# The depressed cubic as area

We investigated the area property of our very own triangle, and guess what! The depressed cubic showed up.

Ww!!! How???

Consider with vertices at , where . PROVE that its area can be given by .

Easy-peasy.

We were given the coordinates . We simply use the well-known area formula

and, by taking , , , we obtain

Observe that the numerator contains three consecutive factors, namely , and . Thus, if is an integer, then at least one of the three will be even, and so the resulting product will be even. This means that the area will be an integer — and a multiple of a triangular number.

## The depressed cubic

A cubic equation of the form in which the second degree term is missing, is usually called a depressed cubic. For a discussion on how to solve such equations, see here.

Suppose we re-arrange our area formula as

then we obtain not just a depressed cubic, but a restricted, simplified, specialized, depressed cubic. As such, its solution also takes a relatively simplified form.

PROVE that the area formula obtained previously, namely , is an odd function.

Recall that an odd function satisfies . Let’s write . Then

One implication of this “oddness” is that two opposite values of will yield the “same” area. As such, when we solve the depressed cubic equation

for , we must remember that any solution we obtain has a corresponding opposite, but the associated opposite doesn’t arise as a solution of the depressed cubic. For example, if is a solution of the depressed cubic, then we must also consider , despite the fact that is not a solution of the depressed cubic (see the exercises at the end).

For the depressed cubic , PROVE that can be given in terms of by:

Let’s do this!

Following the standard method of solving a depressed cubic, we seek and such that and . Then, is a solution of . Indeed, we have:

So it remains to find such and that satisfy and . Isolate from the second equation: . Substituting, the equation becomes:

Clear fractions and bring all terms to the left side:

Since this is now a quadratic in , we can use the quadratic formula to isolate . Doing this gives:

Simplifying,

It suffices to take . Then, gives :

Finally,

Using the preceding example, PROVE that if, and only if, .

First suppose that . Then . And so

reduces to

Conversely, if , then from the depressed cubic equation , we obtain

Discuss the nature of the roots of the depressed cubic .

We use the cubic discriminant to this end. For a general cubic , its discriminant is given by the quantity

Putting , , , and , we obtain, for the depressed cubic :

which reduces to . Consequently:

• if , then the depressed cubic has three real roots, one of which is repeated;
• if , then the depressed cubic has three distinct real roots;
• if , the depressed cubic has one real root and two complex roots.

Solve the depressed cubic equation .

Observe that this is the case in which . The corresponding cubic discriminant becomes . As such, we expect a repeated root. From Example 4, one of the roots is . Accordingly, is a factor of . By long division, we get the quadratic factor , which is a perfect square:

Thus, the complete solution to is . Notice how this solution set becomes “increased” in the next example.

#### Example

has an area of . The slopes of its sides form a geometric sequence . Find possible coordinates for the vertices .

Since the slopes of the sides form the geometric sequence , we can place the vertices at , , and . The area will then be , in view of Example 1. But we’re given that the area is :

which is the depressed cubic solved in the preceding example. Recall that the cubic has as solutions.

Being that the area formula is an odd function (Example 2), we should also consider as possible values for the common ratio , even though these are not solutions to the depressed cubic.

• for , the coordinates will be , , ;
• for , the coordinates will be , , ;
• for , the coordinates will be , , ;
• for , the coordinates will be , , .

Since the area is so small, each case produces a rare-listic diagram when drawn.

#### Example (“Fixed point”)

Find coordinates for the vertices of a whose slopes are and whose area is sq. units.

We solve the cubic equation using :

the quadratic part has complex roots. Again, we should consider , despite the fact that it’s not a solution to the current cubic equation.

Next, let’s use the coordinates , , :

In this case, the area and the common ratio are the same.

Give an example of a scalene triangle with non-zero side slopes, irrational side lengths, and an integer area.

The only challenge with this question is the non-zero slopes requirement; otherwise we could just have taken a triangle with one side on the -axis and the rest becomes easy.

Luckily, we can always turn to triangles with slopes in geometric progressions. In this case we take with vertices at . The side slopes are . For the side lengths we have , , and , all irrational and all different, so the triangle is scalene. Its area is sq. units.

Let be a root of the depressed cubic . PROVE that .

Since (an area), it follows that as well. This ensures that the quantity is defined.

Set

Comparing coefficients of like terms, , and . So , , and again . The two expressions for yield , which is valid because is a root of the original cubic .

## Takeaway

If with vertices at is such that the slopes of sides are , respectively, then the area of the triangle can be given by

By restricting the coordinates in a suitable way (see the exercises below), we obtain the depressed cubic as the area.

1. (Special square) For any right triangle with vertices , , and , PROVE that:
(Notice how squaring respects addition in this case. In addition, not all four absolute values on the left are absolutely necessary .)
2. (Special square) For any triangle with centroid , circumcenter , and orthocenter , PROVE that:
(Notice the absolute absence of absolute values in this case — so long as and . Also, any right triangle satisfies this property “TWICE” — with its regular vertices, and also with its three principal centers.)
3. (Opposite roots) PROVE that if a depressed cubic has two opposite solutions, namely and , then .
(In our own case, the right member represents the area of a certain, obscure triangle. Since , this explains why the depressed cubic equation cannot have two opposite solutions.)
4. has vertices at . Assume that the slopes of sides are , in that order.
• PROVE that its area can be given by
• Find an appropriate choice of for which the area reduces to the depressed cubic
5. has vertices at , , . PROVE that:
• the coordinates of the foot of the altitude from vertex is ;
• the length of the altitude from vertex is ;
• the area of is .
6. Consider with vertices at , , . Its side slopes are , a geometric progression in which and .
• Use the area formula in exercise 4 above to compute its area;
• Use the area formula in exercise 5 above to compute its area;
• Compute its area in the regular way, using ;
• Explain why the formula in exercise 5 did not work correctly in this case.
7. For with vertices at , , , PROVE that:
• its centroid is located at ;
• its circumcenter is at ;
• its orthocenter is at ;
• the slope of its Euler line is .
8. (Same ratio) Given , , , PROVE that the centroid, the circumcenter, and the orthocenter of lie on parallel lines , , and , respectively.
(Notice that the line through the centroid is in the middle of the other two lines. Moreover, these lines preserve the usual ratio in which the centroid divides the circumcenter-orthocenter distance.)
9. If , PROVE that .
10. (Unit area) Find coordinates for the vertices of with slopes and an area of sq. unit.