Absolute Value Quadrilateral

Let $P$ be the top vertex, $Q$ be the right vertex, $R$ be the bottom vertex, and $S$ be the left vertex of quadrilateral $PQRS$ .

Since the slopes of $PQ$ and $QR$ multiply to $-\frac{a}{12} \cdot \frac{12}{a} = -1$ , $PQ$ and $QR$ are perpendicular. By symmetry, $PS$ and $SR$ are perpendicular.

Now reflect $\triangle PSR$ to $\triangle PS'R$ as follows:

Since $PQ$ and $S'R$ have the same slope and same length, and since $PS'$ and $QR$ have the same slope and same length, and since $PQ$ and $QR$ are perpendicular, quadrilateral $PQRS'$ is always a rectangle.

Let the sides of rectangle $PQRS'$ be $m$ and $n$ . Then the (fixed) area $A = mn$ and the perimeter $P = 2m + 2n$ . Substituting gives $P = 2m + 2Am^{-1}$ , so $P' = 2 - 2Am^{-2}$ and $P'' < 0$ . This means the maximum perimeter is when $P' = 2m + 2Am^{-1} = 0$ , which solves to $A = m^2$ , so $PQRS'$ (and $PQRS$ ) must be a square.

If $PQRS$ is a square, then $PQ$ and $PS$ are perpendicular, so $-\frac{a}{12} \cdot \frac{a}{12} = -1$ , which solves to $a = \boxed{12}$ .