Tangent Circle

A synthetic solution:

Let X be point of tangency of both circles. Let Y and Z be midpoints of minor arcs $AB$ and $AC$ respectively. Now $YX\cap AB = D$ , $AC\cap ZX = E$ , and $ZB\cap YC = I$ where $I$ is incenter of $\triangle ABC$ . Thus Pascal's on cyclic hexagon $ACYXZB$ implies points $D, I, E$ collinear.

From this it becomes a routine length chase. $AD=AE$ and $AI \perp DE$ so we can construct P as shown in the diagram and we have $r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}=\sqrt{\frac{32}{5}}$ and $AP=\frac{AB+AC-BC}{2}=4$ . so $AI^{2}=r^{2}+AP^{2}=\frac{112}{5}=AP\cdot AE = 4AD$ .

Thus $AD=\frac{28}{5}$ , and $28+5=33$ .

I shall show a quick solution via flip inversion: We invert the diagram wrt circle centered at $A$ with radius $\sqrt{AB\cdot AC}$ , and then reflect it across the $A$ -angle bisector. Then $B$ would map to $C$ , $C$ would map to $B$ , line $AB$ maps to line $AC$ , line $AC$ maps to line $AB$ , and the circumcircle of $ABC$ is mapped to line $BC$ . Tangency is preserved so $\omega_A$ would map to the $A$ -excircle and suppose its tangency point with line $BC, CA, AB$ well be $X$ , $Y$ , $Z$ respectively. Note that $D,E$ map to $Y,Z$ respectively. Then $AB+AX=AB+BZ=AZ=AY=AC+CY=AC+CX,$ so we know the length of $A$ to $X$ along the perimeter of triangle $ABC$ will be half the perimeter i.e. $15$ , which is also the length of line segment $AY$ . Now by inversion $AY\cdot AD=AB\cdot AC=84$ and hence $AD=\frac{28}{5}$ and the answer is $\boxed{33}$ .