For in which sides have slopes , respectively, the equation connecting their side lengths is

(1)

as we saw a post ago. The equation connecting their median lengths is

(2)

and is today’s goal.

Compare the *relationship between the medians* of a right triangle with the medians of a triangle with slopes .

This is the final, knockout round of our *compare-contrast-contest* with the right triangle. The contest started on June 28, with the right triangle currently having an edge.

- In a right triangle, the medians are related via , where is the length of the median to the hypotenuse (see the exercises). The triangle with slopes doesn’t satisfy this relation, unless we put in (2).

Winner: Right triangle; Score is . - The previous score was , in favour of the right triangle.

Final verdict: Right triangle wins on aggregate. Case closed.

We still celebrate the triangle with slopes , despite been rightly trounced by the right triangle.

Consider with vertices , , and . Verify that three medians satisfy equation (2).

Note that the slopes of sides are , respectively. They form a geometric progression with . The midpoints of these sides have been marked. Thus, by the distance formula:

YEA.

Consider with vertices at , , and . Verify that the three medians satisfy equation (2).

Easy-peasy. Observe that the slopes of sides are . They form a geometric progression with common ratio . Let’s draw a diagram:

The median lengths then follow by the distance formula. We have:

There we go. The formula works for both positive and negative common ratios.

In , the median lengths are related via . Find (possible) coordinates for the vertices .

Try solving this problem in a way other than what we present below.

In (2), set and solve for :

The workload reduces to finding coordinates for a triangle whose slopes form a geometric progression with common ratio .

Easy-peasy.

Among the *many* options we have for the vertices of a triangle with side slopes , we can take , , and With , these become , , and . Use these as . That is, take , , and as the triangle’s vertices. A quick calculation shows that , , and (as in, *triple three*).

Watch what happens when in both (1) and (2).

Put in (1). Obtain .

Put in (2). Obtain .

These are well-known, equivalent conditions for the medians from and to be perpendicular.

Is it surprising that it’s our *favourite* -value that saved the day?

For , let . Evaluate , and show that they are all sums of *two consecutive squares*.

Observe that the only integer values of that produce integer values of are .

In addition, defines a bijection .

In , we have and . Find the value of .

You can suspect what will be. From (1) and (2), set

Clear fractions, and simplify to obtain the quartic polynomial equation

The second quadratic factor, , is *real**ly* irreducible (its *discriminant* is ). So the only real roots come from the first one:

Thus, for , we have:

Similarly, for , we have:

The value of is , as you may have correctly guessed.

OK, the above approach was definitely an overill for something that can be obtained in a simpler way, but it just confirms that our theory conforms to what is more established. See the next example for a better strategy.

In any , PROVE that the following two statements are equivalent:

- the relations and hold
*simultaneously*; - .

Consider and let’s default to standard notation. Use the fact that the median lengths are related to the side lengths via:

(3)

from which . Now, since , this becomes . Also, becomes . Since , we have:

In view of , we see that can’t be negative (in fact, : example 10), so discard , and take as the only valid solution. Thus, .

To see that , suppose that . Assume that . We’ll show that as well. Indeed, from (3):

Similarly, if we started with , we’ll again arrive at . This proves .

Bottom line: the two statements are equivalent as claimed.

Kindly bear with us for switching back and forth between the standard notation and . Hopefully this doesn’t cause any confusion.

In any , if and , PROVE that .

The relationship

holds in any triangle. Using the fact that and , we have:

Although we freely cancelled out the common factor , it shouldn’t obscure the fact that the relations and hold *simultaneously* only when .

If , PROVE that .

Compare with exercise 5 below.

For the proof, begin with one of the expressions in (3):

Since the left side is positive, so must the right side be, and we only have to impose this restriction on the non*k*onstant .

, as desired.

## Takeaway

In the equation

the median from vertex and the median from vertex are perpendicular *only* when . If the focus shifts to the medians from and , or the medians from and , then other values of can still ensure perpendicularity.

## Tasks

- Find possible coordinates for the vertices of if its three medians are related via .
- Find possible coordinates for the vertices of if its three medians are related via .
- PROVE that the medians in a right triangle are related via , where is the length of the median to the hypotenuse.
- Derive equation (2).
- Let and . PROVE that both and have the
*same*minimum value, but that their common minimum occurs at*different*values of . - Consider the rational function :
- Verify equation (2) for a triangle with vertices at , , and .
- PROVE that if a right triangle has slopes in geometric progression with common ratio , then both the squares of the median lengths and the squares of the side lengths form arithmetic progressions.
- PROVE that the “mapping” , is not
*well-defined*. What modifications should be made to the domain or range, so that the mapping can be well-defined? - If and , PROVE that is then a sum of two consecutive squares.