Imaginary triangles

As per @Mark Hennings' post, the easiest way to solve this is numerically. I got curious, though: can we find a formula for the area of the triangle in terms of the parameter $p$ ?

Define $f_p(x)=x^3+(1-p)x^2+px+1$ .

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The first thing to establish is, when do we get a triangle? That is, when does $f_p$ have complex roots?

The discriminant of $f_p$ turns out to be $\left(p^2-p+1 \right) \left(p^2-p-31 \right)$ . We get complex roots when this is negative, ie when $\frac12-\frac{5\sqrt5}{2}<p<\frac12+\frac{5\sqrt5}{2}$

We'll assume $p$ is in this range for the rest of the discussion.

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Since the coefficients of $p$ are real, the two complex roots of $f_p$ are conjugate. Let the roots of $f_p$ be $a,u+iv,u-iv$ , where $a,u,v$ are real, and $v>0$ .

We're interested in the area of the triangle formed by these three points in the complex plane. By symmetry, this triangle is isosceles; its area is just $|(a-u)v|$

As absolute functions aren't always differentiable, it's easier to maximise the square of this. Define $T=(a-u)^2 v^2$

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Since we know the roots, we can write $f_p(x)=x^3+(1-p)x^2+px+1=(x-a)(x-u-iv)(x-u+iv)=(x-a) \left(x^2-2ux+u^2+v^2 \right)$

Comparing coefficients, we find $1-p=-a-2u,\;\;\; p=2au+u^2+v^2,\;\;\; 1=a(u^2+v^2)$

With a bit of manipulation, we get $a=p-1-2u$ and $v^2=3u^2+2(1-p)u+p$ , so that

$T=(3u^2+2(1-p)u+p)(p-1-3u)^2$

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By definition, $f_p(u+iv)=0$ .

Plugging in, and considering the real part, we find $u^3-3uv^2 + (1-p) \left(u^2-v^2 \right) + pu + 1 = 0$

But we already have an expression for $v^2$ ; substituting this in, and tidying up, we find $8 u^3 + 8(1-p)u^2 + \left(p^2-p+1 \right) (2u-1) = 0$

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All that's left is to eliminate $u$ , and we'll have a formula for $T$ in terms of $p$ .

Except...that involves working out the solution of a cubic and then manipulating it in an already fairly tricky expression. This might be possible, and things might cancel out neatly, but I decided on a different approach.

Consider the equation $8 u^3 + 8(1-p)u^2 + \left(p^2-p+1 \right) (2u-1) = 0$ again. It's a cubic in $u$ , but it's also a quadratic in $p$ ! This is much easier to deal with.

Things get messy here, but after solving this quadratic, substituting in to the expression for $T$ and tidying up, we find the faintly ridiculous $T=\frac{-\left(8 u^6 - 24 u^5 + 10 u^4 - 4 u^3 + 18 u^2 - 4 u - 1\right) \pm \sqrt{-3 + 24 u - 44 u^2 - 40 u^3 + 84 u^4 + 80 u^5 + 16 u^6}}{2(2u-1)^2}$

The $\pm$ in there should be taken as negative when $u<\frac12$ and positive when $u>\frac12$ . When $u=\frac12$ we get a special case; we find $p=\frac32$ and that $T=\frac74$ .

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We now have a formula for $T$ in terms of $u$ , and $p$ in terms of $u$ .

Unfortunately, the formula I have for $T$ is still pretty horrific; maximising it involves solving an equation numerically anyway.

We find the overall maximum area is $2.73606$ at $u=1.87103$ and $p=4.55206$ .

If we define $F_p(X) = X^3 + (1-p)X^2 + pX + 1$ , then we note that $F_p(0) = 1 > 0$ , $F_p(1) = 3 > 0$ , while $F_p(2) = 13 - 2p$ and $F_p(\tfrac12) = \tfrac18(11 + 2p)$ . Thus $F_p(2) < 0$ if $p \ge 7$ , and $F_p(\tfrac12) < 0$ if $p \le -6$ . Thus we deduce that the three roots of the cubic $F_p(X)$ are all real if $p \ge 7$ or $p \le -6$ , and hence form a triangle of area $0$ .

The area of a triangle with complex vertices $a,b,c$ is (to within a sign) equal to $\frac{i}{4}\left| \begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^\star & b^\star & c^\star \end{array}\right|$ At this point it is simplest to find the roots of $F_p(X)$ numerically, and calculate the areas of the triangles formed by the roots, in the finite number of cases $p=-5,-4,-3,-2,-1,0,1,2,3,4,5,6$ , to discover that the area is maximized at $2.64932$ when $p = 5$ .

To add to what the others have already said, here is a Desmos visualization of the triangles in the complex plane as $p$ varies from $-8$ to $8$ . The roots and the area are calculated numerically from Cardano's version of the cubic formula.

Over integer values of $p$ , the area attains a maximum of approximately $2.64931954905$ when $\boxed{p = 5}$ .