Dynamic Geometry: P79

Let the centers of the large, yellow, and cyan semicircle, and the green circle be $O$ , $A$ , $B$ , and $C$ , and the radii of the yellow and cyan semicircles, and the green circle be $r_1$ , $r_2$ , and $r_3$ respectively. Then $r_1 + r_2 = 1$ . Since $AO = 1-r_1 = r_2$ and $OB = 1 - r_2 = r_1$ . By cosine rule ,

$\begin{aligned} BC^2 & = OB^2 + OC^2 - 2\cdot OB \cdot OC \cdot \cos \blue{\angle BOC} & \small \blue{\text{Let }\angle BOC = \theta.} \\ (r_2+r_3)^2 & = r_1^2 + (1-r_3)^2 - 2 r_1 (1-r_3) \cos \blue \theta \quad ...(1) & \small \blue{\text{Similarly for }CA^2} \\ (r_1+r_3)^2 & = r_2^2 + (1-r_3)^2 + 2 r_2 (1-r_3) \cos \theta \quad ...(2) \end{aligned}$

From $r_2 \cdot (1) + r_1 \cdot (2)$ :

$\begin{aligned}r_2(r_2+r_3)^2 + r_1(r_1+r_3)^2 & = r_2r_1^2 + r_1r_2^2 + (r_2+r_1)(1-r_3)^2 \\ r_1^3 + r_2^3 + 2r_3(r_1^2+r_2^2) + \blue{(r_1+r_2)}r_3^2 & = r_1r_2\blue{(r_1+r_2)} + \blue{(r_1+r_2)}(1 - 2r_3 + r_3^2) \\ (r_1+r_2)((r_1+r_2)^2 - 3r_1r_2) + 2r_3((r_1+r_2)^2 - 2r_1r_2) + r_3^2 & = r_1r_2 + 1 - 2r_3 + r_3^2 \quad \quad \small \blue{\text{Note that }r_1+r_2 = 1} \\ 1 - 3r_1r_2 + 2r_3 - 4r_1r_2r_3 + r_3^2 & = r_1r_2 + 1 - 2r_3 + r_3^2 \\ 4r_3 - 4r_1r_2r_3 & = 4r_1r_2 \\ \implies r_3 & = \frac {r_1r_2}{1-r_1r_2} \end{aligned}$

Let the radius of the pink circle be $r$ . By Descartes' theorem ,

$\begin{aligned} \frac 1r & = \frac 1{r_1} + \frac 1{r_2} + \frac 1{r_3} + 2 \sqrt{\frac 1{r_1r_2} + \frac 1{r_2r_3} +\frac 1{r_3r_1}} \\ & = \frac {r_1+r_2}{r_1r_2} + \frac 1{r_1r_2} - 1 + 2\sqrt{\frac 1{r_1r_2}+ \left(\frac 1{r_1r_2} -1\right) \frac {r_1+r_2}{r_1r_2}} \\ & = \frac 1{r_1r_2} + \frac 1{r_1r_2} - 1 + \frac 2{r_1r_2} = \frac 4{r_1r_2} - 1 \\ \implies r_1r_2 & = \frac {4r}{1+r} \end{aligned}$

Let the center of the pink circle be $D$ , the two tangent with the semicircles be $E$ and $F$ , and $\angle EDF = \phi$ . Note that $\phi$ is obtuse. When the ratio of the area of pink circle to the area of the triangle is $\frac {2929}{1200}\pi$ , we have:

$\frac {\pi r^2}{\frac 12 r^2 \sin \phi} = \frac {2929 \pi}{1200} \implies \sin \phi = \frac {2400}{2929} \implies \cos \phi = - \frac {1679}{2929}$

By cosine rule ,

$\begin{aligned} AD^2 + BD^2 - 2 \cdot AD \cdot BD \cdot \cos \phi & = AB^2 \\ (r_1+r)^2 + (r_2+r)^2 + 2(r_1+r)(r_2+r) \cdot \frac {1679}{2929} & = 1^2 \\ r_1^2 + r_2^2 + 2\blue{(r_1+r_2)}r + 2r^2 + 2(r_1r_2 + 2\blue{(r_1+r_2)}r + r^2) \cdot \frac {1679}{2929} & = 1 & \small \blue {\text{Note that }r_1+r_2 = 1} \\ (r_1+r_2)^2 - 2r_1r_2 + 2r + 2r^2 + 2(r_1r_2 + 2r + r^2) \cdot \frac {1679}{2929} & = 1 \\ 1 - 2r_1r_2 + 2r + 2r^2 + 2(r_1r_2 + 2r + r^2) \cdot \frac {1679}{2929} & = 1 \\ - r_1r_2 + r + r^2 + (r_1r_2 + 2r + r^2) \cdot \frac {1679}{2929} & = 0 \\ 2304r(1+r) & = 625r_1r_2 = \frac {2500r}{1+r} \\ (1+r)^2 & = \frac {625}{576} \\ \implies r & = \frac {25}{24} - 1 = \frac 1{24} \\ r_1r_2 & = \frac {4r}{1+r} = \frac 4{25} \\ \implies r_3 & = \frac {r_1r_2}{1-r_1r_2} = \frac 4{21} \end{aligned}$

Therefore $\sqrt{p+q} = \sqrt{4+21} = \boxed 5$ .