Dynamic Geometry: P10

Let the green point be $P(u,1-u^2)$ . Then the sum of coordinates is $S = u + 1 - u^2 = \dfrac 54 - \left(u-\dfrac 12 \right)^2$ . Therefore $S_{\max} = \dfrac 54$ , when $u=\dfrac 12$ and $1-u^2 = \dfrac 34$ . Therefore the green point is $P \left(\dfrac 12, \dfrac 34 \right)$ .

The gradient at a point of $y=1-x^2$ is $\dfrac {dy}{dx} = -2x$ and the gradient of the normal at that point is $\dfrac 1{2x}$ . Then the normal through $P$ has a gradient of $\dfrac 1{2u} = 1$ or $\tan 45^\circ$ and its equation is given by $\dfrac {y-\frac 34}{x-\frac 12} = 1 \implies y = x + \dfrac 14$ (the cyan line).

We note that the cyan point or the center of the red circle $O(x_o, y_o)$ is on this cyan line and that $y_o = r$ , the radius of the red circle. Since $y_o = x_o + \dfrac 14 \implies x_o = r - \dfrac 14$ .

$\begin{aligned} r + r \sin 45^\circ & = \frac 34 \\ r & = \frac 3{4\left(1+\frac 1{\sqrt 2}\right)} = \frac {6-3\sqrt 2}4 \\ y_o & = r = \frac {6-3\sqrt 2}4 \\ x_o & = r - \frac 14 = \frac {5-3\sqrt 2}4 \\ \implies \frac {x_o}{y_o} & = \frac {5-3\sqrt 2}{6-3\sqrt 2} = \frac {4-\sqrt2}6 \end{aligned}$

Therefore $a+b+c = 4+2+6 = \boxed{12}$ .