Difference between revisions of "2010 AMC 10A Problems/Problem 22"
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<math>\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 70 \qquad \textbf{(D)}\ 84 \qquad \textbf{(E)}\ 140</math> | <math>\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 70 \qquad \textbf{(D)}\ 84 \qquad \textbf{(E)}\ 140</math> | ||
| − | ==Solution | + | ==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 | + | 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 <math>1</math> 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 <math>{{8}\choose{6}}</math>, which is equivalent to <math>\boxed{\textbf{(A) }28}</math>. |
| − | ==Solution | + | ==Video Solution by TheBeautyofMath== |
| − | + | https://youtu.be/8aBePLkFdss | |
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==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}} | ||
Latest revision as of 08:27, 26 August 2024
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 
.
Video Solution by TheBeautyofMath
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
