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

Revision as of 17:41, 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?

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

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 $\boxed{(A)}$ .

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

@@ Line 10: / Line 10: @@
 ===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>.
+==See Also==
+{{AMC10 box|year=2010|ab=A|num-b=20|num-a=22}}

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