Circle Meets Parabola at the Triangle Bar

We note that diameter of a circumcircle is given by $2r = \dfrac c{\sin C}$ . In this case $c = AB = \sqrt 2$ . We need to find the largest $\sin C$ to find the smallest $r$ .

Let $C(x,x^2)$ and $\angle C = \theta$ . Then

$\begin{aligned} \theta & = \tan^{-1} \frac {x^2-1}{x-1} - \tan^{-1} \frac {x^2}x \\ & = \tan^{-1} (x+1) - \tan^{-1} x \\ \tan \theta & = \frac {x+1-x}{1+x(x+1)} = \frac 1{x^2+x+1} \\ \implies \sin \theta & = \frac 1{\sqrt{(x^2+x+1)^2+1}} \\ & = \frac 1{\sqrt{\left(\blue{\left(x+\frac 12\right)^2}+\frac 34\right)^2+1}} \\ \implies \max (\sin \theta) & = \frac 1{\sqrt{\left(\blue 0+\frac 34\right)^2+1}} = \frac 45 & \small \blue{\text{when }x = -\frac 12} \\ \implies r_{\min} & = \frac {\sqrt 2}{2 \max(\sin \theta)} = \frac {5\sqrt 2}8 \end{aligned}$

Therefore $a+b+c = 5+2+8 = \boxed {15}$ .

Let $C$ have coordinates $(t,t^2)$ .

The triangle's circumcentre lies at the intersection of the perpendicular bisectors of its sides.

The perpendicular bisector of $AB$ is the line $y=1-x$

The perpendicular bisector of $AC$ is the line $\frac{y-\frac12 t^2}{x-\frac12 t}=-\frac{1}{t}$

The intersection of these is the point $P\left(-\frac12 t(t+1),\frac12 \left(t^2+t+2 \right)\right)$

The circumradius is the distance from any of the triangle vertices to $P$ ; it's easiest to look at $AP$ . After going through the paperwork, we find $R^2=\frac12 (t^2 + 1) (t^2 + 2 t + 2)$

(as is often the case, it's easier to work with $R^2$ than $R$ .)

This achieves a minimum value of $\frac{25}{32}$ when $t=-\frac12$ . Taking the square root, we find $R=\frac{5\sqrt2}{8}$

for an answer of $\boxed{15}$ .

Let the coordinates of $C$ be $(k, k^2)$ . Then by the distance formula:

$a = BC = \sqrt{(k - 1)^2 + (k^2 - 1)^2} = \sqrt{k^4 - k^2 - 2k + 2}$

$b = AC = \sqrt{(k - 0)^2 + (k^2 - 0)^2} = \sqrt{k^4 + k^2}$

$c = AB = \sqrt{(1 - 0)^2 + (1 - 0)^2} = \sqrt{2}$ .

By Heron's formula , the area of $\triangle ABC$ is:

$T = \frac{1}{4}\sqrt{4a^2b^2 - (a^2 + b^2 - c^2)^2} = \frac{1}{4}\sqrt{4(k^4 - k^2 - 2k + 2)(k^4 + k^2) - ((k^4 - k^2 - 2k + 2) + (k^4 + k^2) - 2)^2} = \frac{1}{2}k(k - 1)$ .

The square of the circumradius is:

$R^2 = \frac{a^2b^2c^2}{16T^2} = \frac{(k^4 - k^2 - 2k + 2)(k^4 + k^2)(2)}{16 \cdot \frac{1}{4}k^2(k - 1)^2} = \frac{1}{2}k^4 + k^3 + \frac{3}{2}k^2 + k + 1$

By implicit differentiation, $2R \cdot \frac{dR}{dk} = 2k^3 + 3k^2 + 3k + 1$ . The minimum occurs when $\frac{dR}{dk} = 0$ (and when $\frac{d^2R}{dk^2} > 0$ ), so $0 = 2k^3 + 3k^2 + 3k + 1$ , which has one real solution of $k = -\frac{1}{2}$ .

When $k = -\frac{1}{2}$ , $R^2 = \frac{1}{2}k^4 + k^3 + \frac{3}{2}k^2 + k + 1 = \frac{1}{2}(-\frac{1}{2})^4 + (-\frac{1}{2})^3 + \frac{3}{2}(-\frac{1}{2})^2 + (-\frac{1}{2}) + 1 = \frac{25}{32}$ , so $R = \sqrt{\frac{25}{32}} = \frac{5\sqrt{2}}{8}$ .

Therefore, $a = 5$ , $b = 2$ , $c = 8$ , and $a + b + c = \boxed{15}$ .