Let $A=(p^2,-p), B=(q^2,-q), C=(r^2,-r)$ be the vertices of a $\Delta ABC$. A paralellogram $AFDE$ is drawn with $D, E, F$ on line segment $BC, CA, AB$ respectively. Show that the maximum area of such a parallelogram is $\frac{(p+q)(q+r)(p-r)}{4}$.
I have tried bashing with coordinates and exploiting the midpoint theorem of triangles but none of them have led to this elegant result. Our teacher says that an approach using vector geometry may work. How do I go about this problem?
I think the key here is that you have specified that $AFDE$ must be a parallelogram. So once you have decided on, say, the point $F$, the other points $D$ and $E$ are uniquely determined (draw a picture to see why).
Lets do this problem with vectors.
If $F$ must be on the edge $AB$, then we have the vector
$$\vec{AF} = t\vec{AB} = t(q^2-p^2,-q+p)$$
for some $t \in [0,1]$. Similarly, $E$ lies on $AC$, so
$$\vec{AE} = u\vec{AC} = u(r^2-p^2,-r+p)$$
for some $u \in [0,1]$.
Here, it might be easier to switch back to cartesian coordinate system for a moment. We know that $D$ lies on $BC$, so
$$D = vB+(1-v)C = (vq^2+(1-v)r^2,-vq-(1-v)r)$$
for some $v \in [0,1]$. We then obtain
$$\vec{AD} = (vq^2+(1-v)r^2-p^2,-vq-(1-v)r+p)$$
But note that we can express the vector $\vec{AD}$ in a number of different ways. The most important observation here is that $\vec{AD} = \vec{AE} + \vec{AF}$ due to the parallelogram law. Plugging in what he had:
\begin{align} & \vec{AD} = \vec{AE} + \vec{AF} \\ \implies & (vq^2+(1-v)r^2-p^2,-vq-(1-v)r+p) = u(r^2-p^2,-r+p) + t(q^2-p^2,-q+p) \\ \end{align}
$$ \implies \begin{cases} vq^2+(1-v)r^2-p^2 = u(r^2-p^2) + t(q^2-p^2) \\ -vq-(1-v)r+p = u(-r+p) + t(-q+p) \end{cases} $$
$$ \implies \begin{pmatrix} r^2-p^2 & q^2-p^2 \\ -r+p & -q+p \end{pmatrix} \begin{pmatrix} u \\ t \end{pmatrix} = \begin{pmatrix} vq^2+(1-v)r^2-p^2 \\ -vq-(1-v)r+p \end{pmatrix}$$
$$ \implies u = 1-v \qquad t=v$$
Plug these back into the expressions for $\vec{AF}$ and $\vec{AE}$:
$$\vec{AF} = t(q^2-p^2,-q+p) = v(q^2-p^2,-q+p)$$
$$\vec{AE} = u(r^2-p^2,-r+p) = (1-v)(r^2-p^2,-r+p)$$
Finally, the area of the parallelogram is given by the magnitude of the cross product:
\begin{align} \text{Area} & = |\vec{AF} \times \vec{AE}| \\ & = v(1-v) \Bigg|\begin{pmatrix} q^2-p^2 \\ -q+p \\ 0 \end{pmatrix} \times \begin{pmatrix} r^2-p^2 \\ -r+p \\ 0 \end{pmatrix} \Bigg| \\ & = v(1-v)|(q^2-p^2)(-r+p)-(-q+p)(r^2-p^2)| \\ & = v(1-v)|(q-p)(p-r)(r-q)| \end{align}
and this expression is maximized when $v=\frac 12$, giving
$$\text{Area} = \frac{|(q-p)(p-r)(r-q)|}{4}$$
hmmmmm...