20.110 Three circles with centers at $A, B, C$ have radii $r$, where $1<r<2$. The distance between each pair of centers is 2. If $B$ is a point of intersection of circles $A$ and $C$ and lies outside circle $B$, and $C'$ is a point of intersection of circles $A$ and $B$ and lies outside circle $C$, then the length of $B'C'$ is
(A) $3 r-2$.
(B) $r^{2}$.
(C) $r+\sqrt{3(r-1)}$.
(D) $1+\sqrt{3\left(r^{2}-1\right)}$.
(E) None of these.
(30th American High School Mathematics Examination, 1979)
\[\text{Solution}] Let $D, E$ be the intersections of $B' C'$ with $AB, AC$, respectively, and let $B' F$ be the perpendicular from $B'$ to $AC$.

By symmetry, we have $ED \parallel CB$, thus $\angle AED = \angle B'EF = 60^\circ$. Applying the Pythagorean theorem in the right triangle $\triangle B'FC$, we get
\[
B'F = \sqrt{r^2 - 1}.
\]

Furthermore,
\[
\begin{aligned}
B'C' &= B'E + ED + DC' \\
&= 2(B'E) + ED \\
&= 2(B'E) + EA,
\end{aligned}
\]
so
\[
\begin{aligned}
B'C' &= 2\left(\frac{2}{\sqrt{3}}\right) B'F + (1 - EF) \\
&= 2\left(\frac{2}{\sqrt{3}}\right) \sqrt{r^2 - 1} + \left(1 - \frac{1}{\sqrt{3}} \sqrt{r^2 - 1}\right) \\
&= 1 + \frac{3}{\sqrt{3}} \sqrt{r^2 - 1} \\
&= 1 + \sqrt{3(r^2 - 1)}.
\end{aligned}
\]

Therefore, the answer is $(D)$.