Rectangle on a Cubic

Cubic's equation becomes $k(x^3-p^2x)=0$ , where k is a constant.

The four points of the rectangle are the two roots and the points where derivative of the cubic = 0.

$3x^2-p^2=0$

$x=±\frac {p}{√3}$

After putting these x-coordinates in the expression we get the four points as $(p,0),(\frac {p}{√3},\frac {-2p^3k}{3√3}),(-p,0),(\frac {-p}{√3},\frac {2p^3k}{3√3})$

Taking the slopes of any two adjacent sides and making their product = -1, we get $p=(\frac {9}{2k^2})^¼$

The ratio asked for is the value of the slope of the line joining $(-p,0)$ and $(\frac {-p}{√3},\frac {2p^3k}{3√3})$ , which, after putting in the value of $p$ , comes out to be $\frac {√6+√2}{2}$

$m(x) = x(x - p)(x + p) = x^3 - p^2x \implies \dfrac{dm}{dx}|_{x = x_{0}} = 3x_{0}^2 - p^2 = 0 \implies x_{0} = \dfrac{p}{\sqrt{3}}$

$B:(-\dfrac{p}{\sqrt{3}},\dfrac{2p^3}{3\sqrt{3}}), \:\ D:(\dfrac{p}{\sqrt{3}},-\dfrac{2p^3}{3\sqrt{3}})$

$AC = BD = 2p = \dfrac{2p}{3\sqrt{3}}\sqrt{9 + 4p^4} \implies 27 = 9 + 4p^4 \implies p^4 = \dfrac{9}{2} \implies p = (\dfrac{9}{2})^{\frac{1}{4}}$

$\implies AB = \dfrac{1}{2^{\frac{1}{4}}}\sqrt{6 - 2\sqrt{3}} \:\$ and $\:\ BC = \dfrac{1}{2^{\frac{1}{4}}}\sqrt{6 + 2\sqrt{3}}$

$\implies \dfrac{BC}{AB} = \dfrac{\sqrt{3} + 1}{\sqrt{2}} = \dfrac{\sqrt{6} + \sqrt{2}}{2} = \dfrac{\sqrt{a} + \sqrt{b}}{c} \implies$ $a + b + c = \boxed{10}$ .

The polynomial must be $f(x)=a(x)(x-p)(x+p)=ax(x^2-p^2)$ Its derivative is f'(x)=a(3x^2-p^2), so its extremes are at $x_1= -\frac{p}{\sqrt{3}}$ and $x_2= \frac{p}{\sqrt{3}}$ . Without altering the problem, we may assume that a and p both are positive. The maximum value then is

$y=f(x_1)= \frac{2a}{3\sqrt{3}}p^2$

The shape with the zeroes and extrema as its vertices is a parallelogram, in general. Divide up the part of the x-axis between -p and p at $x_1$ , into lengths

$d=p-\frac{p}{\sqrt{3}}$ and

$e=p+\frac{p}{\sqrt{3}}$ Name the short and long sides of the parallelogram b and c.

We have two known rectangular triangles for which $b^2=d^2+y^2$ $c^2=e^2+y^2$ The condition for our parallelogram to be a rectangle now is

$b^2+c^2=(d+e)^2$ which rewrites into $y^2 = de$

(Substituting the given expressions for y, d and e into this results in $a= \frac{3}{p\sqrt{2}}$ , but we won't even need this.)

Now we are ready to calculate the ratio c:b

$\frac{c^2}{b^2}=\frac{y^2+e^2}{y^2+d^2}=\frac{\frac{2}{3}p^2+(1+\frac{1}{3}\sqrt{3})^2p^2}{\frac{2}{3}p^2+(1-\frac{1}{3}\sqrt{3})^2p^2} = \frac{2+(\sqrt{3}+1)^2}{2+(\sqrt{3}-1)^2}=\frac{3+\sqrt{3}}{3-\sqrt{3}}=2+\sqrt{3}$ , so $\frac{c}{b} = \sqrt{2+\sqrt{3}}$

Note: From this point I will use a,b and c as in the problem.

We want this value $\sqrt{2+\sqrt{3}}$ in the form $\frac{\sqrt{a}+\sqrt{b}}{c}$ , so we have to solve $\frac{a+2\sqrt{ab}+b}{c^2} = 2+\sqrt{3}$ which splits up into $a+b=2c^2$ $4ab=3c^4$ From which it is clear that c has to be even. Try the smallest value $c=2$ , then $a+b=8$ and $ab=12$ , leading to $a=2, b=6,c=2$ , which indeed is a solution, because it satisfies $(\frac{\sqrt{2}+\sqrt{6}}{2})^2=2+\sqrt{3}$ . The requested answer then is $a+b+c=2+6+2=10$ .

Taking equation as y=x^3-(p^2)x, first determine,

p^2=3/√2

with condition of perpendicular lines, among three points {-p/√3, 2(p/√3)^3)}, {p,0} and {p/√3, -2(p/√3)^3}.

Now, determine the ratio of side lengths=(1/2)(√6+√2)

Answer=10