Dynamic Geometry: P84

Let the square be $ABCD$ , the center of the circle be $O$ , its radius $r$ and two contact points $E$ and $F$ . Let $\angle OAB = \theta$ .

We note that

$\begin{aligned} AO \cdot \cos \theta + OF & = AB \\ (1+r) \cos \theta + r & = 1 \\ \implies r & = \frac {1-\cos \theta}{1+\cos \theta} & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ & = \frac {1-\frac {1-t^2}{1+t^2}}{1+\frac {1-t^2}{1+t^2}} = t^2 \end{aligned}$

The inradius of a triangle is given by $r_i = \dfrac As$ , where $A$ is the area and semi-perimeter of the triangle. Since $\angle OAB = \theta$ , $\angle EOF = 180^\circ - \theta$ , and since $\triangle EOF$ is isosceles, $\angle OEF=OFE = \dfrac \theta 2$ and its inradius is:

$\begin{aligned} r_i & = \dfrac As = \frac {r^2 \sin \frac \theta 2 \cos \frac \theta 2}{r + r \cos \frac \theta 2} = \frac {r \sin \frac \theta 2 \cos \frac \theta 2}{1 + \cos \frac \theta 2} & \small \blue{\text{Note that }r = t^2 = \tan^2 \frac \theta 2} \\ & = \frac {\sin^2 \frac \theta 2}{\cos^2 \frac \theta 2} \cdot \frac {\sin \frac \theta 2 \cos \frac \theta 2}{1 + \cos \frac \theta 2} = \frac {1-\cos^2 \frac \theta 2}{\cos \frac \theta 2} \cdot \frac {\sin \frac \theta 2}{1 + \cos \frac \theta 2} \\ & = \frac {\sin \frac \theta 2}{\cos \frac \theta 2} \left(1 - \cos \frac \theta 2 \right) = \tan \frac \theta 2 \left(1 - \blue{\cos \frac \theta 2} \right) & \small \blue{\text{Since }1 + \tan^2 \phi = \sec^2 \phi} \\ \implies r_i & = t \left( 1 - \frac 1{\sqrt{1+t^2}} \right) \end{aligned}$

When $r_i = \dfrac 7{600}$ :

$\begin{aligned} t \left( 1 - \frac 1{\sqrt{1+t^2}} \right) & = \frac 7{600} \\ 600t - 7 & = \frac {600t}{\sqrt{1+t^2}} \\ (1+t^2)(600t-7)^2 & = 600^2t^2 \\ 360000t^4 - 8400t^3 + 49t^2 - 8400t + 49 & = 0 \\ (24t-7)(15000t^3+4025t^2+1176t-7) & = 0 & \small \blue{\text{The other irrational root}} \\ \implies t & = \frac 7{24} & \small \blue{\approx 0.0058334 \text{ is too small.}} \\ \implies r & = t^2 = \frac {7^2}{24^2} \end{aligned}$

Therefore $\sqrt p + \sqrt q = 7 + 24 = \boxed{31}$ .

Label the diagram as follows, and let the yellow circle have a radius of $R$ :

Then $AC = 1 + R$ and $AB = 1 - R$ , and letting $\theta = \angle ACB$ , from $\triangle ABC$ , $\sin \theta = \cfrac{AB}{AC} = \cfrac{1 - R}{1 + R}$ , so $\cos \theta = \sqrt{1 - \sin^2 \theta} = \cfrac{2\sqrt{R}}{1 + R}$ .

Let $t = \angle DCF = \angle ECF$ . Then from $\triangle CFE$ , $CF = R \cos t$ and $FE = R \sin t$ , so that the inradius of $\triangle CDE$ is $r = \cfrac{2A_{\triangle CDE}}{P_{\triangle CDE}} = \cfrac{2 \cdot \frac{1}{2} \cdot DE \cdot CF}{CD + CE + DE} = \cfrac{2R \sin t \cdot R \cos t}{R + R + 2R\sin t} = \cfrac{R \sin 2t}{2(\sin t + 1)}$ .

But $2t = \angle DCE = \angle ACB + \angle BCE = \theta + 90°$ , so $\sin 2t = \sin (\theta + 90°) = \cos \theta = \cfrac{2\sqrt{R}}{1 + R}$ , and $\sin t = \sin \frac{1}{2}(\theta + 90°) = \sqrt{\frac{1}{2}(1 - \cos(\theta + 90°))} = \sqrt{\frac{1}{2}(1 + \sin \theta)} = \sqrt{\frac{1}{2}\bigg(1 + \cfrac{1 - R}{1 + R}\bigg)} = \cfrac{1}{\sqrt{1 + R}}$ .

Substituting $r = \cfrac{7}{600}$ , $\sin 2t = \cfrac{2\sqrt{R}}{1 + R}$ , and $\sin t = \cfrac{1}{\sqrt{1 + R}}$ into $r = \cfrac{R \sin 2t}{2(\sin t + 1)}$ gives $\cfrac{7}{600} = \cfrac{R \cdot \frac{2\sqrt{R}}{1 + R}}{2(\frac{1}{\sqrt{1 + R}} + 1)} = \cfrac{R\sqrt{R}}{\sqrt{1 + R} + 1 + R}$ , which solves to $R = \cfrac{49}{576}$ .

Therefore, $p = 49$ , $q = 576$ , and $\sqrt{p} + \sqrt{q} = 7 + 24 = \boxed{31}$ .