Dynamic Geometry: P102

Let the center of the unit circle be $O$ , its diameter perpendicular to the red chord $CD$ be $AB$ , $AB$ and $CD$ intersect at $E$ , the radii of the cyan, yellow, and purple circles be $r_1$ , $r_2$ , and $r_3$ respectively, $\angle COB = \theta$ then $\angle CAO = \dfrac \theta 2$ , and $t = \tan \dfrac \theta 2$ .

Then $r_1 = \dfrac {BE}2 = \dfrac {1-\cos \theta}2 = \dfrac {1-\frac {1-t^2}{1+t^2}}2 = \dfrac {t^2}{1+t^2}$ .

Let $P$ be the center of the yellow circle and $PF$ be perpendicular to $AC$ . Then

$\begin{aligned} AE & = AP + PE = FP \cdot \csc \frac \theta 2 + PE \\ 2 - 2r_1 & = r_2 \csc \frac \theta 2 + r_2 \\ 2 - \frac {2t^2}{1+t^2} & = r_2 \left(\frac {\sqrt{1+t^2}}t + 1 \right) \\ \implies r_2 & = \frac {2t(\sqrt{1+t^2}-t)}{1+t^2} \end{aligned}$

Let $G$ and $H$ be the tangent points between the top purple circle and the unit circle, and the triangle respectively. Then

$\begin{aligned} GH+HO & = GO \\ GH + AO \cdot \sin \frac \theta 2 & = GO \\ 2r_3 + \sin \frac \theta 2 & = 1 \\ \implies r_3 & = \frac {\sqrt{1+t^2}-t}{2\sqrt{1+t^2}} \end{aligned}$

When the three radii are in a geometric progression, we have

$\begin{aligned} r_1r_3 & = r_2^2 \\ \frac {t^2}{1+t^2} \cdot \frac {\sqrt{1+t^2}-t}{2\sqrt{1+t^2}} & = \frac {4t^2(\sqrt{1+t^2}-t)^2}{(1+t^2)^2} \\ \frac 1{2\sqrt{1+t^2}} & = \frac {4(\sqrt{1+t^2}-t)}{1+t^2} \\ 1 + t^2 & = 8(1+t^2) - 8t\sqrt{1+t^2} \\ 8 t \sqrt{1+t^2} & = 7(1+t^2) \\ 8t & = 7\sqrt{1+t^2} \\ 64 t^2 & = 49 + 49 t^2 & \small \blue{\text{Since }\theta > 180^\circ} \\ \implies t & = - \frac 7{\sqrt{15}} \\ \implies r_2 & = \frac {\frac {14}{\sqrt{15}}\left(\sqrt{1+\frac {49}{15}} - \frac 7{\sqrt{15}}\right)}{1+\frac {49}{15}} = \frac 7{32} \end{aligned}$

Therefore $\sqrt{q-p} = \sqrt{32-7} = \boxed 5$ .

Label the diagram as follows:

Let $k = AH$ . Then $HD = AD - AH = 2 - k$ , so that the radius of the blue circle is $R_b = \frac{1}{2}HD = 1 - \frac{1}{2}k$ .

$\triangle AHC \sim \triangle CHD$ , so $\cfrac{AH}{CH} = \cfrac{CH}{HD}$ , or $\cfrac{k}{CH} = \cfrac{CH}{2 - k}$ , which solves to $CH = \sqrt{k(2 - k)}$ .

By the Pythagorean Theorem on $\triangle AHC$ , $AC = \sqrt{AH^2 + CH^2} = \sqrt{k^2 + k(2 - k)} = \sqrt{2k}$ .

As an incircle of $\triangle ACF$ , the radius of the yellow circle is $R_y = \cfrac{2A_{\triangle ACF}}{P_{\triangle ACF}} = \cfrac{2 \cdot \frac{1}{2} \cdot AH \cdot CF}{AC + AF + CF} = \cfrac{k \cdot 2\sqrt{k(2 - k)}}{\sqrt{2k} + \sqrt{2k} + 2\sqrt{k(2 - k)}} = k - 2 + \sqrt{4-2k}$ .

If $R_p$ is the radius of the purple circle, then by the intersecting chords theorem, $BG \cdot GE = AG \cdot GC$ , or $2R_p (2 - 2R_p) = \cfrac{\sqrt{2k}}{2} \cdot \cfrac{\sqrt{2k}}{2}$ , which rearranges to $R_p = \frac{1}{2} - \frac{1}{4}\sqrt{4-2k}$ .

When the radii of the blue, yellow, and purple circles are in a geometric progression, $R_b \cdot R_p = R_y^2$ , or $(1 - \frac{1}{2}k)(\frac{1}{2} - \frac{1}{4}\sqrt{4-2k}) = (k - 2 + \sqrt{4-2k})^2$ , which solves to $k = \cfrac{15}{32}$ for $0 < k < 2$ .

Therefore, the radius of the yellow circle is $R_y = \cfrac{15}{32} - 2 + \sqrt{4-2 \cdot \cfrac{15}{32}} = \cfrac{7}{32}$ , so $p = 7$ , $q = 32$ , and $\sqrt{q - p} = \boxed{5}$ .