Dynamic Geometry: P113

From the compiled calculations in Dynamic Geometry: P96 Series , we know that the radius of the circle is $6.5$ , the dividing chord $AB=12$ , and if the upper and lower moving points $C$ and $C'$ respectively, then $\angle ACB = \pi - tan^{-1} \frac {12}5$ and $\angle AC'B = \tan^{-1} \frac {12}5$ .

Let the midpoint of $AB$ , $M$ , be the origin of the $xy$ -plane, an arbitrary point on the upper locus or the center of the upper rectangle be $P(x,y)$ , $\angle CAB = \theta$ , the height of $\triangle ABC$ , $CD$ , be $h$ , and the side length of the square be $s$ .

By sine rule ,

$\begin{aligned} AC & = \frac {AB \cdot \sin \angle ABC}{\sin \angle ACB} = \frac {12 \sin \left(\tan^{-1} \frac {12}5-\theta \right)}{\sin \left(\pi - \tan^{-1} \frac {12}5\right)} = 12\cos \theta - 5\sin \theta \\ \implies h & = AC \cdot \sin \theta = (12\cos \theta - 5 \sin \theta) \sin \theta = \frac {12\sin 2\theta + 5 \cos 2\theta - 5}2 & \small \blue{\text{Let }t = \tan \theta} \\ & = \frac {24t + 5(1-t^2)-5(1+t^2)}{2(1+t^2)} = \frac {12t - 5t^2}{1+t^2} \end{aligned}$

From the compiled calculations, we have:

$\begin{aligned} s & = \frac {12h}{12+h} = \frac {12(12t-5t^2)}{12(1+t^2)+12t-5t^2} = \frac {12(12t-5t^2)}{7t^2+12t+12} \end{aligned}$

We note that $y = \dfrac s2 = \dfrac {6(12t-5t^2)}{7t^2+12t+12}$ and

$x = s \cot \theta + \frac s2 - 6 = \frac {12(6-5t-6t^2)}{7t^2+12t+12}$

Then the area under the locus:

$\begin{aligned} A & = 2 \int_0^\frac 32 x \ \ce dy \\ & = 2 \int_0^\frac 32 \frac {12(6-5t-6t^1)}{7t^2+12t+12}\ \ce d \left(\frac {6(12t-5t^2)}{7t^2+12t+12} \right) \\ & = 3456 \int_0^\frac 23 \frac {(6-5t-6t^1)^2}{(7t^2+12t+12)^3}\ \ce d t \\ & = \frac 34 \left(169\sqrt 3 \cot^{-1} (2\sqrt 3) -66 \right) \\ & \approx 12.1976827871 \end{aligned}$

Similarly for the lower triangle, $x = \dfrac {12(6+5t-6t^2}{17t^2+12t+12}$ , $y=\dfrac {6(12t+5t^2)}{17t^2+12t+12}$ , and the area under the locus:

$\begin{aligned} A' & = 2 \int_0^\frac {36}7 x \ \ce dy \\ & = 2 \int_0^\frac {36}7 \frac {12(6+5t-6t^1)}{17t^2+12t+12}\ \ce d \left(\frac {6(12t+5t^2)}{17t^2+12t+12} \right) \\ & = 3456 \int_0^\frac 32 \frac {(6+5t-6t^1)^2}{(17t^2+12t+12)^3}\ \ce d t \\ & = \frac 3{98} \left(169\sqrt{42} \tan^{-1} \left(\sqrt{\frac 67} \right) -42 \right) \\ & \approx 23.7562372205 \end{aligned}$

Therefore the area bounded by the two loci $A+A' \approx 35.9539200076 \approx 36$ . And the required answer is $\boxed 6$ .