10. Given that an equilateral triangle and a regular heptagon are inscribed in a circle with a circumference of 84, their vertices divide the circumference into 10 arcs. The maximum possible length of the smallest arc is $\qquad$ (Canada)
10. 2 .

Notice that, the three vertices of the triangle divide the circumference into three arcs.

By the Pigeonhole Principle, we know that one of these arcs must contain at least three vertices of the heptagon. Let the vertices $D$, $E$, and $F$ of the heptagon be on the arc $\overparen{A B}$, as shown in Figure 6.
According to the problem, the length of arc $\overparen{A B}$ is 28, and the lengths of arcs $\overparen{D E}$ and $\overparen{E F}$ are both 12.
Thus, the sum of the lengths of arcs $\overparen{F A}$ and $\overparen{D B}$ is
$$
28 - 2 \times 12 = 4 \text{.}
$$

By the Pigeonhole Principle, at least one of the arcs $\overparen{F A}$ and $\overparen{D B}$ has a length no greater than $4 \div 2 = 2$. If point $E$ is located at the midpoint of arc $\overparen{A B}$, then the maximum possible length of the smallest arc among the 10 segments can reach 2.