Two circles, both with the same radius $r$, are placed in the plane without intersecting each other. A line in the plane intersects the first circle at the points $A, B$ and the other at the points $C, D$ so that $|A B|=|B C|=|C D|=14 \mathrm{~cm}$. Another line intersects the circles at points $E, F$ and $G, H$ respectively, so that $|E F|=|F G|=|G H|=6 \mathrm{~cm}$. Find the radius $r$.
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 $\left|O_{1} O_{2}\right|=2 \cdot\left|O_{1} P\right|=|A C|=28 \mathrm{~cm}$. Let $h=\left|O_{1} T\right|$ 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 $r=13 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_38aae83e469137ba6b44g-4.jpg?height=258&width=442&top_left_y=1981&top_left_x=684)

Figure 4

![](https://cdn.mathpix.com/cropped/2024_04_17_38aae83e469137ba6b44g-4.jpg?height=237&width=297&top_left_y=1983&top_left_x=1185)

Figure 5