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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:

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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:

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    \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}


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.


  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.

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.

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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


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


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):

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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:

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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:

  1. median through vertex A and median through vertex B are perpendicular;
  2. a^2+b^2=5c^2;
  3. m^2_{a}+m^2_{b}=m^2_{c};
  4. 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:

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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.


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.


  1. 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)
  2. (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 A(1,7+4\sqrt{3}), B(0,0), and C(-1-\sqrt{3},5+3\sqrt{3}) be the vertices of \triangle ABC. 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}.
  3. (Nice Nine)For any triangle ABC 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.
  4. 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)
  5. 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)
  6. Let \triangle ABC be such that the side lengths a,b,c are related via 2a^2=b^2+c^2. 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).
  7. 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)
  8. Suppose that m_{c}=kc for some constant k. What are the admissible values of k?
  9. 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).
  10. 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.
  11. PROVE that m_{a}\geq \frac{(b+c)\sin B\sin C}{\sin B+\sin C}.
  12. PROVE that m_{a}=\frac{\sqrt{a^2+2bc}}{2} if, and only if, \angle A=60^{\circ}.
  13. PROVE that an equilateral triangle can never contain two perpendicular medians.
    (Not everything goes the way of equilateral triangles after all.)
  14. PROVE that the medians to the “two legs” in a right triangle can never be perpendicular.