2021 AMC Fall 10B Problem 21

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 $\boxed{\textbf{(E)}\ 68}$.

The count depends only on convexity and the arc lengths, not on the exact placement: a side of the coarser polygon always spans at least one vertex of the finer one, forcing exactly two crossings. Matching intersection points to sides of the smaller polygon (two each) converts a messy picture into a sum of $2m$ over pairs.

Check the claim on the smallest pair: a pentagon and a hexagon in general position. Each pentagon side subtends $72^\circ$, exceeding the hexagon's $60^\circ$ spacing, so every pentagon side is crossed twice, $10$ points; drawing it confirms. Then trust the pattern for the other pairs.

• Assuming every side of each polygon crosses every side of the other, or using 2n (the larger count) instead of 2m for a pair.
• Counting only the $\binom42 = 6$ pairs of polygons or adding the side counts ($5 + 6 + 7 + 8$) without the factor $2$ per side.