Escalating $x$ -er

The purple line has equation $y=a(x-a)+a=-a^{2}+(x+1)a$ where $(a,a)$ is the marked point.

To find the highest point at a given x-coordinate note the equation looks like a negative quadratic in terms of $a$ so its vertex is given by $a=\frac{-(x+1)}{2\cdot -1}=\frac{x+1}{2}$ .

This is enough to find the equation of the dotted curve by subbing $a$ into the original equation again:

$y=\frac{1+x}{2}(x-\frac{1+x}{2})+\frac{1+x}{2}=(\frac{1+x}{2})^{2}$

It's just a parabola. The quickest way of finding the area is the integral

$\int_{0}^{1} (\frac{1+x}{2})^{2} dx = \frac{7}{12} \approx 0.5833$ so the solution is $\boxed{583}$

Let $t$ denote the parametric variable of the purple line. Then, its equation is $y(x,t) = t(x - t) + t = tx + (-t^2 + t) = -t^2 + t(x + 1)$ Since the intersection points of both orange boundary and purple line are the highest peak points (thus global maxima) of all parabolas with respect to $t$ , quadratic formula shows that $t = \dfrac{x + 1}{2}$ which gives $y(x) = -\left(\dfrac{x + 1}{2}\right)^2 + \left(\dfrac{x + 1}{2}\right) \cdot (x + 1) = \dfrac{1}{4}(x + 1)^2$ Thus, since $A = \int\limits_0^1 \dfrac{1}{4}(x + 1)^2\,dx = \dfrac{7}{12}$ the answer to the problem is $\lfloor 1000A \rfloor = \boxed{583}$