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

Revision as of 12:03, 21 August 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

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

Solution 2

We first figure out how many triangles can be created. This is done by choosing $3$ lines out of the $8$ , which is equivalent to $\binom{8}{3}=56$ . However, some of these triangles have vertices on the circle. Therefore, the answer choice must be less than $56$ . The only one that is so is $\boxed{(A)}$ .

@@ Line 5: / Line 5: @@
 ==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 <math>{{8}\choose{6}}</math> which is equivalent to 28, <math>\boxed{(A)}</math>
+===Solution 2===
+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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