3-Way Tangent Circle Center Coordinates

Let the radius of the red semi-circle be $r_1 = \frac{1}{3}$ , the blue semi-circle be $r_2$ , the orange circle $r_3$ , and the black semi-circle $r_4 = 1$ .

Since the diameters of the red and blue semi-circles make up the diameter of the black circle, the radius of the blue circle is $r_2 = \frac{1}{2}(2\cdot r_1 - 2 \cdot r_2) = \frac{2}{3}$ .

By Descartes' Theorem :

$k_4 = k_1 + k_2 + k_3 -2\sqrt{k_1 k_2 + k_2 k_3 + k_3 k_1}$

and after substituting the curvatures of $k_1 = 3$ , $k_2 = \frac{3}{2}$ , $k_3 = \frac{1}{r_3}$ , and $k_4 = -1$ this becomes:

$-1 = 3 + \frac{3}{2} + \frac{1}{r_3} -2\sqrt{3 \cdot \frac{3}{2} + \frac{3}{2} \cdot \frac{1}{r_3} + \frac{1}{r_3} \cdot 3}$

which solves to $r_3 = \frac{2}{7}$ , and makes $k_3 = \frac{7}{2}$ .

Let the center of the orange circle be (x, y) so that its complex number is $z_3 = x + yi$ . Since the center of the red semi-circle is at $(-\frac{2}{3}, 0)$ its complex number is $z_1 = -\frac{2}{3}$ , since the center of the blue semi-circle is at $(\frac{1}{3}, 0)$ its complex number is $z_2 = \frac{1}{3}$ , and since the center of the black semi-circle is at $(0, 0)$ its complex number is $z_4 = 0$

By Complex Descartes' Theorem :

$z_4k_4 = z_1k_1 + z_2k_2 + z_3k_3 -2\sqrt{k_1 k_2z_1z_2 + k_2 k_3z_2z_3 + k_1 k_3 z_1 z_3}$

and after substituting the curvatures and complex numbers from above, this becomes:

$0 \cdot -1 = -\frac{2}{3} \cdot 3 + \frac{1}{3} \cdot \frac{3}{2} + (x + yi)\frac{7}{2} -2\sqrt{3 \cdot \frac{3}{2} \cdot -\frac{2}{3} \cdot \frac{1}{3} + \frac{3}{2} \cdot \frac{7}{2} \cdot \frac{1}{3} \cdot (x + yi) + 3 \cdot \frac{7}{2} \cdot -\frac{2}{3} \cdot (x + yi)}$

which can be rearranged to:

$49x^2 + 42x - 49y^2 + 25 + (98xy + 42y)i = 0$

equating the imaginary part to zero gives:

$98xy + 42y = 0$

which solves to $x = -\frac{3}{7}$ for $y \neq 0$ , and equating the real part to zero and substituting $x = -\frac{3}{7}$ gives:

$49(-\frac{3}{7})^2 + 42(-\frac{3}{7}) - 49y^2 + 25 = 0$

which solves to $y = \frac{4}{7}$ for $y > 0$ .

Therefore, $x + y = -\frac{3}{7} + \frac{4}{7} = \frac{1}{7}$ , so $p = 1$ , $q = 7$ , and $p + q = \boxed{8}$ .

• To solve this problem, we shall denote the radius of the cyan semi-circle as $\boxed{x}$ , the radius of the green semi-circle as $\boxed{1-x}$ and the radius of the blue circle as $\boxed{R(x)}$ , $\boxed{l(x)}$ is the $x$ coordinate of the red circle's center and $\boxed{h(x)}$ is the $y$ coordinate of the red circle's center.

• We shall use the Pythagorean theorem to express $\boxed{R(x)}$ , $\boxed{l(x)}$ and $\boxed{h(x)}$ in terms of $\boxed{x}$

• We shall write three equations using the diagram:

• $\boxed{(1-R(x))^{2}=l(x)^{2}+h(x)^{2}}$
• $\boxed{(1-x+R(x))^{2}=(x-l(x))^{2}+h(x)^{2}}$
• $\boxed{(x+R(x))^{2}=(1-x+l(x))^{2}+h(x)^{2}}$

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• After combining the equation and using minor algebra skills we find:
• $\boxed{l\left(x\right)=\frac{2x-1}{x^2-x+1}}$

• $\boxed{h\left(x\right)=2\cdot \frac{\left(x^2-x\right)}{-x^2+x-1}}$

• Now we calculate $\boxed{l(\dfrac{1}{3})=\dfrac{-3}{7}}$ and $\boxed{h(\dfrac{1}{3}=\dfrac{4}{7})}$

• Finally $\boxed{\dfrac{-3}{7}+\dfrac{1}{3}=\dfrac{1}{7}}$ and $\boxed{1+7=8}$