}\geq \frac{(b+c)\sin B\sin C}{\sin B+\sin C}$ .
PROVE that $m_{a}=\frac{\sqrt{a^2+2bc}}{2}$ if, and only if, $\angle A=60^{\circ}$ .
PROVE that an equilateral triangle can never contain two perpendicular medians.
(Not everything goes the way of equilateral triangles after all.)
PROVE that the medians to the “two legs” in a right triangle can never be perpendicular.