Anyone who’s taken a stat course before may inadvertently misconstrue the title as “*mean, median, mode*” — and thereby get a slight shock, because we actually called “*mean median slope*”. After one recovers from this shock, one discovers that this post covers a leftover from our first post in this October. There, we went over some things hurriedly, which made this carryover necessary.

## Excursion

As we hit the road, have your pencil, paper, and patience ready.

Find the slopes of the three medians of with vertices at , , .

The midpoints of sides , , and are , , and , respectively. Thus, the slopes of the medians are:

What do you notice about the reciprocals ?

Find the slopes of the three medians of with vertices at , , .

The midpoints of sides are , , , respectively. Thus, the slopes of the medians are:

Re-arrange them as . What do you notice?

PROVE that if a triangle has __one side parallel to the -axis__, then the __reciprocals__ of the slopes of the three medians form an arithmetic progression.

In the above diagram, let’s require that , so that side is parallel to the -axis.

Normally, the slopes of the medians from vertices are:

If we now put and consider the reciprocals

then

Thus, the reciprocals of the median slopes form an arithmetic progression, and the arithmetic mean of this progression is the slope of the median from the vertex opposite the side parallel to the -axis.

In the special case when , we have that the median from vertex is __vertical__, and so its slope is __undefined__. Nevertheless, since we’re concerned with the reciprocals of the slopes, the conclusion still holds. In fact, we obtain a special arithmetic progression , where .

Express the common difference of the arithmetic progression in the previous example in terms of the slopes of the parent triangle.

The arithmetic progression is . The common difference is the difference between consecutive terms; for example, :

(1)

Consider the difference between the reciprocals of the slopes of sides and :

(2)

It follows from (1) and (2) that .

## converse results: click to read

The converse of example 3 and example 5 below will be shown in example 9 and example 8.

## clockwise rotation: click to read

Having established the two preceding examples, the next two can be proved using rotations — specifically, a ninety degree (clockwise or counter-clockwise) rotation about a convenient center. Nevertheless, we’ll still use a direct method.

PROVE that if a triangle has __one side parallel to the -axis__, then the slopes of the three medians form an arithmetic progression.

Let the vertices be placed at , , . Here, side is parallel to the -axis. The slopes of the three medians are:

and so

Easy stuff.

Express the common difference of the arithmetic progression in the previous example in terms of the slopes of the parent triangle.

Let the common difference be . Using the median slopes in the preceding example, we have:

Consider the difference between the slopes of sides and :

Thus, .

## Exception

In fact, exceptional. In the case of a right triangle with legs parallel to the coordinate axes, something interesting happens.

PROVE that if a *right triangle* has legs parallel to the coordinate axes, then the slopes of the three medians form a *geometric progression* with , and when re-arranged, form an *arithmetic progression*.

Suppose the situation is pictured below:

and let’s apply, as an informal argument, an exercise from our immediate past post. That exercise says that if form an arithmetic progression and at the same time the reciprocals form another arithmetic progression (in the same order), then the relation holds — which is a condition for to be a geometric progression.

In the present case, because side is parallel to the -axis, we have (by example 3) that the reciprocals of the median slopes form an arithmetic progression. Similarly, since side is parallel to the -axis, we have (by example 5) that the median slopes form an arithmetic progression. Together, we arrive at a scenario described in the previous paragraph. This sheds a little light on why the median slopes behave specially in this case.

We can continue this informal argument to show that the common ratio of the geometric progression is , but there’s a simpler, more direct alternative:

The geometric progression is and the common ratio is . At the same time, we have an arithmetic progression because .

## Extensions

Suppose that the slopes of the medians of form an arithmetic progression with common difference . PROVE that side is parallel to the -axis.

As you’ll notice, this is the *converse* of example 5. The proof is simple, but involves some algebraic manipulations. So we present a snapshot below.

Let the vertices be , , . Since , we have:

(3)

After some simplifications, this reduces to:

The expression in the first square bracket, namely

cannot be zero (it actually represents twice the triangle area). This forces

(4)

(5)

Simplifying, we obtain

(6)

Thus, , showing that side is parallel to the -axis.

Suppose that the *reciprocals* of the slopes of the medians of form an arithmetic progression with common difference . PROVE that side is parallel to the -axis.

Because the proof is similar to the previous one, we omit it. Just take our word for its validity; better still, try it as an activity.

Consider two right triangles in which the legs have the same slopes, but the *hypotenuse of one is parallel to the -axis*, while the *hypotenuse of the other is parallel to the -axis*. PROVE that in both cases, the common differences of the arithmetic progressions formed by the slopes of the medians are equal.

One may think that this has something to do with similarity, but it’s more than that.

The reason it happens is simple. In example 4, the common difference of the *reciprocals* of the median slopes when one side is parallel to the -axis was given by

In In example 6, the common difference of the median slopes when one side is parallel to the -axis was

If these two common differences should be equal, then

So the triangle must be a *right* triangle. Similarity is not enough. To see this, consider with vertices at , , . The *reciprocals of the slopes of the three medians * are:

Now consider with vertices at , , . Notice that is similar to . However, the slopes of the three medians of are

## Takeaway

For any , the following two conditions are *equivalent*:

- side is parallel to the -axis;
- the
*reciprocals*of the median slopes form an arithmetic progression , , with common difference .

Similarly, the following two conditions are *equivalent* for any :

- side is parallel to the -axis;
- the median slopes form an arithmetic progression , , with common difference .

Beautiful, simple things.

## Tasks

- (Iso
*six*less triangle) Let be*isosceles*such that .- if the base is
*parallel to the -axis*, PROVE that ; - if the base is
*parallel to the -axis*, PROVE that .

(Beautiful scenes. More in exercise 2 below.)

- if the base is
- (Iso
*six*less triangle) Consider with vertices at , , , where . (Side is parallel to the -axis.)- PROVE that .
- Deduce that if, and only if, .

(Among all triangles with*one side parallel to the -axis*,*only*the isosceles triangle has the property that . Hence the name iso*six*less.)

- (Equilateral triangle) Let be
*equilateral*with side-slopes , , and median-slopes , , .- PROVE that .
- Using the above and exercise 1, deduce that if side is parallel to the -axis, then the slopes of the other two sides are .
- PROVE that the side-slopes form an
*arithmetic progression*if, and only if, - PROVE that the side-slopes form a
*geometric progression*if, and only if,

(Beautiful scenes.)

- Verify that the solutions to the polynomial equation are , , .
- (Reversible sequence) PROVE that any
*three-term*geometric progression with common ratio can always be re-arranged to form a*three-term*arithmetic progression. - Let be distinct numbers and suppose that form an arithmetic sequence while form a geometric sequence. PROVE that the common ratio of the geometric sequence is necessarily . (Compare with the preceding exercise.)
- Let , , and denote the slopes of medians from vertices , , and . Find a triangle for which the following three relations hold
*simultaneously*:- ;
- ;
- . (Implied by the first two.)

- If the “base” of an
*isosceles*triangle is parallel to the -axis, PROVE that the slopes of the two equal sides are opposites of each other. Thus, the side-slopes will form an arithmetic progression of the form . Furthermore, PROVE that:- .

(If the “base” of an*isosceles*triangle is parallel to the -axis, the slopes of the three medians form an arithmetic progression of the form .)

- .
- Consider with vertices at , , , where is fixed.
- Verify that the median slopes are , , , and that they form an arithmetic progression with common difference .

(It’s somewhat interesting that this common difference stays the same as varies.) - Find the value of for which the arithmetic sequence above becomes a geometric sequence with common ratio .

- Verify that the median slopes are , , , and that they form an arithmetic progression with common difference .
- Find coordinates for the vertices of in which:
- the slopes of the three medians are .
- the slopes of the three medians are .
- the slopes of the three medians are .
- the slopes of the three medians are .
- the slopes of the three medians are .

(Since the median slopes form an arithmetic progression with common difference , one can use the preceding exercise and just substitute suitable values of . However, there other possibilities and procedures that yield the same results.)