G4.3 $A$ and $B$ ran around a circular path with constant speeds. They started from the same place and at the same time in opposite directions. After their first meeting, $B$ took 1 minute to go back to the starting place. If $A$ and $B$ need 6 minutes and $c$ minutes respectively to complete one round of the path, find the value of $c$.
In one minute, $A$ and $B$ ran $\frac{1}{6}+\frac{1}{c}=\frac{c+6}{6 c}$ of the total distance.
They will meet at the first time after $\frac{6 c}{c+6}$ minutes.
After 1 more minute, (i.e. total time elapsed $=\frac{6 c}{c+6}+1$ minutes), $B$ retuned to the starting point. So $\left(\frac{6 c}{c+6}+1\right) \times \frac{1}{c}=1$
$$
\begin{array}{l}
6 c+c+6=c^{2}+6 c \\
c^{2}-c-6=0 \\
(c-3)(c+2)=0 \\
c=3
\end{array}
$$