Dynamic Geometry: P124

Since the system is symmetrical about the $y$ -axis, we need consider only the positive quadrant. Let the radius of the $n$ th circle be $r_n$ , its lowest point be $(0, h_{n-1})$ , and the point it is tangent the positive side of the parabola be $P(x_n, y_n)$ . From the previous problem Dynamic Geometry P123 , we have found that:

$\begin{cases} r_n = \dfrac {2n-1}{2k} \\ h_n = 2 \displaystyle \sum_{j=1}^n r_j \\ y_n = h_{n-1} + r_n - \dfrac 1{2k} \end{cases}$

$\begin{aligned} \implies y_n & = 2 \sum_{j=1}^{n-1} \frac {2j-1}{2k} + \frac {2n-1}{2k} - \frac 1{2k} \\ & = \frac {(n-1)n - (n-1)}k + \frac {n-1}k \\ & = \frac {n(n-1)}k & \small \blue{\text{Since }y_n = kx_n^2} \\ x_n & = \frac {\sqrt{n(n-1)}}k \end{aligned}$

Then the area of the $100$ th quadrilateral is given by:

$\begin{aligned} 2r_{101}x_{101} & = \frac {201\sqrt{101}}{10} \\ 2 \cdot \frac {2(101)-1}{2k} \cdot \frac {10\sqrt{101}}k & = \frac {201\sqrt{101}}{10} \\ \frac {2010\sqrt{101}}{k^2} & = \frac {201\sqrt{101}}{10} \\ \implies k & = 10 \\ r_1 & = \frac {2(1)-1}{2k} = \frac 1{20} \end{aligned}$

Therefore $p+q = 1+20 = \boxed{21}$ .