Triangle, Circle, and Hexagon

Our circle has the triangle incentre as its centre. Radius of the incircle is 1. Points of contact of incircle with triangle sides are midpoints of segments $A_1 A_2, B_1 B_2, C_1 C_2$ . This makes segment lengths $A C_1 = A B_2 = (3-1)-\dfrac{x}{2} = 2-\dfrac{x}{2}$ , $B A_1 = B C_2 = 1-\dfrac{x}{2}$ , $C A_2=C B_1 = 3-\dfrac{x}{2}$ .

$\triangle A C_1 B_2 = \dfrac{1}{2}(2-\dfrac{x}{2})^2 \sin(A) = \dfrac{1}{2} (2-\dfrac{x}{2})^2 \dfrac{4}{5}$ $\triangle B A_1 C_2 = \dfrac{1}{2}(1-\dfrac{x}{2})^2 \sin(B) = \dfrac{1}{2}(1-\dfrac{x}{2})^2$ $\triangle C A_2 B_1 = \dfrac{1}{2}(3-\dfrac{x}{2})^2 \sin(C) =\dfrac{1}{2} (3-\dfrac{x}{2})^2 \dfrac{4}{5}$

$\triangle A C_1 B_2+\triangle B A_1 C_2+\triangle C A_2 B_1 =\triangle ABC-4=2$

This gives value of $x=\dfrac{11-\sqrt{37}}{3} \approx 1.64$