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 $B{B}_{1}D$ 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 $C_1{B}_1+C_{1}D=D{B}_1=4a$. Therefore the required area is equal to $\frac{1}{2}\cdot 2a\cdot 4a=\frac{1}{4}(p^2-4)$.