Quadrilateral in a Cubic Polynomial

Since a cubic has order $2$ rotational symmetry about its point of inflection, the quadrilateral is a parallelogram, and hence has area $\left| \overrightarrow{BF} \times \overrightarrow{BG} \right|$ Suppose that the polynomial has distinct integer roots $P < Q < R$ , and hence is $(X-P)(X-Q)(X-R)$ . It is easy to evaluate the coordinates of $F$ and $G$ , and to solve the equation of the cubic simultaneously with that of the normal to the cubic at $I$ (where $x = \tfrac13(P+Q+R)$ ) to find the coordinates of $B$ - both of these calculations simply require us to solve quadratics. After much algebra and simplification, we obtain the surprising result that the area of the quadrilateral is $A(X) \; =\; \frac{2(27 + 2X^2)\sqrt{9 + X^2}}{27\sqrt{3}X} \hspace{1cm}\mbox{where} \hspace{1cm} X = P^2 + Q^2 + R^2 - PQ - PR - QR$ We note that $2X = (Q-P)^2 + (R-Q)^2 + (R-P)^2$ ; since $Q-P,R-Q \ge 1$ and $R-P \ge 2$ we see that $X \ge 3$ for all allowed values of $P,Q,R$ . Moreover $\frac{dA}{dX} \; = \; \frac{2(4X^4 + 18X^2 - 243)}{27\sqrt{3}X^2\sqrt{9+X^2}}$ and hence $\frac{dA}{dX} > 0$ for all $X \ge 3$ . Thus the smallest possible value of $A$ occurs when $X=3$ , which happens when $P,Q,R$ are consecutive integers. The minimum value of $A$ is thus $\tfrac{10}{3}\sqrt{\tfrac23} = \tfrac{10}{9}\sqrt{6}$ , making the answer $10 + 9 + 6 = \boxed{25}$ .

Here is my computer assisted solution heavily relying on sympy module for python.

I consider a polynomial $f_{L, R}(x) = \left(L + x\right) x \left(x - R\right),$ where $L$ and $R$ are some natural numbers (i.e. I shift the polynomial from the problem in such a way, that $0$ is its middle root).

After few calculations (cf. code below) it turns out, that the area of the quadrilateral of our interest expressed as a function of $L$ and $R$ has the form: $A(L, R) = \frac{2 \sqrt{3} \sqrt{L^{4} + 2 L^{3} R + 3 L^{2} R^{2} + 2 L R^{3} + R^{4} + 9} \left(2 L^{4} + 4 L^{3} R + 6 L^{2} R^{2} + 4 L R^{3} + 2 R^{4} + 27\right)}{81 \left(L^{2} + L R + R^{2}\right)}.$ It attains its minimum for $L = R = 1$ . The minimal value is $\frac{10 \sqrt{6}}{9}$ , which gives the answer 25.

And the code:

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"""
solution of:
    https://brilliant.org/problems/quadrilateral-in-a-cubic-polynomial/
"""
import itertools
import sympy
def get_area_expr():
    """
    returns an expression for the area of the quadrilateral described in:
        https://brilliant.org/problems/quadrilateral-in-a-cubic-polynomial/
    """
    x_sym = sympy.symbols('x', real=True)
    l_sym = sympy.symbols('L', positive=True)
    r_sym = sympy.symbols('R', positive=True)
    poly_fun = (x_sym+l_sym)*x_sym*(x_sym-r_sym)
    def get_point(in_x):
        """
        returns the coordinates of a point on the graph of the function poly_fun for given in_x
        """
        return sympy.Point2D(in_x, sympy.simplify(poly_fun.subs(x_sym, in_x)))

    #we determine point_f and point_g
    poly_der = sympy.simplify(sympy.diff(poly_fun, x_sym))
    tmp_sol_data = list(sympy.solveset(poly_der, x_sym))
    assert len(tmp_sol_data) == 2
    assert tmp_sol_data[0] < tmp_sol_data[1]
    point_f = get_point(tmp_sol_data[0])
    point_g = get_point(tmp_sol_data[1])

    #we determine point_i
    poly_dder = sympy.simplify(sympy.diff(poly_der, x_sym))
    tmp_sol_list = list(sympy.solveset(poly_dder, x_sym))
    point_i = get_point(tmp_sol_list[0])

    #we determine the slope of the straigth line passing through point_i
    slope_i = poly_der.subs(x_sym, point_i[0])
    line_i = sympy.simplify(-1/slope_i*(x_sym-point_i[0])+point_i[1])

    #we determine point_b and point_e
    tmp_sol_data = sympy.solveset(poly_fun-line_i, x_sym)
    assert point_i[0] in tmp_sol_data
    assert len(tmp_sol_data) == 3
    tmp_sol_data = list(tmp_sol_data-sympy.FiniteSet(point_i[0]))
    tmp_sol_data = [sympy.simplify(_) for _ in tmp_sol_data]
    assert tmp_sol_data[0] < tmp_sol_data[1]
    point_b = get_point(tmp_sol_data[0])
    point_e = get_point(tmp_sol_data[1])

    poly_data = sympy.geometry.polygon.Polygon(point_b, point_g, point_e, point_f)
    return sympy.simplify(poly_data.area), l_sym, r_sym

AREA_EXPR, L_SYM, R_SYM = get_area_expr()
MIN_VAL = sympy.oo

for (l, r) in itertools.product(range(1, 37), repeat=2):
    cur_val = AREA_EXPR.subs([(L_SYM, l), (R_SYM, r)])
    if cur_val < MIN_VAL:
        MIN_VAL = cur_val
        MIN_ARG = (l, r)

print(f"Minimal area: {MIN_VAL} for l={MIN_ARG[0]}, r={MIN_ARG[1]}")

As Mark Hennings has already pointed out in his solution, the cubic has order 2 rotational symmetry about its point of inflection $Ι$ , hence, the quadrilateral is a parallelogram. Working in 2D, its area $A$ can be given by $A = \left| {\det \left( {\overrightarrow {FE} ,\overrightarrow {FB} } \right)} \right|$ .

We notice that a translation of the cubic by vector $\overrightarrow {IO}$ , ( $O$ is the origin) does not affect the compound shape of the cubic and the parallelogram, but it simplifies quite a bit the algebraic manipulations needed.

Due to the symmetry, the zeros of the translated cubic $g$ are $0$ , $\pm c$ for some positive real number $c$ .
Hence, $g\left( x \right) = x\left( {{x^2} - {c^2}} \right) = {x^3} - {c^2}x$ .
Consequently, $g'\left( x \right) = 3{x^2} - {c^2} \Rightarrow g'\left( 0 \right) = - {c^2} \Rightarrow {\text{slope of }}{{\text{B}}_1}{{\text{E}}_1}{\text{ = }}\frac{1}{{{c^2}}} \Rightarrow {\text{ equation of }}{{\text{B}}_1}{{\text{G}}_1}{\text{: }}y = \frac{1}{{{c^2}}}x$ .
To find points ${E_1}$ and ${B_1}$ , we solve simultaneously $\left\{ {\begin{matrix} {y = \frac{1}{{{c^2}}}x} \\ {y = {x^3} - {c^2}x} \end{matrix}} \right.$
and we easily get ${E_1}\left( {\frac{{\sqrt {{c^4} + 1} }}{c},\;\frac{{\sqrt {{c^4} + 1} }}{{{c^3}}}} \right)$ , ${B_1}\left( { - \frac{{\sqrt {{c^4} + 1} }}{c},\; - \frac{{\sqrt {{c^4} + 1} }}{{{c^3}}}} \right)$ .

For the point $F_1$ ,
$\left. {\begin{matrix} {g'\left( x \right) = 0 \Leftrightarrow 3{x^2} - {c^2} = 0\mathop \Leftrightarrow \limits^{x < 0} x = - \frac{c}{{\sqrt 3 }}} \\ {g\left( { - \frac{c}{{\sqrt 3 }}} \right) = {{\left( { - \frac{c}{{\sqrt 3 }}} \right)}^3} - {c^2}\left( { - \frac{c}{{\sqrt 3 }}} \right) = \frac{{2{c^3}}}{{3\sqrt 3 }}} \end{matrix}} \right\} \Rightarrow {F_1}\left( { - \frac{c}{{\sqrt 3 }},\;\frac{{2{c^3}}}{{3\sqrt 3 }}} \right)$

Thus, $\overrightarrow {{F_1}{E_1}} = \left( {\frac{{\sqrt {{c^4} + 1} }}{c} + \frac{c}{{\sqrt 3 }},\frac{{\sqrt {{c^4} + 1} }}{{{c^3}}} - \frac{{2{c^3}}}{{3\sqrt 3 }}} \right)$ and $\overrightarrow {{F_1}{B_1}} = \left( { - \frac{{\sqrt {{c^4} + 1} }}{c} + \frac{c}{{\sqrt 3 }}, - \frac{{\sqrt {{c^4} + 1} }}{{{c^3}}} - \frac{{2{c^3}}}{{3\sqrt 3 }}} \right)$ .

Finally, $A = \left| {\det \left( {\overrightarrow {FE} ,\overrightarrow {FB} } \right)} \right| = \left| {\det \left( {\overrightarrow {{F_1}{E_1}} ,\overrightarrow {{F_1}{B_1}} } \right)} \right| = \frac{2}{\sqrt 3}\sqrt {{c^4} + 1} \left( {\frac{{2{c^2}}}{3} + \frac{1}{{{c^2}}}} \right)$ .

Now, due to the concavity of the cubic, the distance between $P$ and $R$ cannot exceed the length of segment $MN$ , which is parallel to the x-axis, through the point of inflection $I$ .
Hence, $2 \leqslant R - P \leqslant \overline {MN} = \overline {LK} = 2c$ . Thus, $c \geqslant 1$ .

If we label ${c^2} = x$ , we have a function for the area of the parallelogram: $A\left( x \right) = \frac{2}{{\sqrt 3 }}\sqrt {{x^2} + 1} \left( {\frac{{2x}}{3} + \frac{1}{x}} \right), x \geqslant 1$ . For $x \geqslant 1$ , $A'\left( x \right) = \frac{{2\left( {4{x^4} + 2{x^2} - 3} \right)}}{{3\sqrt 3 {x^2}\sqrt {{x^2} + 1} }} > 0$ . Thus, $A\left( x \right)$ is increasing, so ${A_{\min }} = A\left( 1 \right)$ , which occurs when $c = \sqrt {{1^2}} = 1$ .

Since $2 \leqslant R - P \leqslant 2c = 2$ , it turns out that the zeros $P$ , $Q$ , $R$ of the initial cubic are three consecutive integers, i.e. $\left( {P,Q,R} \right) \in \left\{ {\left( {0,1,2} \right),{\text{ }}\left( {1,2,3} \right),{\text{ }} \ldots ,{\text{ }}\left( {34,35,36} \right)} \right\}$ No matter which triad is the actual zeros of the cubic, the minimum area of the parallelogram is $A\left( 1 \right) = \frac{{10\sqrt 6 }}{9}$ .

For the answer, $a=10$ , $b=6$ , $c=9$ , hence, $a+b+c=\boxed{25}$ .

the area asked will be minimum when the three zeros of the polynomial are the closest. So, let us choose the polynomial $y=x(x-1)(x-2)=x^3-3x^2+2x$ .

So the coordinates of $I$ are $(1,0)$ obtained from the condition $\dfrac{d^2y}{dx^2}=0$ and the equation of the curve.

Coordinates of $F$ and $G$ are $(\dfrac{3-\sqrt 3}{3},\dfrac{2\sqrt 3}{9})$

and $(\dfrac{3+\sqrt 3}{3},\dfrac{-2\sqrt 3}{9})$ respectively, obtained from the condition $\dfrac{dy}{dx}=0$ and the equation of the curve.

Slope of $\overline {BE}$ is $1$ , and it's equation is $y=x-1$ . Solving this and the equation of the curve we get the coordinates of $B$ and $E$ as

$(1-\sqrt 2,-\sqrt 2 )$ and $(1+\sqrt 2, \sqrt 2)$ .

Hence the required area is $\dfrac{10\sqrt 6}{9}$ , $a=10,b=6,c=9$ and $a+b+c=10+6+9=\boxed {25}$ .