2010 AMC 10A Problem 22

P21-25~5 minPrint

Basic Counting Circles Map the objects to something easier to count

Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle are created?

Show solution without answering

Key insight

Each inside triangle comes from three chords that all cross, using six different points; each group of 6 points gives exactly one triangle.

Solution

Instead of counting triangles directly, find what each triangle is made of.

1. Three chords per triangle. Each side of a triangle inside the circle lies along one of the chords. Its corners are points where two of those chords cross. So a triangle needs three chords, and every two of them must cross inside the circle.
2. When do two chords cross? Two chords cross inside the circle only if they have no shared endpoint and their endpoints take turns around the circle (A, then B, then A, then B). So three chords that all cross each other use $3 \times 2 = 6$ different points.
3. Going the other way. Pick any $6$ of the $8$ points and number them $1$ to $6$ going around the circle. Is there a way to join them into three chords that all cross each other?
- Join each point to the one directly across: $1$ – $4$ , $2$ – $5$ , $3$ – $6$ . Each pair of these crosses, so they form a triangle. (No three chords meet at one point, so the three crossings really are three different corners.)
- Any other pairing uses a chord that skips over at most one point, like $1$ – $2$ or $1$ – $3$ . A chord $1$ – $2$ has no points on one side, so nothing crosses it. A chord $1$ – $3$ has only point $2$ on one side, so only the chord from $2$ can cross it, not both of the others. So joining across is the only way.
4. So each group of 6 points gives exactly one triangle, and each triangle comes from exactly one group of 6 points. Count the groups of $6$ points out of $8$ . Choosing $6$ to keep is the same as choosing $2$ to leave out:
$\frac{8 \times 7}{2} = 28.$

The answer is $\boxed{\textbf{(A)}\ 28}$ .

Why this works

A hard-to-count shape (a small triangle inside the circle) matches up perfectly with an easy-to-count thing (a group of six points on the circle). When you can pair up each object with exactly one simple choice, counting the choices counts the objects. The same idea shows there is one inside crossing point for every group of $4$ points.

Concise solution

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

A triangle with all three vertices inside the circle has its three sides lying on three chords, and each pair of those chords must cross inside the circle (the crossing points are the triangle's vertices).

Two chords cross inside the circle exactly when their four endpoints alternate around the circle, so they share no endpoint. Three chords that pairwise cross therefore use $3 \cdot 2 = 6$ distinct points on the circle.

Conversely, pick any $6$ of the $8$ points. Label them in circular order $P_1, \dots, P_6$ . The only way to pair them into three mutually crossing chords is the three "long diagonals" $P_1P_4$ , $P_2P_5$ , $P_3P_6$ (any other pairing includes a chord like $P_1P_2$ or $P_1P_3$ that some other chord in the pairing fails to cross). Those three diagonals pairwise cross at three distinct points, forming exactly one interior triangle.

So interior triangles correspond one-to-one with $6$ -element subsets of the $8$ points:
$\binom{8}{6} = \binom{8}{2} = 28.$

The answer is $\boxed{\textbf{(A)}\ 28}$ .

Review in the AoPS books

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