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The Cubic Equation and the Equilateral Triangle

Whoa! Or Duh!

I watched this YouTube video that shows how to solve a cubic equation with a triangle. It was amazing to see an equilateral triangle’s vertices line up with the roots of the cubic. Although the goal was to find the roots of the cubic rather than the triangle itself, I wanted recreate the proof on my own but it seems daunting. However, after trying to create a Geogebra activity, it came to light that it was a duh moment. The proof is quite trivial and we do not even need to involve a cubic polynomial. But, on this page, I’ll extend the triangle and its properties.

First, let’s take a look at all the amazing properties about the cubic and the equilateral triangle. The following image is quite complicated, but we will break it down. The cubic curve is in red. Its derivative is the parabola in green. The equilateral triangle is in blue, and the incircle is in yellow. Figure 1 shows the triangle in the midst of the curves. Figure 2 shows the triangle above the curves. The y-coordinate of the circle really do not matter - it can be shifted up and down the vertical line.

1. When vertical lines are drawn from the roots, \( x_{1} \), \( x_{2} \), and \( x_{3} \), of the cubic equation (in red), they intersect the vertices of the equilateral triangle ABC.

2. Lines extended from the extrema \( E_{1} \) and \( E_{2} \) are the farthest points on the circle inscribed in the equilateral triangle. Hence, they form the diameter and pass through the center of the circle (in yellow). These lines also pass through the roots of the derivative of the cubic (the parabola in green), \( e_{1} \) and \(e_{2} \).

3. Line extended from the extremum of the derivative (the parabola in green) passes through the center of the circle and the inflection point I.

4. There are actually 2 equilateral triangles and are mirror images. The other triangle in light blue is A'B'C'. We will mainly focus on one of the triangles but all these properties apply to both.

The Equilateral Triangle

First, let’s omit the cubic. Suppose there are three points on the x-axis and we draw vertical lines through these points. Actually, let’s just say there are three vertical lines that intersect the x-axis at three points. It really doesn’t matter. Then, one can place three points on the three lines and create an infinite number of equilateral triangles. The only caveat is that the center of the equilateral triangle lies on another vertical line that is the average distance of the three lines from the y-axis.

The proof is quite simple. In the figure above, we see an equilateral triangle with vertical segments drawn to meet the vertices of the equilateral triangle. The center of the equilateral triangle lies on the line in gray. The triangle can be slid up and down this line for an infinite number of equilateral triangles.

So, if we draw the equilateral triangle first and then draw segments from the vertices to the x-axis, then we will have 3 points on the x-axis. In the same way, if we start with three points on the x-axis first, then we can conclude there is an equilateral triangle that we can draw. So, the proof is simply intuitive.

Now, if these three points are \( (x_{1}, 0) \), \( (x_{2}, 0) \), and \( (x_{3}, 0) \), then the x-coordinates of the vertices of the equilateral triangle are \( x_{1} \), \( x_{2} \), and \( x_{3} \), and the equation of the cubic is \( y = (x - x_{1})(x - x_{2})(x - x_{3}) \) or \( y = x^{3} - (x_{1} + x_{2} + x_{3})x^{2} + (x_{1}x_{2} + x_{1}x_{3} + x_{2}x_{3})x - x_{1}x_{2}x_{3} \).

There is always a cubic equation for any three real roots.

The Center of The Equilateral Triangle

We need to prove the center of the equilateral triangle lies on anywhere along a specific line and we need to find the equation of the line. In Figure 4, midpoint \( M_{A} \) has be drawn for side BC. The segment AM_A is the median. The center of the equilateral triange (and the incircle and the circumcircle) lies two-thirds of the way from the vertex at A (or any vertex). We can use similar trianges to find the length \( \Delta x \), then find the x-coordinate. First, the midpoint of BC is \( \left( \frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2} \right) \).

Now, we can use similar triangles to find the two-thirds distance of the x position of the center. Let d be the length of AM_A. Therefore, we have:

(i) \( \dfrac{\Delta x}{\frac{2}{3}d} = \dfrac{\frac{x_{2}+x_{3}}{2}-x_{1}}{d} \)

(ii) \( \Delta x = \dfrac{2}{3}\left( \dfrac{x_{2}+x_{3}}{2} - x_{1} \right) \)

(iii) \( \Delta x = \dfrac{1}{3}x_{2} + \dfrac{1}{3}x_{3} - \dfrac{2}{3}x_{1} \)

The equation in (iii) gives us the absolute distance between point A and the center. To find the coordinate, we add \( x_{1} \) to get \( x_{c} \).

(iii) \( x_{c} = \dfrac{1}{3}x_{2} + \dfrac{1}{3}x_{3} - \dfrac{2}{3}x_{1} + x_{1} = \dfrac{1}{3}(x_{1} + x_{2} + x_{3}) \)

Therefore, we can state the x-coordinate of the center of the equilateral triangle as below (and the y-coordinate by similar method) as follows:

Note that if we know the roots of the cubic equation, we can determine the x-coordinate of center simply from the roots. However, since the center of the triangle can be anywhere on the line \( x = \frac{1}{3}(x_{1}+x_{2}+x_{3}) \), we get to choose the y-coordinate of the center. However, if the coordinates of the triangle have been determined, then we can use \( y_{c} = \dfrac{1}{3}(y_{1} + y_{2} + y_{3}) \) to pinpoint the center.

The Inflection Point and Center of the Circle

The inflection point (shown as I in red) lies on the same line that passes through the center of the incircle and the circumcircle. The inflection point can be found by finding the root of the second derivative.

(i) \( f'(x) = 3x^{2} - 2(x_{1}+x_{2}+x_{3})x + (x_{1}x_{2} + x_{1}x_{3} + x_{2}x_{3}) \)

(ii) \( f''(x) = 6x - 2(x_{1}+x_{2}+x_{3}) \)

(iii) \( 6x - 2(x_{1}+x_{2}+x_{3}) = 0 \)

(iv) \( x = \frac{1}{3}(x_{1}+x_{2}+x_{3}) \)

The Radius and the Diameter

The radius of the incircle can be found by finding the roots of the derivative of the cubic. The cubic has the equation \( f(x) = x^3 - (x_{1}+x_{2}+x_{3})x^{2} + (x_{1}x_{2} + x_{1}x_{3} + x_{2}x_{3})x + x_{1}x_{2}x_{3} \)

(i) \( f'(x) = 3x^{2} - 2(x_{1}+x_{2}+x_{3})x + (x_{1}x_{2} + x_{1}x_{3} + x_{2}x_{3}) \)

(ii) \( 3x^{2} - 2(x_{1}+x_{2}+x_{3})x + (x_{1}x_{2} + x_{1}x_{3} + x_{2}x_{3}) = 0 \)

(iii) \( x = \frac{2(x_{1}+x_{2}+x_{3}) \pm \sqrt{4(x_{1}+x_{2}+x_{3})^{2} - 4(3)(x_{1}x_{2} + x_{1}x_{3} + x_{2}x_{3})}}{6} \)

(iv) \( x = \frac{(x_{1}+x_{2}+x_{3}) \pm \sqrt{(x_{1}+x_{2}+x_{3})^{2} - 3(x_{1}x_{2} + x_{1}x_{3} + x_{2}x_{3})}}{3} \)

If we subtract the x-coordinate of the center of the circle, we get the length of the radius. The circumradius of an equilateral triangle is twice the incircle radius.

The Extrema and the Diameter

The farthest horizontal points on the circle correspond to the extrema of the cubic, which are also the zeros of the quadratic derivative. We found the radius above already by finding the roots of the quadratic already. The roots are \( x = \frac{(x_{1}+x_{2}+x_{3}) \pm \sqrt{(x_{1}+x_{2}+x_{3})^{2} - 3(x_{1}x_{2} + x_{1}x_{3} + x_{2}x_{3})}}{3} \).

If you add or subtract the length of the radius of the center of the circle, then we get the above.

The Y-Coordinates of the Equilateral Triangle

The most difficult of all of these characteristics is finding the simplest way to locate the y-coordinates of the vertices. I read an article that gave simple equations that only involved the roots of the cubic (i.e. x-coordinates of the vertices). I was able to find complex formulas but was not able to state them using only the roots. Proving these formulas was a task, but I finally succeeded with some geometry. The proof is quite satisfying.

The y-coordinates of the equilateral triangle can be determined solely based on the roots of the cubic, \( x_{1} \), \( x_{2} \), and \( x_{3} \), if the center of the incircle lies on the x-axis, meaning the circle’s center is at \( \left( \frac{1}{3}(x_{1} + x_{2}+ x_{3}), 0 \right) \). If not, then y-coordinates only need to be shifted up or down based on the position of the center of the circle.

Consider the equilateral triangle above with its center on the x-axis to make it simpler. To prove the location of the vertices is as stated above, we need to show that AX₁V and BQP are congruent 30-60-90 triangles. Let \( BQ = x_{3} - x_{1} \). If BQP is a 30-60-90 triangle and is congruent to AX₁V, then \( PQ = \frac{x_{3} - x_{1}}{\sqrt{3}} \) and AX₁ is also equal to \( \frac{x_{3} - x_{1}}{\sqrt{3}} \). AX₁ gives the position of \( y_{1} \).

Steps of Proof

First, let’s draw circumradius of ABC.

Then, draw a line from C parallel to the y-axis and a line from B parallel to the x-axis. They meet at Q and CQ intersects the circumcircle at P.

APB = 60° since they share a common chord AB. APC = 60° since they share a common chord AC. Hence, BPQ = 60° and that makes BQP a 30-60-90 right triangle.

This makes AX₁V a 30-60-90 right triangle by similar triangles. So we know AX₁V and BQP are similar, but we need to prove they are congruent.

Let’s extend AX₁ to T on the circumcircle. Now draw TC. TX₁V is congruent to AX₁V. Moreoever, CTB = 60° because of the common chord CB. AVT and CVP are both 60°, which means AVC and TVP are 120°. Since VTB = 60° and TVP = 120°, TV and BP have to be parallel. That means TV = BP. That shows TX₁V, AX₁V and BQP are all congruent.

This all means AX₁ = \( \frac{x_{3}-x_{1}}{\sqrt{3}} \). This is profound because the y position of vertex A is solely based on the horizontal distance between B and C. Whoever discovered this is quite a genius!

Note: The y-coordinate for y₂ seems to not match the patter of the other 2 coordinates. For y₁ and y₃, the smaller root is subtracted from the larger root. If we do the same for y₂, then the coordinate is of the other equilateral triangle. It is best to plot them all and see which belongs to which triangle because I have not determined this is always the case.

The Other Triangle

Remember that there were 2 equilateral triangles, one is a horizontal reflection of the other. If we subtract the x-coordinates in reverse order, we get the y-coordinates of the other triangle.

For example, \( \left( x_{1}, \frac{x_{3}-x_{2}}{\sqrt{3}} \right) \) gives the y-coordinate of one triangle, while \( \left( x_{1}, \frac{x_{2}-x_{3}}{\sqrt{3}} \right) \) gives the y-coordinate of the other triangle.

The Angle of Rotation

Now, we want to find the angle of rotation of the equilateral triangle, assuming that the triangle has been rotated from a baseline position where one side is parallel to the x-axis.

To make this easiest, we need to pick one side of the three sides and mark how much it has rotated using the tangent function. In the image above, we can pick side opposive vertex A since it produces an acute angle with the x-axis and the tangent function range is within the acute angle range.

For the image above, best side is BC.

(i) \( \tan \theta = \frac{y_{3}-y_{2}}{x_{3}-x_{2}} \)

(ii) \( \theta = \tan^{-1}\left( \frac{y_{3}-y_{2}}{x_{3}-x_{2}}\right) \)

Since we know the y-coordinates in terms of x, we can state the above with the substitutions as follows.

Notes

If a point is rotated and we know the difference in the horizontal distance \( \Delta x\) but not the angle of rotation, we can determine \( \Delta y \), and hence \( y' \) - the new y-value.

\( y + \Delta y = \sqrt{r^{2} - (x-\Delta x)^{2}} \), where r is the distance of the point from the origin. We already know \( r^{2} = x^{2} + y^{2} \). So, \( y + \Delta y = \sqrt{y^{2} + 2x\Delta x - (\Delta x)^{2}} \) and \( \Delta y = -y + \sqrt{y^{2} + 2x\Delta x - (\Delta x)^{2}} \)

In the same way, if we know the change in y, we can find the new x-coordinate: \( x + \Delta x = \sqrt{x^{2} + 2y\Delta y - (\Delta y)^{2}} \). So, \( \Delta x = -x + \sqrt{x^{2} + 2y\Delta y - (\Delta y)^{2}} \)

Anirdesh.com Last Updated: May 31 2025, 00:15:22.