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Difference between revisions of "2010 AMC 10A Problems/Problem 22"

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==Solution==
 
==Solution==
  
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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 (this is due to the fact that we want the triangles to be nondegenerate, meaning that the 3 points are non-collinear- ~Williamgolly) need to be chosen. We also know that for any six points we pick, there is only 1 way to connect the points such that a triangle is formed in the circle's interior. Therefore, the answer is <math>{{8}\choose{6}}</math>, which is equivalent to <math>\boxed{\textbf{(A) }28}</math>.
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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 1 way to connect the points such that a triangle is formed in the circle's interior. Therefore, the answer is <math>{{8}\choose{6}}</math>, which is equivalent to <math>\boxed{\textbf{(A) }28}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2010|ab=A|num-b=21|num-a=23}}
 
{{AMC10 box|year=2010|ab=A|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:15, 25 January 2020

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?

$\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 70 \qquad \textbf{(D)}\ 84 \qquad \textbf{(E)}\ 140$

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 1 way to connect the points such that a triangle is formed in the circle's interior. Therefore, the answer is ${{8}\choose{6}}$, which is equivalent to $\boxed{\textbf{(A) }28}$.

See Also

2010 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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