Dynamic Geometry: P100

Let the curvatures of the purple circle, green semicircle, and cyan semicircle be $k_0$ , $k_1$ , and $k_2$ respectively. This means that their radii are $\dfrac 1{k_0}$ , $\dfrac 1{k_1}$ , and $\dfrac 1{k_2}$ respectively. By Descarte's theorem :

$\begin{aligned} k_0 & = k_1 + k_2 - 1 + 2 \sqrt{k_1k_2 - k_1 - k_2} \\ & = k_1 + k_2 - 1 + 2 \sqrt{(k_1-1)k_2 - k_1} \\ & = k_1 + k_2 - 1 + 2 \sqrt{k_1 - k_1} \\ & = k_1 + k_2 - 1 \end{aligned}$

Let the curvatures of the $n$ th blue circle and yellow circle be $b_n$ and $y_n$ respectively, and $b_0 = y_0 = k_0$ . Through observation it is found that

$b_n = (k_1-1)(n+1)^2 + k_2$

Let us prove this claim by induction for all $n \ge 0$ .

When $n=0$ , $b_0 = k_1 - 1 + k_2$ , Therefore the claim is true for $n=0$ . Assuming that the claim is true for $n$ , then

$\begin{aligned} b_{n+1} & = b_n + k_1 - 1 + 2\sqrt{b_nk_1 - b_n-k_1} \\ & = (k_1-1)(n+1)^2 + k_2 + k_1 - 1 + 2 \sqrt{(k_1-1)b_n - k_1} \\ & = (k_1-1)\left((n+1)^2 + 1\right) + k_2 + 2 \sqrt{(k_1-1)^2(n+1)^2 + k_1 - k_1} \\ & = (k_1-1)\left(n^2 + 2n + 2 \right) + k_2 + 2 (k_1-1)(n+1) \\ & = (k_1-1)\left(n^2 + 4n + 4 \right) + k_2 \\ & = (k_1-1)(n + 2)^2 + k_2 \end{aligned}$

The claim is also true for $n+1$ and it is true for all $n \ge 0$ . Since $k_1$ and $k_2$ are interchangeable and $k_0$ is common,

$y_n = (k_2-1)(n+1)^2 + k_1$

When $\dfrac {1/b_{25}}{1/y_{20}} = \dfrac {1779}{6094}$ , we have:

$\begin{aligned} \frac {y_{20}}{b_{25}} & = \frac {1779}{6094} \\ \frac {21^2(k_2-1)+k_1}{26^2(k_1-1)+k_2} & = \frac {1779}{6094} \\ 2687454k_2 - 2687454 + 6094k_1 & = 1202604k_1 - 1202604 + 1779k_2 \\ 537135k_2 - 239302k_1 & = 296970 & \small \blue{\text{Note that }\frac 1{k_1} + \frac 1{k_2} = 1} \\ \frac {537135k_1}{k_1-1} - 239302k_1 & = 296970 & \small \blue{\implies k_2 = \frac {k_1}{k_1-1}} \\ 239302k_1^2 -479467k_1 - 296970 & = 0 \\ (2k_1 -5)(119651k_1+59394) & = 0 & \small \blue{\text{Since }k_1 > 0} \\ k_1 & = \frac 52 \\ \implies \frac {1/k_1}{1/k_2} & = \frac {k_2}{k_1} = \frac 1{k_1-1} = \frac 23 \end{aligned}$

Therefore $p+q = 2 + 3 = \boxed 5$ .

Let $AB$ , $BC$ be the diameters of the green and the cyan semicircles respectively.

The sequence of the blue circles together with the purple circle form a Pappus chain $\left\{ {{C}_{1}},\ {{C}_{2}},\ {{C}_{3}},\ \ldots \right\}$ . Using the ratio $r=\dfrac{AB}{AC}$ of the green and the black semicircles’ diameters, in which the chain is inscribed, the radius of the ${{n}^{th}}$ circle of the chain is given by the expression: ${{r}_{n}}=\dfrac{\left( 1-r \right)r}{2\left[ {{n}^{2}}{{\left( 1-r \right)}^{2}}+r \right]}$ The 25th blue circle is the 26th circle of the chain, thus, its radius ${{r}_{b}}$ is ${{r}_{b}}={{r}_{26}}=\dfrac{\left( 1-r \right)r}{2\left[ {{26}^{2}}{{\left( 1-r \right)}^{2}}+r \right]}$ Likewise, the set of the yellow circles together with the purple circle form a Pappus chain as well. The corresponding ratio of the diameters of the cyan and black semicircles of the arbelos is $\dfrac{BC}{AC}=\dfrac{AC-AB}{AC}=1-r$ Using this ratio in the formula for the radii of the circles in the chain we get ${{r}_{n}}=\frac{r\left( 1-r \right)}{2\left[ {{n}^{2}}{{r}^{2}}+\left( 1-r \right) \right]}$ and for the radius of the 20th yellow circle we have ${{r}_{y}}={{r}_{21}}=\dfrac{r\left( 1-r \right)}{2\left[ {{21}^{2}}{{r}^{2}}+\left( 1-r \right) \right]}$ According to the condition given in the problem we have the equation $\begin{aligned} d\frac{\dfrac{r\left( 1-r \right)}{2\left[ {{26}^{2}}{{r}^{2}}+\left( 1-r \right) \right]}}{\dfrac{r\left( 1-r \right)}{2\left[ {{21}^{2}}{{r}^{2}}+\left( 1-r \right) \right]}}=\dfrac{1779}{6094} & \Leftrightarrow \dfrac{{{21}^{2}}{{r}^{2}}+\left( 1-r \right)}{{{26}^{2}}{{r}^{2}}+\left( 1-r \right)}=\dfrac{1779}{6094} \\ & \Leftrightarrow 296970{{r}^{2}}+479467r-239302=0 \\ & \overset{r>0}{\mathop{\Leftrightarrow }}\,r=\dfrac{2}{5} \\ \end{aligned}$ Consequently, the ratio of the green semicircle's radius to the radius of the cyan semicircle is $\dfrac{{{r}_{green}}}{{{r}_{purple}}}=\dfrac{AB}{BC}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{r}{1-r}=\dfrac{\dfrac{2}{5}}{1-\dfrac{2}{5}}=\dfrac{2}{3}$ For the answer, $p=2$ , $q=3$ , thus, $p+q=\boxed{5}$ .