Dynamic Geometry: P112

Let the dividing chord of the circle be $AB$ , the upper and lower variable triangles be $\triangle ABC$ and $\triangle ABC'$ respectively, and the measure of the orange angle be $\theta$ . Referring the calculations in Dynamic Geometry: P96 Series , the radius of the circle is $6.5$ , $AB=12$ , $\angle ACB = \pi - \tan^{-1} \frac {12}5$ , and $\angle AC'B = \tan^{-1} \frac {12}5$ .

Now let us consider the relationship between the side length $s$ of a square $KLMN$ to the height $CD=h$ and width $AB=w$ of the triangle $\triangle ABC$ that inscribes it. Note that $\triangle KLC$ and $\triangle ABC$ are similar. Let the height of $\triangle KLC$ be $CE$ , then $CE = \dfrac swh$ . From $CD = CE + ED$ , we have

$h = \frac swh + s \implies s = \frac h{1+\frac hw} = \frac {wh}{h+w} = \frac {12h}{h+12} \quad \small \blue{\text{Since }w=AB=12}$

Similarly for the square in $\triangle ABC'$ , $s' = \dfrac {12h'}{h'+12}$ .

Since $ACBC'$ is a cyclic quadrilateral , $\angle BCC' = \angle BAC' = \theta$ and $\angle BC'C = \angle BAC = \theta$ . Therefore $\triangle BCC'$ is isosceles and $BC = BC' = b$ . By sine rule ,

$\frac {BC}{\sin \angle CAB} = \frac {AB}{\sin \angle ACB} \implies b = \frac {12 \sin \theta}{\sin \left(\pi - \tan^{-1} \frac {12}5\right)} = \frac {12 \sin \theta}{\sin \left(\tan^{-1} \frac {12}5 \right)} = 13 \sin \theta$

Then $\begin{cases} h = b \sin \angle CBA = b \sin \left(\tan^{-1} \frac {12}5 - \theta \right) & = \sin \theta (12 \cos \theta - 5 \sin \theta ) \\ h' = b \sin \angle C'BA = b \sin \left(\tan^{-1} \frac {12}5 + \theta \right) & = \sin \theta (12 \cos \theta + 5 \sin \theta) \end{cases}$

The ratio of areas of the square $\dfrac {s^2}{s'^2} = \dfrac {961}{2116} \implies \dfrac s{s'} = \dfrac {12h(h'+12)}{12h'(h+12)} = \dfrac {h(h'+12)}{h'(h+12)} = \dfrac {31}{46}$ . Therefore,

$\begin{aligned} \frac {(12\cos \theta - 5\sin \theta)(\sin \theta(12\cos \theta + 5\sin \theta) + 12)} {(12\cos \theta + 5\sin \theta)(\sin \theta(12\cos \theta - 5\sin \theta) + 12)} & = \frac {31}{46} \\ \sin \theta(144 \cos^2 \theta - 25 \sin^2 \theta) + 144 \cos \theta - 308 \sin \theta & = 0 \\ 169 \sin \theta \cos^2 \theta + 144 \cos \theta - 333 \sin \theta & = 0 \\ 169 \cos^2 \theta + 144 \cot \theta - 333 & = 0 & \small \blue{\text{Let }t = \tan \theta} \\ \frac {169}{1+t^2} + \frac {144}t - 333 & = 0 \\ (3t-2)(111t^2 + 26t+72) & = 0 & \small \blue{\text{For real value of }t} \\ \implies t & = \tan \theta = \frac 23 \end{aligned}$

Then the distance between the two moving points $CC' = 2b\cos \theta = 26 \sin \theta \cos \theta = 26 \cdot \dfrac 2{\sqrt{13}} \cdot \dfrac 3{\sqrt{13}} = \boxed{12}$ .