Dynamic Geometry: P123

Let the radius of the $n$ th circle be $r_n$ . First find $r_1$ , the radius of the largest inscribed lowest circle. The lowest circle satisfies

$\begin{cases} x^2 + (y-r_1)^2 = r_1^2 \\ y = kx^2 \end{cases} \\ \begin{aligned} \implies x^2 + (kx^2 - r_1)^2 & = r_1^2 \\ x^2 + k^2x^4 - 2kr_1x^2 + r_1^2 & = r_1^2 \\ x^2(1 + k^2 x^2 - 2kr_1) & = 0 \\ \implies x & = \begin{cases} 0 \\ \pm \dfrac {\sqrt{2kr_1-1}}k \end{cases} \end{aligned}$

When the circle is inscribable by the parabola, there is only one root $x=0$ . This means that $x = \dfrac {\sqrt{2kr_1 -1}}k$ is $0$ or imaginary. This means that $2kr_1 - 1 \le 0 \implies r_1 \le \dfrac 1{2k}$ .

Now consider $r_n$ . Let the lowest point of the $n$ th circle be $(0, h_{n-1})$ and the point it is tangent to the right side of $y=kx^2$ be $P(x,y)$ . Then $P(x,y)$ satisfies:

$\begin{cases} x^2 + (y-h_{n-1} - r_n)^2 = r_n^2 & ...(1) \\ y = kx^2 & ...(2) \end{cases}$

At $P(x,y)$ , the gradient of the circle is the same as that of the parabola, $\dfrac {dy}{dx} = 2kx$ . Differentiating the equation of the circle w.r.t. $x$ :

$\begin{aligned} 2x + 2(y - h_{n-1} - r_n) \blue{\frac {dy}{dx}} & = 0 & \small \blue{\text{Note that }\frac {dy}{dx} = 2kx} \\ 2x + 2(y - h_{n-1} - r_n)(2kx) & = 0 & \small \blue{\text{Since }x \ne 0} \\ 1 + 2k(y - h_{n-1} - r_n) & = 0 \\ \implies y - h_{n-1} - r_n & = - \frac 1{2k} \end{aligned}$

Then the equation of circle $(1)$ becomes:

$\begin{aligned} x^2 + \left(-\frac 1{2k}\right)^2 & = r_n^2 & \small \blue{\text{Since }y=kx^2 \text{ and }y = h_{n-1}+r_n-\frac 1{2k}} \\ \frac {h_{n-1}+r_n}k - \frac 1{2k^2} + \frac 1{4k^2} & = r_n^2 \\ r_n^2 - \frac {r_n}k + \frac 1{4k^2} & = \frac {h_{n-1}}k \\ \left(r_n - \frac 1{2k} \right)^2 & = \frac {h_{n-1}}k \\ \implies r_n & = \sqrt{\frac {h_{n-1}}k} + \frac 1{2k} \end{aligned}$

We can prove by induction that $r_n = \dfrac {2n-1}{2k}$ . When $n=1$ , $y_0 = 0$ , $\implies r_1 = \dfrac 1{2k}$ . The claim is true for $n=1$ . Assuming the claim is true for $n$ , then

$\begin{aligned} r_{n+1} & = \sqrt{\frac {y_n}k} + \frac 1{2k} = \sqrt{\frac {h_{n-1}+2r_n}k} + \frac 1{2k} = \sqrt{\left(r_n - \frac 1{2k} \right)^2 + \frac {2r_n}k} + \frac 1{2k} \\ & = r_n + \frac 1{2k} + \frac 1{2k} = \frac {2n-1}{2k} + \frac 1{2k} + \frac 1{2k} = \frac {2(n+1)-1}{2k} \end{aligned}$

The claim is also true for $n+1$ , hence it is true for all $n \ge 1$ .

When the circumference of the $100$ th circle,

$\begin{aligned} 2\pi r_{100} & = \frac {199 \pi}{25} \\ 2 \cdot \frac {2(100)-1}{2k} & = \frac {199}{25} \\ \implies k & = 25 \end{aligned}$

The sum of all reciprocals of area of circle,

$\begin{aligned} \sum_{n=1}^\infty \frac 1{\pi r_n^2} & = \sum_{n=1}^\infty \frac {2500}{(2n-1)^2\pi} = \frac {2500}\pi \cdot \frac 34 \zeta(2) = \frac {2500}\pi \cdot \frac 34 \cdot \frac {\pi^2}6 = \frac {625\pi}2 \end{aligned}$

Therefore $p+q = 625+2 = \boxed{627}$ .

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Notation: $\zeta (\cdot)$ denotes the Riemann zeta function .

• Let us evaluate the radus of the first circle in terms of $k$ , this circle needs to be the largest possible.
• We can write its equation : $x^2+\left(y-r_1\right)^2-r_1^2=0$
• We can combine the two equations: $\begin{cases}x^2+\left(y-r_1\right)^2-r_1^2=0\\y=k\cdot x^2\end{cases}$
• We get $x²+\left(k\cdot \:x²-r_1\right)^2-r_1^2=0$ $<=>$ $x^4\cdot k^2+x^2\left(1-2kr_1\right)=0$
• We evaluate $\Delta \:_x=\left(1-2kr_1\right)^2$ and set it equal to $0$ : $\left(1-2kr_1\right)^2=0\:<=>r_1=\dfrac{1}{2k}$
• We can use a similar method to get the next radii:

• $r_1=\dfrac{1}{2k}$ ; $r_2=\dfrac{3}{2k}$ ; $r_3=\dfrac{5}{2k}$ ; $r_4=\dfrac{7}{2k}$ ; $r_5=\dfrac{9}{2k}$

• This allows us to identify a pattern, $r_n=\dfrac{2n-1}{2k}\:\:$

• Then $P_{100}=\dfrac{2\cdot 100-1}{2k}\cdot 2\pi =\dfrac{199\pi }{25}$ $<=>$ $k=25$
• Now we can find $\sum _{n=1}^{\infty }\:\left(\dfrac{2\cdot \:25}{2n-1}\right)^2\cdot \dfrac{1}{\pi }$ and we get $\dfrac{625\pi }{2}$
• Finally $625+2=627$