Dynamic Geometry: P72

Disclaimer: This is a lengthy and detailed approach.

Preliminaries
Let $ABCD$ be the square. Denote the centers of the semicircles by $K$ , $L$ , their radii by ${{r}_{1}}$ , ${{r}_{2}}$ respectively and their intersection point by $P$ . 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 figure 1. Let $J$ be the intersection of line $AE$ with the red arc.

Let $\angle KDP=\angle KPD=θ$ and $\angle LBP=\angle LPB=φ$ (figure 2).
Points $K$ , $P$ and $L$ are collinear, and $\angle AKL=2θ$ , $\angle ALK=2φ$ , thus $2\theta +2\varphi =180{}^\circ -\angle KAL=180{}^\circ -90{}^\circ =90{}^\circ$ , hence $\theta +\varphi =45{}^\circ \ \ \ \ \ (1)$ Step 1: To prove that $E$ is the center of the cyan circle (figure 1).
It is easy to see that $Q$ and $R$ belong to the diagonal $BD$ of the square.
Moreover, $QE=KH-KQ-EH=KH-KQ-LB=1-{{r}_{1}}-{{r}_{2}} \ \ \ \ \ (2)$
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, $EJ=AJ-AE=AJ-KL=1-\left( {{r}_{1}}+{{r}_{2}} \right)$ Thus, $E$ is the center of the cyan circle tangent to the semicircles and the red arc.

Step 2: To establish a relation for ${{r}_{1}}$ and ${{r}_{2}}$
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 ${{r}_{1}}+{{r}_{2}}=1-{{r}_{1}}{{r}_{2}} \ \ \ \ \ (3)$

Step 3: To find expressions for the area of $\triangle PQR$ (figure 3)
Figure 3 Let $PM$ be a height of $\triangle PQR$ .
We notice that $PD=2{{r}_{1}}\cos \theta$ and $PB=2{{r}_{2}}\cos \varphi$ .
Hence for the area, $\begin{aligned} \left[ PQR \right] & =\frac{1}{2}QR\cdot PM \\ & =\frac{1}{2}\left( \sqrt{2}QE \right)\cdot PD\sin \left( 45{}^\circ -\theta \right) \\ & \overset{\left( 2 \right)}{\mathop{=}}\,\frac{\sqrt{2}}{2}\left( 1-{{r}_{1}}-{{r}_{2}} \right)\cdot \left( 2{{r}_{1}}\cos \theta \right)\sin \left( 45{}^\circ -\theta \right)= \\ & \overset{\left( 3 \right),\left( 1 \right)}{\mathop{=}}\,\sqrt{2}\left( {{r}_{1}}{{r}_{2}} \right){{r}_{1}}\cos \theta \sin \varphi \\ & \\ & \Rightarrow {{r}_{2}}\left[ PQR \right]=\sqrt{2}{{\left( {{r}_{1}}{{r}_{2}} \right)}^{2}}\cos \theta \sin \varphi \ \ \ \ \ (4)\\ \end{aligned}$

In the same time, $\begin{aligned} \left[ PQR \right] & =\frac{1}{2}\left( \sqrt{2}QE \right)\cdot PB\sin \left( 45{}^\circ -\varphi \right) \\ & \overset{\left( 2 \right),\left( 1 \right)}{\mathop{=}}\,\frac{\sqrt{2}}{2}\left( 1-{{r}_{1}}-{{r}_{2}} \right)\cdot \left( 2{{r}_{2}}\cos \varphi \right)\sin \theta = \\ & \overset{\left( 3 \right)}{\mathop{=}}\,\sqrt{2}\left( {{r}_{1}}{{r}_{2}} \right){{r}_{2}}\cos \varphi \sin \theta \\ & \\ & \Rightarrow {{r}_{1}}\left[ PQR \right]=\sqrt{2}{{\left( {{r}_{1}}{{r}_{2}} \right)}^{2}}\cos \varphi \sin \theta \ \ \ \ \ (5)\\ \end{aligned}$

Step 4: To deduce an equation for $x={{r}_{1}}{{r}_{2}}$ and solve it
Adding $(4)$ and $(5)$ , $\begin{aligned} & \left( {{r}_{1}}+{{r}_{2}} \right)\left[ PQR \right]=\sqrt{2}{{\left( {{r}_{1}}{{r}_{2}} \right)}^{2}}\left( \cos \theta \sin \varphi +\cos \varphi \sin \theta \right) \\ & \overset{\left( 3 \right)}{\mathop{\Rightarrow }}\,\left( 1-{{r}_{1}}{{r}_{2}} \right)\frac{36}{1015}=\sqrt{2}{{\left( {{r}_{1}}{{r}_{2}} \right)}^{2}}\sin \left( \theta +\varphi \right) \\ & \overset{\left( 1 \right)}{\mathop{\Rightarrow }}\,\left( 1-{{r}_{1}}{{r}_{2}} \right)\frac{36}{1015}=\sqrt{2}{{\left( {{r}_{1}}{{r}_{2}} \right)}^{2}}\sin \left( 45{}^\circ \right) \\ & \Rightarrow \left( 1-{{r}_{1}}{{r}_{2}} \right)\frac{36}{1015}=\sqrt{2}{{\left( {{r}_{1}}{{r}_{2}} \right)}^{2}}\frac{\sqrt{2}}{2} \\ & \Rightarrow 1015{{\left( {{r}_{1}}{{r}_{2}} \right)}^{2}}+36{{r}_{1}}{{r}_{2}}-36=0 \\ & \overset{{{r}_{1}}{{r}_{2}}>0}{\mathop{\Rightarrow }}\,{{r}_{1}}{{r}_{2}}=\frac{6}{35} \\ \end{aligned}$

Step 5: To find ${{r}_{1}}$ , ${{r}_{2}}$ and hence their ratio
Since ${{r}_{1}}{{r}_{2}}=\dfrac{6}{35}$ , $\left( 3 \right)\Rightarrow {{r}_{1}}+{{r}_{2}}=\dfrac{29}{35}$ , hence ${{r}_{1}}$ , ${{r}_{2}}$ are the roots of the quadratic equation ${{x}^{2}}-\dfrac{29}{35}x+\dfrac{6}{35}=0$ , i.e. ${{r}_{1}}=\dfrac{3}{7} \ \ \ and \ \ \ {{r}_{2}}=\dfrac{2}{5}$ or vice versa.
Hence the ratio of the larger yellow semicircle’s radius to the smaller yellow semicircle’s radius is $\dfrac{\dfrac{3}{7}}{\dfrac{2}{5}}=\dfrac{15}{14}$ For the answer, $p=15$ , $q=15$ , thus, $p-q=\boxed{1}$ .

$\ \ \$

Fun facts about the purple triangle:

• base $QR$ always lie on one of the square's diagonals.
• The angle $\angle QPR$ remains constantly $45{}^\circ$ .
• Vertex $P$ and the point of tangency of the cyan circle with the red quarter of a circle are symmetrical with respect to a square's diagonal.
• $P$ is moving along a quarter of a circle with center one vertex of the square and radius 1.
• $P$ and the centers of the two yellow semicircles are collinear points.

Let the radii of the right semicircle, left semicircle, and cyan circle be $r_0$ , $r_1$ , and $r_2$ respectively. By Pythagorean theorem , we note that:

$\begin{aligned} (1-r_0)^2 + (1-r_1)^2 & = (r_0+r_1)^2 \\ 1 - 2r_0 + r_0^2 + 1 - 2r_1 + r_1^2 & = r_0^2 + 2r_0r_1 + r_1^2 \\ 1 - r_0 - r_1 & = r_0r_1 \\ \implies r_1 & = \frac {1-r_0}{1+r_0} \end{aligned}$

We also note that:

$\begin{aligned} r_1 + r_2 & = 1 - r_0 \\ r_2 & = \blue{1 - r_0 - r_1} & \small \blue{\text{Note that }1-r_0-r_1 = r_0r_1} \\ \implies r_2 & = \blue{r_0r_1} \end{aligned}$

If $A_c$ is the area of triangle of centers (blue) and $A_t$ is the area of triangle of tangent points \(purple) of three mutually externally tangent circles, then

\[\begin{align} A_t & = \frac {2r_0r_1r_2}{(r_0+r_1)(r_1+r_2)(r_2+r_0)} \blue{A_c} & \small \blue{\text{Note that }A_c = \frac {(1-r_0)(1-r_1)}2} \\ & = \frac {r_0^2r_1^2(1-r_0)(1-r_1)}{(r_0+r_1)(r_1+r_0r_1)(r_0r_1+r_0)} \\ & = \frac {r_0^2r_1^2(1-r_0-r_1+r_0r_1)}{r_0r_1(r_0+r_1)(1+r_0)(1+r_1)} \\ & = \frac {r_0r_1(2r_0r_1)}{(r_0+r_1)(1+r_0+r_1+\blue{r_0r_1)}} & \small \blue{\text{Again }1-r_0-r_1 = r_0r_1} \\ & = \frac {2r_0^2r_1^2}{(r_0+r_1)(2)} = \frac {r_0^2r_1^2}{r_0+r_1} = \frac {r_0^2r_1^2}{1-r_0r_1} \end{align} \]

Putting $A_t = \dfrac {36}{1015}$ and rearrange:

$\begin{aligned} 1015(r_0r_1)^2 + 36r_0r_1 - 36 & = 0 \\ (35r_0r_1 - 6) (29r_0r_1 + 6) & = 0 & \small \blue{\text{Since }r_0r_1 > 0} \\ r_0r_1 & = \frac 6{35} \\ \frac {r_0(1-r_0)}{1+r_0} & = \frac 6{35} \\ 35r_0 - 35r_0^2 & = 6 + 6r_0 \\ 35r_0^2 - 29r_0 + 6 & = 0 \\ (5r_0-2)(7r_0-3) & = 0 \\ \implies r_0 & = \frac 25, \frac 37 \end{aligned}$

Since $r_0$ and $r_1$ are interchangeable, these two values are the radii of both semicircle when $A_t = \dfrac {36}{1015}$ . Therefore the ratio of large radius to the small radius of the semicircles is $\dfrac 37 \cdot \dfrac 52 = \dfrac {15}{14}$ $\implies p - q = 15-14 = \boxed 1$ .