Dynamic Geometry: P110

We can define the variable parabola as $y = h - x^2$ , where $h \ge 0$ is the variable height of the parabola $OV$ . Define the right top vertex of $n$ th rectangle be $(x_n, y_n)$ and its height $h_n$ , then $h_n = y_n - y_{n-1}$ , $y_n = \sum_{k=1}^n h_k$ , and $y_n = h - x_n^2$ .

Now let us consider the largest rectangle inscribed by $y = h - x^2$ . The area of the rectangle is given by:

$\begin{aligned} A & = 2xy = 2x(h-x^2) & \small \blue{\text{To find the maximum area}} \\ \frac {dA}{dx} & = 2h - 6x^2 & \small \blue{\text{Putting }\frac {dA}{dx}= 0} \\ \implies x_1 & = \sqrt{\frac h3} \\ y_1 & = h - \frac h3 = \frac {2h}3 = h_1 \end{aligned}$

As the pattern repeats,

$\begin{aligned} h_2 & = \frac 23 (h-h_1) = \frac {2h}9 \\ h_3 & = \frac 23(h-h_1-h_2) = h_2 - \frac {2h_2}3 = \frac {h_2}3 = \frac {2h}{3^2} \\ h_4 & = \frac {h_3}3 = \frac {2h}{3^4} \\ \implies h_n & = \frac {2h}{3^n} \\ y_n & = \sum_{k=1}^n h_k = \sum_{k=1}^n \frac {2h}{3^k} = h-\frac h{3^n} = h - x_n^2 \\ \implies x_n & = \sqrt{\frac h{3^n}} \end{aligned}$

Given that the perimeter of the $16$ th rectangle is $\dfrac {55}{729}$ ,

$\begin{aligned} 4x_n + 2y_n & = \frac {55}{729} \\ \frac {4\sqrt h}{3^8} + \frac {4h}{3^{16}} & = \frac {55}{729} \\ \frac {h}{3^{16}} + \frac {\sqrt h}{3^8} & = \frac {55}{2916} \\ \left(\frac {\sqrt h}{3^8} + \frac 12\right)^2 & = \frac {55}{2916} + \frac 14 \\ \frac {\sqrt h}{3^8} & = \frac {14}{27} - \frac 12 = \frac 1{54} \\ \sqrt h & = \frac {3^5}2 \\ \implies h & = \frac {59049}4 \end{aligned}$

The sum of pairs of heights of the rectangles

$\sum_{n=1}^\infty 2h_n = \sum_{n=1}^\infty \frac {4h}{3^2} = \frac {4h}3 \cdot \frac 1{1-\frac 13} = 2h = \frac {59049}2$

Therefore $p + q = 59049+2 = \boxed{59051}$ .