The chances that you’ve seen the quadratic equations

(1)

(2)

are not slim. We now show you that they spring from familiar scenes.

## Equilateral triangles

#### Example

If the slopes of the sides of an *equilateral triangle* are , PROVE that .

In a previous post we showed that the side lengths of any triangle with slopes are related via

(3)

In the case of an equilateral triangle we put in (3) and obtain

from which follows after simplification.

That was too simple, aided by (3). Below we provide an alternate derivation.

Let be the common ratio of the slopes of the sides of an *equilateral triangle*, and let be the common ratio of the slopes of the medians. PROVE that .

To see this, first observe that if the slopes of the sides of a triangle are , then the slopes of the medians to these sides are

(4)

respectively. For an equilateral triangle, the three medians are perpendicular to the sides, so:

(5)

The first and third ones give . Divide by the common factor :

(6)

From (4), the common ratio for the slopes of the medians is . Like so:

(7)

where is the original common ratio. Equation (6) may then be re-written as

Square roots: . We claim that is not possible. Otherwise, suppose we have . Use the fact that to obtain .

Now use and to get , whence . From slope consideration, this is impossible for the common ratio of a geometric progression, and so ules out the possibility of .

This forces .

While having is enough for what we want to achieve, it is important to point out that one can also have , depending on the arrangement of the geometric progressions. Hopefully this doesn’t cause any confusion.

Let be the common ratio of the slopes of the sides of an *equilateral triangle*, and let be the common ratio of the slopes of the medians. PROVE that .

From the preceding example, . Use this in (7):

Simple.

Use examples 2 and 3 to deduce that in an *equilateral triangle* with side slopes , the quadratic equation is satisfied by the common ratio of the slopes of the sides and the common ratio of the slopes of the medians.

Since (view this as product of roots) and (sum of roots), we get . Similarly, .

Solve the quadratic equation and interpret the solution(s) in the context of equilateral triangles.

The discriminant of the quadratic is . Since it is positive but not a perfect square, we expect two real, irrational solutions. By the quadratic formula:

Should the slopes of the sides (and medians) of an equilateral triangle form a geometric progression, the common ratio of the progression is restricted to either or . Talk about estictions.

## Right isosceles triangles

If the slopes of the sides of a *right isosceles triangle* are , PROVE that .

The side lengths will satisfy (3):

where is the length of the side with slope , is the length of the side with slope , and is the length of the side with slope .

Since cannot be the hypotenuse, we’ll choose either or as the hypotenuse. To be specific, let be the hypotenuse.

The isosceles condition then forces . And:

By the Pythagorean theorem, . As , this yields . Together with , we obtain

Solve the equation and interpret the solution(s) in relation to right isosceles triangles.

The discriminant is . Since the discriminant is positive but not a perfect square, we expect two real, irrational solutions.

By the quadratic formula:

Should the slopes of the sides of a right isosceles triangle form a geometric progression, there are only two possible values for the common ratio of the progression, namely or .

Let be the slopes of the sides of a right isosceles triangle, with . PROVE that .

The side with slope cannot be the hypotenuse. Let the hypotenuse be the side with slope . Then the slopes of the legs are and . As such :

where was obtained by first setting then squaring both sides to get , and then finally solving the linear-quadratic system to get , (or , ).

Due to its connection to the golden ratio, we’ll have more to say about this example in a later post.

Let and . PROVE that and .

By direct calculation:

Of course, both and yield .

Let and . PROVE that and

As in the previous example, proceed by direct calculation:

Several other nice relations exist between and .

## Takeaway

At the level of slopes, equilateral triangles and right isosceles triangles are very close, as these equations show:

The slopes must form a geometric progression though.

## Tasks

- (T w o nice) Let be the slopes of the sides of a triangle.
- If the triangle is
*equilateral*, PROVE that*the sum of product of the slopes, taken two slopes at a time, is always negative three*. That is, ; - If the triangle is
*right isosceles*, PROVE that*the sum of product of the slopes, taken two slopes at a time, is always twice the common ratio*. That is, .

- If the triangle is
- In , suppose that the slopes of sides are , respectively. PROVE that:
- the slope of the median from is ;
- the slope of the median from is ;
- the slope of the median from is .

- In , let be the slopes of sides , in that order.
- If , PROVE that (Use exercise 2 above together with the fact that the median from vertex to side is actually an altitude due to the fact that );
- If and , deduce that . (This provides another means of deriving equation (2).)

- PROVE that if the slopes of the
*sides*of a triangle are (), then the slopes of the*medians*do not form a geometric progression. (Use exercise 2 above.) - Suppose that the slopes of the
*sides*of a triangle form a geometric progression with common ratio , and that the slopes of the*medians*form a geometric progression with common ratio . PROVE that:- ;
- ;
- and cannot have the same
*signs*; - and cannot both be integers.

- In , suppose that sides have slopes , respectively. PROVE that:
- the median from vertex and the median from vertex are
*perpendicular*if and only if , ; - the median from vertex and the median from vertex are
*perpendicular*if and only if . Since this equation has*no real*solutions, does it imply that the median from vertex and the median from vertex can*never*be perpendicular?

- the median from vertex and the median from vertex are
- Consider with vertices at , , and . PROVE that:
- the slopes of the sides form a geometric progression in which ;
- the median from vertex and the median from vertex are perpendicular.

- Find coordinates for the vertices of in which:
- the slopes of sides form a geometric progression with ;
- the median from vertex and the median from vertex are perpendicular.

- Give an example of an
*isosceles*triangle with slopes in which equation (1), namely , is satisfied. (There are isosceles triangles (that are not equilateral) that satisfy equation (1).) - Let . PROVE that is a perfect square for . Are these the only for which is a perfect square?