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

(Problem)
(Solution)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
To choose 3 points on a circle with 8 points, we simply have <math>{{8}\choose{3}}</math> to get the answer <math>\boxed{56}</math>
+
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 <math>{{8}\choose{6}}</math> which is equivalent to 28, <math>\boxed{(A)}</math>

Revision as of 21:38, 2 January 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?

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

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. Therefore, the answer is ${{8}\choose{6}}$ which is equivalent to 28, $\boxed{(A)}$