Dynamic Geometry: P121

Let us first check what is the radius when all three moving circles coincide or their radius is the largest. We can use the Descartes; theorem by treating the straight base line as a circle of infinite radius or zero curvature. Then the largest radius is:

$\frac 1{r_{\max}} = \frac 11 + \frac 14 + 2 \sqrt{\frac 14} = \frac 94 \implies r_{\max} = \frac 49$

Let the center of this largest circle be $O(0, \frac 49)$ , the centers of red circle, blue circle, and orange circle be $A(x_0, y_0)$ , $B(x_1, y_1)$ , and $C(x_2,y_2)$ . $A$ has to move from $D(\frac 8{15}, \frac 85)$ , where the green circle is tangent to the cyan circle and its radius is $0$ , to $O$ , a horizontal distance of $\frac 8{15}$ . $B$ has to move from $E(-\frac 83, 0)$ , where the green circle is tangent to the base line and its radius is $0$ , a horizontal distance of $\frac 83$ . $C$ has to move from $F(\frac 43, 0)$ , where the cyan circle is tangent to the base line and its radius is $0$ , a horizontal distance of $\frac 43$ . Therefore $B$ moves $5$ times faster than $A$ , and $C$ moves $\frac 52$ faster than $A$ . Using $x_0$ as reference, then

$x_1 = - 5x_0, \qquad x_2 = \frac 52 x_0$

We know that the horizontal distance between two circles tangent to each other and tangent to a horizontal base line is given by $d = 2\sqrt{r_1r_2}$ , where $r_1$ and $r_2$ are the radii of the two circles. Note that the radius in this case is equal to the $y$ -coordinate. Then we have:

$\begin{cases} \dfrac 83 + x_1 = \dfrac 83 - 5x_0 = 2\sqrt {4y_1} = 4\sqrt{y_1} & \implies y_1 = \dfrac {(8-15x_0)^2}{144} \\ \dfrac 43 - x_2 = \dfrac 43 - \dfrac 52 x_0 = 2\sqrt {y_2} & \implies y_2 = \dfrac {(8-15x_0)^2}{144} \end{cases}$

Note that $y_1=y_2$ ; this means that the radii of the blue and orange circles are the same all the time, as confirmed by the perfect animation of @Valentin Duringer . There the equations of the circles are:

$\begin{cases} (x+5x_0)^2 + (y-y_1)^2 = y_1^2 & ...(1) \\ \left(x - \dfrac 52 x_0 \right)^2 + (y-y_1)^2 = y_1^2 & ...(2) \end{cases}$

Using the fact the fact that for three tangent circles , the incircle of the triangle, formed by the centers of the three circles, is tangent to the triangle sides at the points where the circles are tangent to each other, we derive that:

$(x-x_0)^2 + (y-y_0)^2 = r_0^2 \quad ...(3) \text{, where} \quad y_0 = \frac{63x_0+624x_0+64}{72(3x_0+2)}, \quad r_0 = \frac {(8-15x_0)^2}{72(3x_0+2)}$

The black point $P$ , where the three circles intersect, satisfies the three equations. Solving the system of equations, we have:

$\begin{array} {lll} x_0 = \dfrac 6{\sqrt{65}} - \dfrac 23, & \ \ y_0 = \dfrac 52 - \dfrac {29}{2\sqrt{65}}, & \ \ y_1 = \dfrac {9\left(9-\sqrt{65}\right)}{26} \end{array}$

The area of $\triangle ABC$

$A = \frac {(x_2-x_1)(y_0-y_1)}2 = \frac {15x_0(y_0-y_1)}4 = \frac {56}{13} - \frac {88\sqrt{65}}{169}$

Therefore $m+n+l+p+q = 56+88+13+65+169 = \boxed{391}$ .