# 2010 AMC 10A Problems/Problem 22

## 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. We also know that for any six points we pick, there is only $1$ way to connect the points such that a triangle is formed in the circle's interior (this is because we want no two chords to be parallel ~Williamgolly). Therefore, the answer is ${{8}\choose{6}}$, which is equivalent to $\boxed{\textbf{(A) }28}$.

## Solution 2 (Guessing)

To make a triangle, where the $3$ points are arranged on a circle, you just need to choose $3$ points because no $3$ points are arranged in a straight line on a circle, meaning that to count the number of triangles we get $\binom{8}{3}=56$. However, this is incorrect because of the restriction that no three chords intersect in a single point in the circle meaning that the answer is most likely less than $56$, and the only answer choice that satisfies this condition is $\boxed{\textbf{(A)}28}$ ~Batmanstark

## Video Solution by TheBeautyofMath

 2010 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 21 Followed byProblem 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