2021 AMC Fall 10B Problem 21

Handle the polygons two at a time. There are $6$ pairs, and each crossing point belongs to exactly one pair (no three sides meet at one point).

Start with a pentagon and a hexagon.

1. Each side of the pentagon cuts off an arc of the circle of $\frac{360^\circ}{5} = 72^\circ$.
2. The hexagon's corners are spaced $\frac{360^\circ}{6} = 60^\circ$ apart. An arc of $72^\circ$ is longer than that gap, so it must contain at least one hexagon corner. The rest of the circle is even longer, so it also contains a hexagon corner.
3. So the hexagon has corners on both sides of that pentagon side. Travel around the hexagon's outline: you have to cross the pentagon side once to get over and once to get back. Since the hexagon is a convex shape (no dents), it crosses the side exactly twice, no more.
4. Every one of the $5$ pentagon sides is crossed twice, so the pentagon and hexagon meet at $2 \times 5 = 10$ points.

The same argument for any pair. If the polygons have $m$ and $n$ sides with $m$ smaller, each side of the $m$-gon cuts off an arc bigger than the gap between corners of the $n$-gon. So each of its $m$ sides is crossed twice: $2m$ points. Always use the smaller number.

Total: $10 + 10 + 10 + 12 + 12 + 14 = 68$.

The answer is $\boxed{\textbf{(E)}\ 68}$.

The count does not depend on exactly how the polygons are turned. A side of the polygon with fewer sides is always long enough to have a corner of the other polygon on each side of it, and that forces exactly two crossings. Counting crossings by "sides of the smaller polygon, two each" turns a crowded picture into a short sum.

Take two of the polygons, with $m < n$ sides. Look at one side of the $m$-gon; it is a chord cutting off an arc of $\frac{360^\circ}{m}$.

That arc is longer than $\frac{360^\circ}{n}$, so it contains at least one vertex of the $n$-gon; the remaining arc is even longer and also contains one. Thus the $n$-gon has vertices on both sides of the chord's line. A convex polygon with vertices on both sides of a line crosses that line exactly twice, and since the $n$-gon lies inside the circle both crossings are on the chord itself. So each side of the $m$-gon meets the $n$-gon in exactly $2$ points, and the two polygons meet in $2m$ points.

(No point is double counted: no three sides are concurrent and no vertices coincide.)

Sum over the six pairs, using the smaller side count each time:
$$2(5 + 5 + 5) + 2(6 + 6) + 2 \cdot 7 = 30 + 24 + 14 = 68.$$

The answer is $\boxed{\textbf{(E)}\ 68}$.