Call the cubic polynomial
z3 + (1-p)z2 + pz + 1, p ∈ Z
then its solutions are of the form z=(x+yi) with real x,y
If we make this substitution we get
(x+yi)3 + (1-p)(x+yi)2 + p(x+yi) + 1 = 0
a polynomial that can be separated into real and imaginary parts:
Re: x^3-3xy^2+(1-p)(x^2-y^2)+px+1=0
Im: 3x^2y-y^3+(1-p)2xy+py=0
These equations can be plotted, complex plane becomes the (x,y) plane: https://www.desmos.com/calculator/fggwdkex9w
The complex roots of the original polynomial are the intersections of the real (red) and imaginary (green) plus the x-intercept of the real (red).
Now we can at least see the triangle's vertices. From here I'm stuck. call the vertices (a,b), (a,-b), (c,0). I need to find a,b,c in terms of p and the area becomes (a-c)b.
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Posted by Jer
on 2021-09-28 08:38:55 |
A start. Stuck.
