Bulgarian National Mathematical Olympiad

Lines $p$ and $q$ with the given properties exist if and only if $f(x)$ has local minima and maxima, and $p$ and $q$ pass through the point of maxima $D$ and the point of minima $B$ on the graph $f(x)$, respectively. Using transformations of the forms $g(x)=f(x)+a$ and $g(x)=f(x+a)$, we may move this graph such that $D(0,0)$ and $f(x)=x^2(x-t)$, $t>0$.

Then $p$ is the $x$-axis and its second common point with the graph is $C(t,0)$. Since $f^{\prime }(x)=3x^2-2xt$, then $B\left(\frac{2t}{3},-\frac{4t^3}{27}\right)$. The second common point of $q$ and the graph is $A(\alpha ,-\frac{4t^3}{27})$, where $\alpha$ is a root of the equation $f(x)=-\frac{4t^3}{27}$. Since this equation has double root $x_1=x_2=\frac{2t}{3}$, then $x_1{x}_2\alpha =-\frac{4t^3}{27}$ gives $\alpha =-\frac{t}{3}$. Using that $AB=DC=t$, it follows that $ABCD$ is a parallelogram with area $S_{ABCD}=t\cdot \frac{4t^3}{27}=\frac{4t^4}{27}$.

Moreover,

$$BC=\sqrt{{\left(\frac{2t}{3}-t\right)}^2+{\left(-\frac{4t^3}{27}\right)}^2}=\frac{t}{3}\sqrt{1+\frac{16t^4}{81}}.$$

Straightforward calculations show that the conditions $S_{ABCD}=6$ and $BC=t$

are equivalent to $t^4=\frac{81}{2}$ which solves the problem.