12. Given that the radius of the large circle in Figure 1 is $R$, and the three smaller circles inside the large circle are pairwise tangent to each other and to the large circle. Their radii are $2r$, $r$, and $r$. Try to find $\frac{r}{R}$.
12. $\frac{4 \sqrt{2}-5}{2}$.

It is easy to prove that $O$ and $B$ lie on the common tangent of $\odot A$ and $\odot A^{\prime}$, so we only need to consider half of the figure. (See Figure 9 and Figure 10)
Let $x=M O, y=M B$. Then we have
$$
M N=x+R=y+2 r,
$$

which gives $x=y-R+2 r$.
Also, $A O=O P-A P=R-r, A B=A K+K B=3 r$.

By the Pythagorean theorem, we get
$$
\begin{array}{l}
y^{2}=M B^{2}=A B^{2}-A M^{2}=8 r^{2}, \\
y=\sqrt{8} r .
\end{array}
$$

Thus, $x=y-R+2 r=(2+\sqrt{8}) r-R$.
Similarly, $x^{2}=M O^{2}=A O^{2}-A M^{2}=R^{2}-2 R r$.
So, $[(2+\sqrt{8}) r-R]^{2}=R^{2}-2 R r$.
This simplifies to $(2+2 \sqrt{8}) R=(2+\sqrt{8})^{2} r$.
$$
\begin{aligned}
\therefore \frac{r}{R} & =\frac{2+2 \sqrt{8}}{(2+\sqrt{8})^{2}}=\frac{1+\sqrt{8}}{2(\sqrt{2}+1)^{2}} \\
& =\frac{(1+2 \sqrt{2})(\sqrt{2}-1)^{2}}{2}=\frac{4 \sqrt{2}-5}{2} .
\end{aligned}
$$