Dynamic Geometry: P78

Preliminaries Let $ABCD$ be the square. Denote the centers of the semicircles by $K$ , $L$ , their radii by ${{r}_{1}}$ , ${{r}_{2}}$ respectively. Denote by ${{r}_{3}}$ , ${{r}_{4}}$ the radii of the cyan and the blue circles respectively. Denote by $E$ the intersection of the perpendicular bisectors of $AB$ and $AD$ . Denote by $Q$ , $H$ , $R$ and $I$ the intersections of these lines with the semicircles and sides $BC$ and $CD$ , as seen in the figure. Let $J$ be the intersection of line $AE$ with the red arc.

Step 1: To prove that $E$ is the center of the cyan circle.
It is easy to see that $Q$ and $R$ belong to the diagonal $BD$ of the square. Moreover, $QE=KH-KQ-EH=KH-KQ-LB=1-{{r}_{1}}-{{r}_{2}} \ \ \ \ \ (1)$ Similarly, $ER=IL-IE-RL=IL-DK-RL=1-{{r}_{1}}-{{r}_{2}}$ Hence the circle with center $E$ and radius $y=QE=ER=1-{{r}_{1}}-{{r}_{2}}$ is tangent to both semicircles. Furthermore, $EJ=AJ-AE=AJ-KL=1-\left( {{r}_{1}}+{{r}_{2}} \right)=y$ Thus, $E$ is the center of the cyan circle tangent to the semicircles and the red arc.

Step 2: To establish relations for ${{r}_{1}}$ , ${{r}_{2}}$ and ${{r}_{3}}$
From $(1)$ we get ${{r}_{1}}+{{r}_{2}}+{{r}_{3}}=1 \ \ \ \ \ (2)$ By Pythagorean theorem on $\triangle AKL$ we have

$K{{L}^{2}}=A{{K}^{2}}+A{{L}^{2}} \Rightarrow {{\left( {{r}_{1}}+{{r}_{2}} \right)}^{2}}={{\left( 1-{{r}_{1}} \right)}^{2}}+{{\left( 1-{{r}_{2}} \right)}^{2}}$ which gives ${{r}_{1}}+{{r}_{2}}=1-{{r}_{1}}{{r}_{2}} \ \ \ \ \ (3)$ The latter, combined with $(2)$ gives ${{r}_{1}}{{r}_{2}}={{r}_{3}} \ \ \ \ \ (4)$

Step 3: To express ${{r}_{4}}$ in terms of ${{r}_{3}}$
By Descartes’ Circle Theorem ,

$\begin{aligned} {{k}_{4}}={{k}_{1}}+{{k}_{2}}+{{k}_{3}}+{{k}_{4}}+2\sqrt{{{k}_{1}}{{k}_{2}}+{{k}_{2}}{{k}_{3}}+{{k}_{3}}{{k}_{1}}} & \Rightarrow \frac{1}{{{r}_{4}}}=\frac{1}{{{r}_{1}}}+\frac{1}{{{r}_{2}}}+\frac{1}{{{r}_{3}}}+2\sqrt{\frac{1}{{{r}_{1}}}\cdot \frac{1}{{{r}_{2}}}+\frac{1}{{{r}_{2}}}\cdot \frac{1}{{{r}_{3}}}+\frac{1}{{{r}_{3}}}\cdot \frac{1}{{{r}_{1}}}} \\ & \Rightarrow \frac{1}{{{r}_{4}}}=\frac{{{r}_{1}}{{r}_{2}}+{{r}_{2}}{{r}_{3}}+{{r}_{3}}{{r}_{1}}}{{{r}_{1}}{{r}_{2}}{{r}_{3}}}+2\sqrt{\frac{{{r}_{1}}+{{r}_{2}}+{{r}_{3}}}{{{r}_{1}}{{r}_{2}}{{r}_{3}}}} \\ & \overset{\left( 2 \right),\left( 4 \right)}{\mathop{\Rightarrow }}\,\frac{1}{{{r}_{4}}}=\frac{{{r}_{3}}+\left( 1-{{r}_{3}} \right){{r}_{3}}}{{{r}_{3}}^{2}}+2\sqrt{\frac{1}{{{r}_{3}}^{2}}} \\ & \Rightarrow \frac{1}{{{r}_{4}}}=\frac{4-{{r}_{3}}}{{{r}_{3}}} \\ & \Rightarrow {{r}_{4}}=\frac{{{r}_{3}}}{4-{{r}_{3}}} \ \ \ \ \ (5)\\ \end{aligned}$

Step 4: To find ${{r}_{1}}$ , ${{r}_{2}}$ and hence their ratio
Since the sum of all the radii equals $\dfrac{117}{112}$ we have ${{r}_{1}}+{{r}_{2}}+{{r}_{3}}+{{r}_{4}}=\frac{117}{112}\overset{\left( 2 \right)}{\mathop{\Rightarrow }}\,1+{{r}_{4}}=\frac{117}{112}\overset{\left( 5 \right)}{\mathop{\Rightarrow }}\,1+\frac{{{r}_{3}}}{4-{{r}_{3}}}=\frac{117}{112}\Leftrightarrow {{r}_{3}}=\dfrac{20}{117} \ \ \ \ \ (6)$ $\left( 2 \right)\Rightarrow {{r}_{1}}+{{r}_{2}}=1-{{r}_{3}}\overset{\left( 6 \right)}{\mathop{\Rightarrow }}\,{{r}_{1}}+{{r}_{2}}=\frac{97}{117}$ ${{\left( 4 \right)}_{3}}\overset{\left( 6 \right)}{\mathop{\Rightarrow }}\,{{r}_{1}}+{{r}_{2}}=\dfrac{20}{117}$

Hence, ${{r}_{1}}$ , ${{r}_{2}}$ are the roots of the quadratic equation ${{x}^{2}}-\dfrac{97}{117}x+\dfrac{20}{117}=0$ , i.e. ${{r}_{1}}=\dfrac{4}{9} \ \ \ and \ \ \ {{r}_{2}}=\dfrac{5}{13}$ or vice versa.

Finally, the ratio of the larger yellow semicircle’s radius to the smaller yellow semicircle’s radius is $\dfrac{\dfrac{4}{9}}{\dfrac{5}{13}}=\dfrac{52}{45}$ For the answer, $p=52$ , $q=45$ , thus, $p+q=\boxed{97}$ .

Let the centers and radii of the left semicircle, right semicircle and cyan circle be $A$ , $B$ , and $C$ and $r$ , $r_1$ , and $r_2$ respectively and the bottom-left vertex of the unit square be $D$ . By Pythagorean theorem , we note that

$\begin{aligned} AD^2 + BD^2 & = AB^2 \\ (1-r)^2 + (1-r_1)^2 & = (r+r_1)^2 \\ 2 - 2r - 2r_1 & = 2rr_1 \\ 1 - r & = r_1 + rr_1 \\ \implies r_1 & = \frac {1-r}{1+r} \end{aligned}$

Also note that $AC = BD \implies r+r_2 = 1 - r_1 \implies r + r_1+r_2 = 1 \implies r_2 = 1-r-r_1 = \dfrac {r(1-r)}{1+r} = rr_1$ .

Let the radius of the blue circle be $r_3$ . Then by Descarte's theorem ,

$\begin{aligned} \frac 1{r_3} & = \frac 1r + \frac 1{r_1} + \frac 1{r_2} + 2 \sqrt{\frac 1{rr_1} + \frac 1{r_1r_2} + \frac 1{r_2r}} \\ & = \frac {rr_1 + r_1r_2+r_2r}{rr_1r_2} + 2 \sqrt{\frac {r+r_1+r_2+r}{rr_1r_2}} \\ & = \frac {r_2(1+r+r_1}{r_2^2} + 2 \sqrt{\frac 1{r_2^2}} \\ & = \frac {1+1-r_2}{r_2} + \frac 2{r_2} = \frac {4-r_2}{r_2} \\ \implies r_3 & = \frac {r_2}{4-r_2} \end{aligned}$

When the sum of all radii is $\dfrac {117}{112}$ :

$\begin{aligned} \blue{r + r_1 + r_2} + r_3 & = \frac {117}{112} \\ \blue 1 + \frac {r_2}{4-r_2} & = \frac {117}{112} \\ \frac 4{4-r_2} & = \frac {117}{112} \\ \implies r_2 & = \frac {20}{117} \\ \frac {r(1-r)}{(1+r)} & = \frac {20}{117} \\ 117 r^2 - 97 r + 20 & = 0 \\ (13r - 5)(9r - 4) & = 0 & \small \blue{\text{Since }r \text{ and }r_1 \text{ are interchangeable.}} \\ r, r_1 & = \frac 5{13}, \frac 49 \end{aligned}$

Then the ratio of larger radius to the other radius is $\dfrac 49 \cdot \dfrac {13}5 = \dfrac {52}{45}$ and $p+q = 52+45 = \boxed{97}$ .