2010 AMC 10A Problems/Problem 22
Contents
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. We also know that for any six points we pick, there is only way to connect the points such that a triangle is formed in the circle's interior (this is because we want no two chords to be nonintersecting ~Williamgolly). Therefore, the answer is , which is equivalent to .
Solution 2 (Guessing)
To make a triangle, where the points are arranged on a circle, you just need to choose points because no points are arranged in a straight line on a circle, meaning that to count the number of triangles we get . However, this is incorrect because of the restriction that no three chords intersect in a single point in the circle meaning that the answer is most likely less than , and the only answer choice that satisfies this condition is ~Batmanstark
Video Solution by TheBeautyofMath
See Also
2010 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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