Reach for the Summit - M-S4-A2

Putting $p=0$ works easily; but we can show this is the only possibility.

Let the roots be $x_1,x_2,x_3$ . Then, by Vieta, $x_1+x_2+x_3=0$ , $x_2 x_3 + x_3 x_1 + x_1 x_2 = p$ and $x_1 x_2 x_3=-1$ .

Since they form an equilateral triangle in the complex plane, they must be an affine transformation of the points $\{1,\omega,\omega^2\}$ , where $\omega$ is a complex cube root of $1$ . That is, for some complex $a,b$ , we have $x_1=a+b,\;\;\;x_2=a+b\omega,\;\;\; x_3=a+b\omega^2$

The relation $x_1+x_2+x_3=0$ means we need $a=0$ .

Then $x_1 x_2 x_3=-1$ means $b^3=-1$ .

Whatever the choice of $b$ , we end up with a permutation of $\{-1,-w,-w^2\}$ .

Hence $p=0$ , and the side of the equilateral triangle is $\sqrt3$ giving area $\frac{3\sqrt3}{4}$ and answer $\boxed{1299}$ .

The given condition of the problem implies that the roots $z_1,z_2,z_3$ of the equation $x^3+px+1=0$ satisfy

$z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_3z_1$

$\implies p=0\implies$ the given equation reduces to

$x^3+1=0$

The three roots of this equation are

$-1,\dfrac 12+i\dfrac{\sqrt 3}{2},\dfrac 12-i\dfrac{\sqrt 3}{2}$

So the length of each side of the triangle is $\sqrt 3$ and area of the triangle is

$\dfrac {3\sqrt 3}{4}\approx 1.2990381$

Therefore the required answer is $\boxed {1299}$ .