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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 \triangle ABC with vertices at A(1,r^2),B(0,0),C(1+r,r+r^2), where 0,\pm 1\neq r. PROVE that its area \Delta can be given by \Delta=\frac{r^3-r}{2}.


We were given the coordinates A(1,r^2),B(0,0),C(1+r,r+r^2). We simply use the well-known area formula


and, by taking (x_1,y_1)=(1,r^2), (x_2,y_2)=(0,0), (x_3,y_3)=(1+r,r+r^2), we obtain

    \begin{equation*} \begin{split} \Delta&=\frac{r^2(1+r-0)+0(1-(1+r))+(r+r^2)(0-1)}{2}\\ &=\frac{r^2(1+r)+(r+r^2)(-1)}{2}\\ &=\frac{r^2(r+1)-r(r+1)}{2}\\ &=\frac{(r+1)(r^2-r)}{2}\\ &=\frac{(r-1).r.(r+1)}{2}\\ \textrm{Or,}~\Delta&=\frac{r^3-r}{2} \end{split} \end{equation*}

Observe that the numerator contains three consecutive factors, namely r-1,r, and r+1. Thus, if r is an integer, then at least one of the three will be even, and so the resulting product (r-1).r.(r+1) will be even. This means that the area \Big(=\frac{(r-1).r.(r+1)}{2}\Big) will be an integer — and a multiple of a triangular number.

The depressed cubic

A cubic equation of the form \boxed{x^3+Ax=B} in which the second degree term x^2 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 \Delta=\frac{r^3-r}{2} as

    \[r^3-r = 2\Delta,\]

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 \Delta=\frac{r(r^2-1)}{2}, is an odd function.

Recall that an odd function f(x) satisfies f(-x)=-f(x). Let’s write \Delta(r)=\frac{r(r^2-1)}{2}. Then

    \begin{equation*} \begin{split} \Delta(-r)&=\frac{(-r)\Big((-r)^2-1\Big)}{2}\\ &=\frac{(-r)\Big(r^2-1\Big)}{2}\\ &=-\frac{r\Big(r^2-1\Big)}{2}\\ &=-\Delta(r) \end{split} \end{equation*}

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


for r, 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 r=2 is a solution of the depressed cubic, then we must also consider r=-2, despite the fact that r=-2 is not a solution of the depressed cubic (see the exercises at the end).

For the depressed cubic r^3-r=2\Delta, PROVE that r can be given in terms of \Delta by: \boxed{r=\sqrt[3]{\frac{9\Delta+\sqrt{81\Delta^2-3}}{9}}-\sqrt[3]{\frac{-9\Delta+\sqrt{81\Delta^2-3}}{9}}}

Let’s do this!

Following the standard method of solving a depressed cubic, we seek s and t such that 3st=-1 and s^3-t^3=2\Delta. Then, r=s-t is a solution of r^3-r=2\Delta. Indeed, we have:

    \begin{equation*} \begin{split} r^3-r&=(s-t)^3-(s-t)\\ &=(s^3-3s^2t+3st^2-t^3)-(s-t)\\ &=(s^3-t^3)-3st(s-t)-(s-t)\\ &=2\Delta-(-1)(s-t)-(s-t)\quad\textrm{by assumption}\\ &=2\Delta+(s-t)-(s-t)\\ &=2\Delta \end{split} \end{equation*}

So it remains to find such s and t that satisfy s^3-t^3=2\Delta and 3st=-1. Isolate t from the second equation: t=-\frac{1}{3s}. Substituting, the equation s^3-t^3=2\Delta becomes:

    \[s^3-\Big(-\frac{1}{3s}\Big)^3=2\Delta \implies s^3+\frac{1}{27s^3}=2\Delta\]

Clear fractions and bring all terms to the left side:

    \[27s^6-(54\Delta)s^3+1=0 \implies 27(s^3)^2-(54\Delta)s^3+1=0\]

Since this is now a quadratic in s^3, we can use the quadratic formula to isolate s^3. Doing this gives:

    \[s^3=\frac{54\Delta\pm\sqrt{(54\Delta)^2-4\times 27\times 1}}{2\times 27}\]



It suffices to take s=\sqrt[3]{\frac{9\Delta+\sqrt{81\Delta^2-3}}{9}}. Then, s^3-t^3=2\Delta gives t^3=s^3-2\Delta:

    \begin{equation*} \begin{split} t^3&=\frac{9\Delta+\sqrt{81\Delta^2-3}}{9}-2\Delta\\ &=\frac{9\Delta+\sqrt{81\Delta^2-3}}{9}-\frac{18\Delta}{9}\\ &=\frac{-9\Delta+\sqrt{81\Delta^2-3}}{9}\\ \implies t&=\sqrt[3]{\frac{-9\Delta+\sqrt{81\Delta^2-3}}{9}} \end{split} \end{equation*}



Using the preceding example, PROVE that \Delta=\frac{\sqrt{3}}{9} if, and only if, r=\frac{2\sqrt{3}}{3}.

First suppose that \Delta=\frac{\sqrt{3}}{9}. Then 81\Delta^2-3=0. And so


reduces to

    \begin{equation*} \begin{split} r&=\sqrt[3]{\Delta}-\sqrt[3]{-\Delta}\\ &=2\sqrt[3]{\Delta}\\ &=2\sqrt[3]{\frac{\sqrt{3}}{9}}\\ &=\frac{2\sqrt{3}}{3} \end{split} \end{equation*}

Conversely, if r=\frac{2\sqrt{3}}{3}, then from the depressed cubic equation r^3-r=2\Delta, we obtain

    \begin{equation*} \begin{split} \Delta&=\frac{r^3-r^}{2}\\ &=\frac{r(r^2-1)}{2}\\ &=\frac{\frac{2\sqrt{3}}{3}\Big[\Big(\frac{2\sqrt{3}}{3}\Big)^2-1\Big]}{2}\\ &=\frac{2\sqrt{3}}{3}\times\frac{1}{3}\times\frac{1}{2}\\ &=\frac{\sqrt{3}}{9} \end{split} \end{equation*}

Discuss the nature of the roots of the depressed cubic r^3-r=2\Delta.

We use the cubic discriminant to this end. For a general cubic ax^3+bx^2+cx+d, its discriminant is given by the quantity


Putting a=1, b=0, c=-1, and d=-2\Delta, we obtain, for the depressed cubic r^3-r=2\Delta:


which reduces to 4-108\Delta^2. Consequently:

  • if 4-108\Delta^2=0, then the depressed cubic has three real roots, one of which is repeated;
  • if 4-108\Delta^2> 0, then the depressed cubic has three distinct real roots;
  • if 4-108\Delta^2< 0, the depressed cubic has one real root and two complex roots.

Solve the depressed cubic equation r^3-r=2\frac{\sqrt{3}}{9}.

Observe that this is the case in which \Delta=\frac{\sqrt{3}}{9}. The corresponding cubic discriminant 4-108\Delta^2 becomes 4-108\times\frac{1}{27}=0. As such, we expect a repeated root. From Example 4, one of the roots is r=\frac{2\sqrt{3}}{3}. Accordingly, \Big(r-\frac{2\sqrt{3}}{3}\Big) is a factor of r^3-r-2\frac{\sqrt{3}}{9}. By long division, we get the quadratic factor \Big(r^2+\frac{2\sqrt{3}}{3}r+\frac{1}{3}\Big), which is a perfect square:


Thus, the complete solution to r^3-r=2\frac{\sqrt{3}}{9} is r=-\frac{\sqrt{3}}{3},~\frac{2\sqrt{3}}{3}. Notice how this solution set becomes “increased” in the next example.


\triangle ABC has an area of \Delta=\frac{\sqrt{3}}{9}. The slopes of its sides form a geometric sequence 1,r,r^2. Find possible coordinates for the vertices A,B,C.

Since the slopes of the sides form the geometric sequence 1,r,r^2, we can place the vertices at A(1,r^2), B(0,0), and C(1+r,r+r^2). The area will then be \frac{r^3-r}{2}, in view of Example 1. But we’re given that the area is \Delta=\frac{\sqrt{3}}{9}:

    \[\frac{r^3-r}{2}=\frac{\sqrt{3}}{9}\implies r^3-r=2\frac{\sqrt{3}}{9},\]

which is the depressed cubic solved in the preceding example. Recall that the cubic has r=\frac{-\sqrt{3}}{3},~\frac{2\sqrt{3}}{3} as solutions.

Being that the area formula \Delta=\frac{r^3-r}{2} is an odd function (Example 2), we should also consider r=\frac{\sqrt{3}}{3},~-\frac{2\sqrt{3}}{3} as possible values for the common ratio r, even though these are not solutions to the depressed cubic.

  • for r=\frac{-\sqrt{3}}{3}, the coordinates will be A\left(1,\frac{1}{3}\right), B(0,0), C\left(1-\frac{\sqrt{3}}{3},~-\frac{\sqrt{3}}{3}+\frac{1}{3}\right)=\left(\frac{3-\sqrt{3}}{3},\frac{1-\sqrt{3}}{3}\right);
  • for r=\frac{\sqrt{3}}{3}, the coordinates will be A\left(1,\frac{1}{3}\right), B(0,0), C\left(1+\frac{\sqrt{3}}{3},~\frac{\sqrt{3}}{3}+\frac{1}{3}\right)=\left(\frac{3+\sqrt{3}}{3},\frac{1+\sqrt{3}}{3}\right);
  • for r=\frac{-2\sqrt{3}}{3}, the coordinates will be A\left(1,\frac{4}{3}\right), B(0,0), C\left(1-\frac{2\sqrt{3}}{3},~-\frac{2\sqrt{3}}{3}+\frac{4}{3}\right)=\left(\frac{3-2\sqrt{3}}{3},\frac{4-2\sqrt{3}}{3}\right);
  • for r=\frac{2\sqrt{3}}{3}, the coordinates will be A\left(1,\frac{4}{3}\right), B(0,0), C\left(1+\frac{2\sqrt{3}}{3},~\frac{2\sqrt{3}}{3}+\frac{4}{3}\right)=\left(\frac{3+2\sqrt{3}}{3},\frac{4+2\sqrt{3}}{3}\right).

Since the area \Delta=\frac{\sqrt{3}}{9} is so small, each case produces a rare-listic diagram when drawn.

Example (“Fixed point”)

Find coordinates for the vertices of a \triangle ABC whose slopes are 1,r,r^2 and whose area is \sqrt{3} sq. units.

We solve the cubic equation r^3-r=2\Delta using \Delta=\sqrt{3}:

    \[r^3-r=2\sqrt{3}\implies (r-\sqrt{3})(r^2+\sqrt{3}r+2)=0\implies r=\sqrt{3};\]

the quadratic part has complex roots. Again, we should consider r=-\sqrt{3}, despite the fact that it’s not a solution to the current cubic equation.

Next, let’s use the coordinates A(1,r^2), B(0,0), C(1+r,r+r^2):

    \[A(1,3),B(0,0),C(1+\sqrt{3},3+\sqrt{3}),\quad\textrm{OR}\quad A(1,3),B(0,0),C(1-\sqrt{3},3-\sqrt{3}).\]

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 x-axis and the rest becomes easy.

Luckily, we can always turn to triangles with slopes in geometric progressions. In this case we take \triangle ABC with vertices at A(1,4),B(0,0),C(3,6). The side slopes are 1,2,4. For the side lengths we have AB=\sqrt{17}, BC=3\sqrt{5}, and CA=2\sqrt{2}, all irrational and all different, so the triangle is scalene. Its area is 3 sq. units.

Let \alpha be a root of the depressed cubic r^3-r-2\Delta=0. PROVE that r^3-r-2\Delta=(r-\alpha)(r^2+\alpha r+\frac{2\Delta}{\alpha}).

Since \Delta\neq 0 (an area), it follows that \alpha\neq 0 as well. This ensures that the quantity \frac{2\Delta}{\alpha} is defined.


    \[r^3-r-2\Delta=(r-\alpha)(r^2+br+c)=r^3+(b-\alpha)r^2+(c-\alpha b)r-\alpha c.\]

Comparing coefficients of like terms, b-\alpha=0,~c-\alpha b=-1, and \alpha c=2\Delta. So b=\alpha, c=\alpha b-1=\alpha^2-1, and again c=\frac{2\Delta}{\alpha}. The two expressions for c yield \alpha(\alpha^2-1)=2\Delta, which is valid because \alpha is a root of the original cubic r^3-r-2\Delta=0.

\therefore r^3-r-2\Delta=(r-\alpha)(r^2+\alpha r+\frac{2\Delta}{\alpha}).


If \triangle ABC with vertices at A(x_1,y_1),B(x_2,y_2),C(x_3,y_3) is such that the slopes of sides AB,BC,CA are a,ar,ar^2, respectively, then the area \Delta of the triangle can be given by


By restricting the coordinates in a suitable way (see the exercises below), we obtain the depressed cubic \frac{1}{2}r(r^2-1) as the area.


  1. (Special square) For any right triangle with vertices (\alpha_1,\beta_1), (\alpha_2,\beta_2), and (\alpha_3,\beta_3), PROVE that: \Big(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_3|+|\beta_3-\beta_1|}\Big)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}.
    (Notice how squaring respects addition in this case. In addition, not all four absolute values on the left are absolutely necessary \cdots\vdots\cdots.)
  2. (Special square) For any triangle with centroid (\alpha_1,\beta_1), circumcenter (\alpha_2,\beta_2), and orthocenter (\alpha_3,\beta_3), PROVE that: \Big(\frac{(\alpha_2-\alpha_1)+(\beta_2-\beta_1)}{(\alpha_3-\alpha_3)+(\beta_3-\beta_1)}\Big)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}.
    (Notice the absolute absence of absolute values in this case — so long as (\alpha_2-\alpha_1)+(\beta_2-\beta_1)\neq 0 and (\alpha_3-\alpha_1)+(\beta_3-\beta_1)\neq 0. 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 x^3+Ax=B has two opposite solutions, namely x=\alpha and x=-\alpha, then B=0.
    (In our own case, the right member B represents the area \Delta of a certain, obscure triangle. Since \Delta\neq 0, this explains why the depressed cubic equation r^3-r=2\Delta cannot have two opposite solutions.)
  4. \triangle ABC has vertices at A(x_1,y_1),B(x_2,y_2),C(x_3,y_3). Assume that the slopes of sides AB,BC,CA are a,ar,ar^2, in that order.
    • PROVE that its area \Delta can be given by \Delta=\frac{1}{2}\Big(\frac{r-1}{r+1}\Big)\Big(\frac{1}{ar}\Big)(y_2-y_3)^2.
    • Find an appropriate choice of (y_2-y_3) for which the area \Delta reduces to the depressed cubic \Delta=\frac{a}{2}r(r^2-1).
  5. \triangle ABC has vertices at A(x_1+1,y_1+ar^2), B(x_1,y_1), C(x_1+1+r,y_1+ar+ar^2). PROVE that:
    • the coordinates of the foot of the altitude from vertex A is \Big(x_1+\frac{a^2r^3+1}{a^2r^2+1},~y_1+\frac{a^3r^4+ar}{a^2r^2+1}\Big);
    • the length of the altitude from vertex A is h=\frac{ar(r-1)}{a^2r^2+1};
    • the area of \triangle ABC is \frac{a}{2}(r-1)(r)(r+1)=\frac{a}{2}r(r^2-1).
  6. Consider \triangle ABC with vertices at A(2,36)=(x_1,y_1), B(0,0)=(x_2,y_2), C(8,48)=(x_3,y_3). Its side slopes are 2,6,18, a geometric progression in which a=2 and r=3.
    • 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 \frac{y_1(x_3-x_2)+y_2(x_1-x_3)+y_3(x_2-x_1)}{2};
    • Explain why the formula in exercise 5 did not work correctly in this case.
  7. For \triangle ABC with vertices at A(1,r^2), B(0,0), C(1+r,r+r^2), PROVE that:
    • its centroid is located at \Big(\frac{2+r}{3},\frac{r+2r^2}{3}\Big);
    • its circumcenter is at \Big(\frac{r^3+r^2+r-1}{2(r-1)},\frac{r^3-r^2-r-1}{2(r-1)}\Big);
    • its orthocenter is at \Big(\frac{1+r^3}{1-r},\frac{-1-r^3}{1-r}\Big);
    • the slope of its Euler line is m_{E}=-\Big(\frac{r^3+r^2+r+3}{3r^3+r^2+r+1}\Big).
  8. (Same ratio) Given A(1,r^2), B(0,0), C(1+r,r+r^2), PROVE that the centroid, the circumcenter, and the orthocenter of \triangle ABC lie on parallel lines x+y=\frac{2}{3}(1+r+r^2), x+y=1+r+r^2, and x+y=0, respectively.
    (Notice that the line through the centroid is in the middle of the other two lines. Moreover, these lines preserve the usual 1:2 ratio in which the centroid divides the circumcenter-orthocenter distance.)
  9. If \sqrt[3]{a+\sqrt{b}}=c+\sqrt{d}, PROVE that \sqrt[3]{-a+\sqrt{b}}=-c+\sqrt{d}.
  10. (Unit area) Find coordinates for the vertices of \triangle ABC with slopes 1,r,r^2 and an area of 1 sq. unit.