Difference between revisions of "2010 AMC 10A Problems/Problem 22"
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We first figure out how many triangles can be created. This is done by choosing <math>3</math> lines out of the <math>8</math>, which is equivalent to <math>\binom{8}{3}=56</math>. However, some of these triangles have vertices on the circle. Therefore, the answer choice must be less than <math>56</math>. The only one that is so is <math>\boxed{(A)}</math>. | We first figure out how many triangles can be created. This is done by choosing <math>3</math> lines out of the <math>8</math>, which is equivalent to <math>\binom{8}{3}=56</math>. However, some of these triangles have vertices on the circle. Therefore, the answer choice must be less than <math>56</math>. The only one that is so is <math>\boxed{(A)}</math>. | ||
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| + | ==See Also== | ||
| + | {{AMC10 box|year=2010|ab=A|num-b=20|num-a=22}} | ||
Revision as of 18:41, 20 December 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
Solution 1
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, 
Solution 2
We first figure out how many triangles can be created. This is done by choosing 
lines out of the 
, which is equivalent to 
. However, some of these triangles have vertices on the circle. Therefore, the answer choice must be less than 
. The only one that is so is .
See Also
| 2010 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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 | ||
