Dynamic Geometry: P47

In $\triangle OAB,$ $\;\;\;\;\sin \angle AOB=\frac{AB}{OA}\;\;\; \implies \sin \alpha =\frac{r}{1-r} \tag{1}$ In $\triangle OCM,$ $\cos \angle OCM=\frac{CM}{OC}\implies \cos 2\alpha=\frac{x}{1} \tag{2}$ From equation $(1)$ and $(2)$ , $\begin{aligned} x&=\cos 2\alpha \\ &=1-2\sin^2 \alpha \\ &=1-2\left(\frac{r}{1-r}\right)^2 \\ &=\frac{(1-r)^2-2r^2}{(1-r)^2} \end{aligned}$ Since the product of the perimeters of the triangle and circle is $\dfrac{66\pi }{44}$ , we have, $\begin{aligned} (2+2x)\cdot (2\pi r)&=\frac{66\pi }{49} \\ \frac{4(1-r)^2-4r^2}{(1-r)^2}\cdot r &=\frac{33}{49} \\ \frac{4r-8r^2}{(1-r)^2}&=\frac{33}{49} \\ \implies 425r^2-262r+33&=0 \end{aligned}$ Solving the above quadratic equation, we have, $r=\dfrac{3}{17}$ as the only valid solution.

Therefore, $|OA|=1-r=\dfrac{14}{17}\implies p+q=\boxed{31}$

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Note: The quadratic equation, $425r^2-262r+33=0$ , yields two solutions, $r_1=\dfrac{3}{17}$ and $r_2=\dfrac{11}{25}$ . But $r_2$ must be discarded since $\dfrac{11}{25}=r_2>r_{\max}=\sqrt{2}-1$ . $\begin{aligned} (1-r_{\max})^2&=r_{\max}^2+r_{\max}^2 \\ 1-r_{\max}&=r_{\max} \sqrt{2} \\ \therefore \; r_{\max}&=\frac{1}{\sqrt{2}+1}=\sqrt{2}-1 \end{aligned}$

Let the distance between the centers of the circle and the semicircle be $\ell$ and the radius of the circle be $r$ at an instant in time. Let the angle between a leg of the isosceles triangle and the diameter of the semicircle be $\theta$ . Note that the green line bisects the angle $\theta$ and

$\begin{aligned} \ell + r & = 1 \\ \ell + \ell \sin \frac \theta 2 & = 1 \\ \implies \sin \frac \theta 2 & = \frac {1-\ell}\ell \end{aligned}$

Now note that the perimeter of the triangle is $p = 2 + 2 \cos \theta$ and the product of the circumference of the circle and the perimeter of the triangle is $2\pi r p$ and when the product is equal to $\dfrac {66\pi}{49}$ :

$\begin{aligned} 2 \pi \ell \sin \frac \theta 2 (2+2 \cos \theta) & = \frac {66\pi}{49} \\ 2\ell \sin \frac \theta 2 \left(1 + 1 - 2 \sin^2 \frac \theta 2 \right) & = \frac {33}{49} \\ \frac {4\ell(1-\ell)}\ell \left(1 - \left(\frac {1-\ell}\ell\right)^2 \right) & = \frac {33}{49} \\ 196(1-\ell)(2\ell -1) & = 33 \ell^2 \\ 425 \ell^2 - 588\ell + 196 & = 0 \\ (17\ell - 14)(25\ell - 14) & = 0 & \small \blue{\text{See note: }\ell = \frac {14}{25} \text{ is unacceptable.}} \\ \implies \ell & = \frac {14}{17} \end{aligned}$

Therefore $p+q = 14 + 17 = \boxed{31}$ .

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Note: For $\ell = \dfrac {14}{25}$ , the circle is tangent to the left leg of the isosceles triangle.