Belarusian Mathematical Olympiad

Answer: $S = \frac{p^2}{4} - 1$.

Note that if the points with coordinates $(m, m^2)$ and $(n, n^2)$ belong to the parabola $y = x^2$, then the line passing through them has the equation $y = (m + n)x - mn$. Indeed, since the coordinates of each of the two points satisfy this linear equation, the entire line is given by this equation.

Denote the coordinates of the points: $A(-a, a^2)$, $C(a, a^2)$, $B(b, b^2)$, $D(d, d^2)$ (we take into account that $AC \parallel Ox$, so points $A$ and $C$ are symmetric with respect to $Oy$). The line $AD$ has the equation $y = (d-a)x + da$. On the other hand, from the conditions $\angle DAC = 45^\circ$, so the slope of this line equals to $1$, i.e. $d-a = 1$. Similarly the line $AB$ has the equation $y = (b-a)x + ba$ and its slope equals to $-1$, so $b-a = -1$. Denote the points $B_1(d, b^2)$ and $C_1(d, c^2)$ (see the Fig.). From the right triangle $BB_1D$ we get $p^2 = (d-b)^2 + (d^2-b^2)^2$. Since $d-b = 2$ and $d+b = 2a$ it follows that $p^2 = 4 + 16a^2$, whence $a^2 = \frac{1}{16}(p^2 - 4)$.

The required area is equal to $S_{ABCD} = S_{ABC} + S_{ADC}$. The triangles $ABC$ and $ADC$ have common base $AC = 2a$ and the sum of their altitudes equals $C_1B_1 + C_1D = DB_1 = 4a$. Therefore the required area is equal to $\frac{1}{2} \cdot 2a \cdot 4a = \frac{1}{4}(p^2 - 4)$.