Dynamic Geometry: P103

Let the large quadrilateral be $ABCD$ . Then the area of bottom triangle $ABC$ , $[ABC] = \dfrac 12 \cdot 15 \cdot AB = \dfrac 12 \cdot \dfrac {180}{17}\cdot BC = \dfrac 12 \cdot \dfrac {36}5 \cdot CA$ , $\implies 15 \cdot AB = \dfrac {180}{17} \cdot BC = \dfrac {36}5 \cdot CA$ . Since $\triangle ABC$ is Heronian, its side lengths are integers. Assuming $BC=17$ , then we have $CA = 25$ and $AB=12$ . For the top triangle $ACD$ , with $CA=25$ , $\implies CD = 7$ and $DA = 24$ .

Since all four circles are moving at the same rate, the distance between the left vertical side of the internal quadrilateral from $A$ and that of its right vertical side from $C$ are the same. Let that distance at any moment be $x$ . Since the system is symmetrical about the central vertical axis of the large quadrilateral, where the two left circles interchange positions with the right two circles, we need only consider $0\le x \le 12.5$ to cover all cases.

Let the radii of the top left, top right, bottom left, and bottom right circles be $r_1$ , $r_2$ , $r_3$ , and $r_4$ respectively; and $\angle DAC = \alpha$ , $\angle DCA = \beta$ , $\angle BAC = \gamma$ , and $\angle BCA = \delta$ . For small $x$ , we have:

$\begin{cases} r_1 = x \tan \dfrac \alpha 2 = \dfrac {1-\cos \alpha}{\sin \alpha} \cdot x = \dfrac {25-24}7 x & = \dfrac x7 \\ r_2 = x \tan \dfrac \beta 2 = \dfrac {1-\cos \beta}{\sin \beta} \cdot x = \dfrac {25-7}{24} x & = \dfrac {3x}4 \\ r_3 = x \tan \dfrac \gamma 2 = \dfrac {1-\cos \gamma}{\sin \gamma} \cdot x = \dfrac {5-4}3 x & = \dfrac x3 \\ r_4 = x \tan \dfrac \delta 2 = \dfrac {1-\cos \delta}{\sin \delta} \cdot x = \dfrac {85-77}{36} x & = \dfrac {2x}9 \end{cases}$

When $x = 4$ , $r_2 = x \tan \dfrac \beta 2 = 3$ . But $(25-x) \tan \dfrac \alpha 2 = \dfrac {21}7 = 3$ . This means that the top right circle is tangent to both $CD$ and $DA$ , and the base line $CA$ . For $x > 4$ , $r_2 = (25-x) \tan \dfrac \alpha 2 = \dfrac {25-x}7$ . Similarly, when $x=10$ , $r_3 = \dfrac x3 = \dfrac {10}3$ ; but $(25-x) \tan \dfrac \delta 2 = \dfrac {10}3$ . $\implies r_3 = \dfrac {2(25-x)}9$ for $x > 10$ .

Therefore the area of the internal quadrilateral is given by:

$\begin{aligned} A = \frac {(r_1+r_2+r_3+r_4)(25-2x)}2 & = \begin{cases} \dfrac {25-2x}2 \left(\dfrac x7 + \dfrac {3x}4 + \dfrac x3 + \dfrac {2x}9 \right) & \rm for \ 0 \le x \le 4 \\ \dfrac {25-2x}2 \left(\dfrac x7 + \dfrac {25-x}7 + \dfrac x3 + \dfrac {2x}9 \right) & \rm for \ 4 \le x \le 10 \\ \dfrac {25-2x}2 \left(\dfrac x7 + \dfrac {25-x}7 + \dfrac {2(25-x)}9 + \dfrac {2x}9 \right) & \rm for \ 10 \le x \le 12.5 \end{cases} \\ & = \begin{cases} \dfrac {365x(25-2x)}{504} & \rm for \ 0 \le x \le 4 \\ \dfrac {(25-2x)(35x+225)}{126} & \rm for \ 4 \le x \le 10 \\ \dfrac {575(25-2x)}{126} & \rm for \ 10 \le x \le 12.5 \end{cases} \end{aligned}$

Plotting $A$ against $x$ , we find that $A$ is maximum, when $x = 4$ . That is

$A_{\max} = \frac {365(4)(25-2(4))}{504} = \frac {6205}{126}$

Therefore $p+q = 6205+126 = \boxed{6331}$ .