2010 AMC 10A Problems/Problem 22: Difference between revisions
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==Solution== | ==Solution== | ||
To choose | To choose a chord, we know that two points must be chosen. This implies that for three chords to create a triangle and not intersect at a single point, six points need to be chosen. Therefore, the answer is <math>{{8}\choose{6}}</math> which is equivalent to 28, <math>\boxed{(A)}</math> | ||
Revision as of 21:38, 2 January 2011
Problem
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?
Solution
To choose a chord, we know that two points must be chosen. This implies that for three chords to create a triangle and not intersect at a single point, six points need to be chosen. Therefore, the answer is 
which is equivalent to 28, 
