Center-Mark the Spot

The radial length of that isosceles triangle from the incenter = circumcenter are 1, 0.5 and 0.5.

By applying Pythagoras, we have

(Half base)² = 0.5² – r²
and
(Longer part of slant side)² = 1² – r²

Taking one right triangle half of the isosceles into account, we have

[(Longer part of slant side) + (Shorter part of slant side)]²
= [(Longer part of slant side) + (Half base)]²
= [√(1 – r²) + √(0.25 – r²)]²
= (Half base)² + (Height)²
= (0.25 – r²) + (1 + r)²
= (0.25 – r²) + 1 + 2r + r²
= 2r + 1.25
= 1.25 – 2r² + 2√[(1 – r²)(0.25 – r²)]

Rearranging, dividing by a factor of 2 and squaring, we have
(r² + r)² = (1 – r²)(0.25 – r²)
r⁴ + 2r³ + r² = r⁴ – 1.25r² + 0.25
8r³ + 9r² = 1
0 = (r + 1)(8r² + r – 1)
r > 0, so r = (√33 – 1) / 16

Answer = 1 + 16 + 1 + 33 + 1 = 52

Let $O$ be the center of the three circles, $\triangle ABC$ , $\triangle AEF$ the equilateral and the isosceles triangle respectively, $N$ , $K$ the contact points of the smaller circle with the sides of $\triangle AEF$ as seen in the figure. If we denote by $r$ the inradius of $\triangle AEF$ , then on $\triangle ΑΟΝ$ we have $ON=OA\cdot \sin \theta \Rightarrow r=1\cdot \sin \theta \Rightarrow \cos \theta =\sqrt{1-{{r}^{2}}}$ $OE$ is the inradius of the equilateral triangle, thus it half its circumradius, i.e. $OE=\dfrac{1}{2}$ .

Since $OE$ is also the angle bisector of $\angle AEK$ , $\begin{aligned} & \dfrac{ON}{OE}=\sin \left( \angle OEN \right)=\sin \left( \dfrac{\angle AEK}{2} \right)=\sin \left( \dfrac{90{}^\circ -\theta }{2} \right)=\sin \left( 45{}^\circ -\dfrac{\theta }{2} \right) \\ & \Rightarrow ON=OE\cdot \sin \left( 45{}^\circ -\dfrac{\theta }{2} \right) \\ & \Rightarrow r=\dfrac{1}{2}\cdot \left( \dfrac{\sqrt{2}}{2}\cos \theta -\dfrac{\sqrt{2}}{2}\sin \theta \right) \\ & \Rightarrow 2\sqrt{2}r=\cos \dfrac{\theta }{2}-\sin \dfrac{\theta }{2} \\ & \Rightarrow 2\sqrt{2}r=\sqrt{\dfrac{1+\cos \theta }{2}}-\sqrt{\dfrac{1-\cos \theta }{2}} \\ & \Rightarrow 4r=\sqrt{1+\cos \theta }-\sqrt{1-\cos \theta } \\ & \Rightarrow 4r=\sqrt{1+\sqrt{1-{{r}^{2}}}}-\sqrt{1-\sqrt{1-{{r}^{2}}}} \\ \end{aligned}$ After squaring and rearranging, the latter becomes $8{{r}^{2}}+r-1=0$ which gives one positive solution $r=\dfrac{1}{16}\left( 1\sqrt{33}-1 \right)$ For the answer, $A=1$ , $B=16$ , $C=1$ , $D=33$ , $E=1$ , thus, $A+B+C+D+E=\boxed{52}$ .