20. In an isosceles right $\triangle ABC$, $AB = AC = 3$, $\angle A = 90^{\circ}$, $M$ is the midpoint of side $BC$, points $I$ and $E$ are on sides $AC$ and $AB$ respectively, and $AI > AE$, points $A, I, M, E$ are concyclic. If $S_{\triangle EMI} = 2$, $CI = \frac{a - \sqrt{b}}{c}$ ($a, b, c$ are positive integers, and $b$ is not a perfect square), then the value of $a + b + c$ is ( ).
(A) 9
(B) 10
(C) 11
(D) 12
(E) 13
20. D.

As shown in Figure 9.
Since points $A, I, M, E$ are concyclic, $\angle A=90^{\circ}$, therefore,
$$
\begin{array}{l}
\angle I M E=90^{\circ}, \\
\angle M I E=\angle M A E \\
=\angle M A I=\angle M E I \\
=45^{\circ} .
\end{array}
$$

Given $S_{\triangle E M I}=2$, we know
$$
M I=M E=2, C M=\frac{3 \sqrt{2}}{2} \text {. }
$$

In $\triangle C I M$, by the cosine rule,
$$
\begin{array}{l}
C I^{2}+C M^{2}-2 C I \cdot C M \cos \frac{\pi}{4}=I M^{2} \\
\Rightarrow C I=\frac{3 \pm \sqrt{7}}{2} .
\end{array}
$$

Since $A I>A E$, then $C I=\frac{3-\sqrt{7}}{2}$.
Thus, $a+b+c=3+7+2=12$.