A monic quartic polynomial with 4 distinct real roots, such that if we were to construct a triangle whose vertices are the polynomials' 3 turning points, the triangle formed is equilateral Find the area of a triangle.
Can I take $(0,0)$ as a turning point?
On taking $(0,0)$ as a turning point I am getting $3^{0.83}$
Am I right? Please provide a precise solution or hint.
Suppose the polynomial is $p(x)$. If the turning points (the roots of $p'(x)$) are at $x=a,b,c$, we have $p'(x) = 4 (x-a)(x-b)(x-c)$. Since the formulation is invariant under translation, we may assume wlog $a=0 < b < c$. Then $p'(x) = 4 x (x-b)(x-c)$, and $p(x) = {x}^{4}+4/3\, \left( -b-c \right) {x}^{3}+2\,bc{x}^{2}+d$ for some $d$. The three turning points are $(0,d)$, $(b, -{b}^{4}/3 +2{b}^{3}c/3+d)$ and $(c,-{c}^{4}/3+2b{c}^{3}/3+d)$. Setting these to be equidistant, I get two equations in $b,c$. The solution that is real and has $0 < b < c$ turns out to be $$ b = 3^{1/6}, c = 2 \cdot 3^{1/6}$$ And now you should be able to figure out the area of the triangle.