2021 AMC Fall 10B Problem 21

P21-25~5 minPrint

Basic Counting Angles & Polygons Circles Map the objects to something easier to count Set up the equation/formula and compute; no special trick needed

Regular polygons with $5,6,7,$ and $8$ sides are inscribed in the same circle. No two of the polygons share a vertex, and no three of their sides intersect at a common point. At how many points inside the circle do two of their sides intersect?

Show solution without answering

Key insight

Each side of the polygon with fewer sides crosses the other polygon exactly twice, so each pair meets at twice the smaller side count.

Solution

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.

Each side of the pentagon cuts off an arc of the circle of $\frac{360^\circ}{5} = 72^\circ$ .
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.
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.
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.

pair	smaller count $m$	crossings $2m$
5 and 6	5	10
5 and 7	5	10
5 and 8	5	10
6 and 7	6	12
6 and 8	6	12
7 and 8	7	14

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

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

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Geometry, Chapter 2: Angles
Introduction to Geometry, Chapter 11: Circles
Intermediate Counting & Probability, Chapter 4: Constructive Counting and 1-1 Correspondences (§4.4, §4.5, §4.6)