Dynamic Geometry: P86

Label the centers of large, purple, and orange circles $O$ , $P$ , and $Q$ respectively, and the two tangent points of the orange circle with the other two circles $A$ and $B$ , Let the radius of of the orange circle be $r$ and $\angle OPQ = \theta$ . By cosine rule ,

$\begin{aligned} OQ^2 & = OP^2 + PQ^2 - 2\cdot OP \cdot PQ \cdot \cos \angle OPQ \\ (25-r)^2 & = 9^2 + (16+r)^2 - 2\cdot 9 (16+r) \cos \theta \\ \implies r & = \frac {144(1+\cos \theta)}{41-9 \cos \theta} \end{aligned}$

We note that the largest sine value of $\triangle ABO$ comes from the largest angle $\angle AOB$ , which is $180^\circ - \angle PQO$ . Let $\angle PQO = \phi$ . Then we need to find the maximum value of $\sin (180^\circ - \phi) = \sin \phi$ . By sine rule , we have:

$\begin{aligned} \frac {\sin \phi}9 & = \frac {\sin \theta}{25-r} \\ \implies \sin \phi & = \frac {9 \sin \theta}{25 - \frac {144(1+\cos \theta)}{41-9\cos \theta}} \\ \max (\sin \phi) & = \frac {720}{1681} \end{aligned}$

Therefore $\sqrt{q-p} = \sqrt{1681-720} = \boxed{31}$ .

Label the diagram as follows, with $P$ , $Q$ , and $O$ as the centers of the circles, and let $x$ be the radius of the orange circle and $\theta = \angle PQO$ :

Then $PQ = AQ + PQ = x + 16$ , $OQ = OB - BQ = 25 - x$ , and $OP = 25 - 16 = 9$ .

By the law of cosines on $\triangle OPQ$ , $\cos \theta = \cfrac{PQ^2 + OQ^2 - OP^2}{2 \cdot PQ \cdot OQ} = \cfrac{(x + 16)^2 + (25 - x)^2 + 9^2}{2(x + 16)(25 - x)} = \cfrac{x^2 - 9x + 400}{-x^2 + 9x + 400}$ .

So $\sin \angle AQB = \sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \bigg(\cfrac{x^2 - 9x + 400}{-x^2 + 9x + 400}\bigg)^2} = \cfrac{40\sqrt{x(9 - x)}}{-x^2 + 9x + 400}$ .

The maximum sine occurs when $\cfrac{dy}{dx} = \cfrac{20(9 - 2x)(x^2 - 9x + 400)}{\sqrt{x(9 - x)}(-x^2 + 9x + 400)^2} = 0$ , which is when $9 - 2x = 0$ or $x = \cfrac{9}{2}$ .

At this value of $x = \cfrac{9}{2}$ , $\sin \theta = \cfrac{40\sqrt{x(9 - x)}}{-x^2 + 9x + 400} = \cfrac{40\sqrt{\frac{9}{2}(9 - \frac{9}{2})}}{-(\frac{9}{2})^2 + 9 \cdot \frac{9}{2} + 400} = \cfrac{720}{1681}$ , so $p = 720$ , $q = 1681$ , and $\sqrt{q - p} = \sqrt{961} = \boxed{31}$ .