2010 AMC 10A Problem 22

Below is the professionally curated solution for Problem 22 of the 2010 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 10A solutions, or check the answer key.

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Concepts:counting intersectionscombinationsbijection

Difficulty rating: 2160

22.

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?

2828

5656

7070

8484

140140

Solution:

An interior triangle is formed by three chords that pairwise intersect inside the circle. Such a triangle uses six distinct endpoints on the circle.

Conversely, for any six chosen points in circular order, exactly one set of three chords pairs opposite endpoints so that the three chords intersect pairwise inside the circle.

Therefore the number of triangles is (86)=(82)=28.\binom{8}{6}=\binom{8}{2}=28.

Thus, A is the correct answer.

Problem 22 in Other Years