The unique rectangle on a cubic

Since the roots of the cubic are $0$ , $p$ and $-p$ , its equation can be expressed in the form $y=ax(x-p)(x+p)=ax(x^2-p^2)$ where $a$ is a constant.

Now, the angle subtended by the diameter of a circle at the circumference is always a right angle. Therefore, since the angles in a rectangle are right angles, the vertices of the rectangle at C and D must lie on a circle with diameter AB, i.e. $x^2+y^2=p^2$ .

Equating the equation of the circle with the equation of the cubic:

$\begin{aligned}x^2+y^2&=p^2\\ y^2&=p^2-x^2\\ \left[ax\left(x^2-p^2\right)\right]^2&=p^2-x^2\\ a^2x^2\left(x^2-p^2\right)^2&=-\left(x^2-p^2\right)\\ a^2x^2\left(x^2-p^2\right)&=-1\\ a^2x^4-a^2p^2x^2+1&=0\end{aligned}$

This is a quadratic equation in terms of $x^2$ . Now, notice how the circle intersects the cubic at C and D and also another pair of points. For there to be only one possible rectangle, there must be only two intersections (which by the symmetry of the rectangle must be at $x=n$ and $x=-n$ for some number $n$ and are therefore the solutions to $x^2=n^2$ ), and so the quadratic above must have equal roots. This means the discriminant equals zero.

$\begin{aligned}\Delta=B^2-4AC&=0\\ \therefore\left(a^2p^2\right)^2-4\left(a^2\right)(1)&=0\\ a^4p^4&=4a^2\\ a^2&=\frac4{p^4}\\ a&=\frac2{p^2}&(a,p>0\ \text{WLOG})\end{aligned}$

Now we can complete the quadratic equation and find the coordinates of C and D.

$\begin{aligned}\frac4{p^4}x^4-\frac4{p^4}p^2x^2+1&=0\\ 4\frac{x^4}{p^4}-4\frac{x^2}{p^2}+1&=0\\ \left(2\frac{x^2}{p^2}-1\right)^2&=0\\ 2\frac{x^2}{p^2}&=1\\ x^2&=\frac{p^2}2\\ x&=\pm\frac p{\sqrt2}\\ \therefore y&=\pm\frac2{p^2}\frac p{\sqrt2}\left(\frac{p^2}2-p^2\right)\\ &=\mp\frac p{\sqrt2}\\ \therefore C\left(-\frac p{\sqrt2},\frac p{\sqrt2}\right),&D\left(\frac p{\sqrt2},-\frac p{\sqrt2}\right)\end{aligned}$

Finally, we simply need to find the ratio between the length and width of the rectangle. Draw a perpendicular from C to meet the x-axis at M, as in the diagram. Now, using the coordinates of C, $CM=\frac p{\sqrt2}$ and $AM=p+p{\sqrt2}$ . The right triangle ACM is similar to triangle ABC, since both have a right angle and also a common angle at B. Therefore, the ratio of the length to the width of the rectangle, i.e. $\frac{AC}{BC}$ , equals $\frac{AM}{CM}=\frac{p+p\sqrt2}p=\frac{1+\sqrt2}1$ . Therefore the answer is $1+2+1=\boxed{4}$