Balti Tee 1993 · Ülesanne 16

Solution:

First, note that the centres $O_{1}$ and $O_{2}$ of the two circles lie on different sides of the line $E H$—otherwise we have $r<12$ and $A B$ cannot be equal to $14$. Let $P$ be the intersection point of $E H$ and $O_{1} O_{2}$ (see Figure 4).

Points $A$ and $D$ lie on the same side of the line $O_{1} O_{2}$ (otherwise the three lines $A D$, $E H$ and $O_{1} O_{2}$ would intersect in $P$ and $|A B|=|B C|=|C D|$, $|E F|=|F G|=|G H|$ would imply $|B C|=|F G|$, a contradiction).

It is easy to see that $|O_{1} O_{2}|=2 \cdot |O_{1} P|=|A C|=28~\mathrm{cm}$.

Let $h=|O_{1} T|$ be the height of triangle $O_{1} E P$. Then we have $h^{2}=14^{2}-6^{2}=160$ from triangle $O_{1} T P$ and $r^{2}=h^{2}+3^{2}=169$ from triangle $O_{1} T F$. Thus $r=13~\mathrm{cm}$.

Figure 4