2010 AMC 10A Problem 22

A triangle with all three vertices inside the circle has its three sides lying on three chords, and each pair of those chords must cross inside the circle (the crossing points are the triangle's vertices).

Two chords cross inside the circle exactly when their four endpoints alternate around the circle, so they share no endpoint. Three chords that pairwise cross therefore use $3 \cdot 2 = 6$ distinct points on the circle.

Conversely, pick any $6$ of the $8$ points. Label them in circular order $P_1, \dots, P_6$. The only way to pair them into three mutually crossing chords is the three "long diagonals" $P_1P_4$, $P_2P_5$, $P_3P_6$ (any other pairing includes a chord like $P_1P_2$ or $P_1P_3$ that some other chord in the pairing fails to cross). Those three diagonals pairwise cross at three distinct points, forming exactly one interior triangle.

So interior triangles correspond one-to-one with $6$-element subsets of the $8$ points:
$$\binom{8}{6} = \binom{8}{2} = 28.$$

The answer is $\boxed{\textbf{(A)}\ 28}$.

The condition "no three chords concurrent" guarantees each triple of crossing chords gives a genuine triangle. The key step is a bijection: a complicated geometric object (a triangle inside the circle) is determined by a simple combinatorial one (six points on the circle). The same idea shows $\binom{n}{4}$ interior intersection points for $n$ points, one per quadruple.

• Counting triangles by choosing 3 of the C(8,2) chords, or choosing 3 intersection points, without requiring the chords to pairwise cross.
• Answering $\binom{8}{3} = 56$, which counts triangles with vertices on the circle rather than inside it.