Difference between revisions of "2010 AMC 10A Problems/Problem 22"
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There are a total of <math>\dbinom{8}{3}=56</math> total triangles that can be made out of these chords. We know that the amount of triangles which have all their vertices inside the circle has to be less than this, to the answer can only be <math>\boxed{\textbf{(A) }28}</math>. | There are a total of <math>\dbinom{8}{3}=56</math> total triangles that can be made out of these chords. We know that the amount of triangles which have all their vertices inside the circle has to be less than this, to the answer can only be <math>\boxed{\textbf{(A) }28}</math>. | ||
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| + | proposed by mumpu2k16 | ||
==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 13:37, 23 December 2017
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 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. We also know that for any six points we pick, there is only 1 way to connect the chords such that a triangle is formed in the circle's interior. Therefore, the answer is 
, which is equivalent to 
.
Solution 2
There are a total of 
total triangles that can be made out of these chords. We know that the amount of triangles which have all their vertices inside the circle has to be less than this, to the answer can only be .
proposed by mumpu2k16
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. 
