Triangles in a circle

We break this into cases based on the number of vertices on the circle. Since Brilliant doesn't yet support Asymptote or another geometry diagram drawing program, try to draw pictures as you read this solution.

.

Case 3: $\Delta ABC$ has three vertices on the circle.

There are $\binom{8}{3} = 56$ ways to choose $3$ points $A,B,C$ on the circle, and each choice yields exactly $1$ triangle $\Delta ABC$ . Thus, there are $56$ triangles in this case.

.

Case 2: $\Delta ABC$ has two vertices on the circle.

Let $A,B$ be on the circle and $C$ be inside the circle. Extend $AC$ and $BC$ to meet the circle at $D$ and $E$ respectively. Connecting all the chords between $A,B,D,E$ yields $4$ triangles with two vertices on the circle. There are $\binom{8}{4} = 70$ ways to choose $4$ vertices $A,B,D,E$ on the circle, and each choice yields $4$ triangles with two vertices on the circle. Thus, there are $70 \cdot 4 = 280$ triangles in this case.

.

Case 1: $\Delta ABC$ has one vertex on the circle.

Let $A$ be on the circle and $B,C$ be inside the circle. Extend $AB$ and $AC$ to meet the circle at $D$ and $E$ . Extend $BC$ to meet the circle at $F$ and $G$ . Connecting all the chords between $A,D,E,F,G$ yields $5$ triangles with one vertex on the circle. There are $\binom{8}{5} = 56$ ways to choose $5$ vertices $A,D,E,F,G$ on the circle, and each choice yields $5$ triangles with two vertices on the circle. Thus, there are $56 \cdot 5 = 280$ triangles in this case.

.

Case 0: $\Delta ABC$ has no vertices on the circle.

Let $A,B,C$ be inside the circle. Extend $AB$ to meet the circle at $D$ and $E$ . Extend $BC$ to meet the circle at $F$ and $G$ . Extend $AC$ to meet the circle at $H$ and $I$ . Connecting all the chords between $D,E,F,G,H,I$ yields $1$ triangle with all vertices inside the circle. There are $\binom{8}{6} = 28$ ways to choose $6$ vertices $D,E,F,G,H,I$ on the circle. Most of these yield $1$ triangle with all vertices inside the circle. However, there are $4$ ways in which $DE$ , $FG$ , $HI$ are concurrent at the center of the circle, and $8$ ways in which $DE$ , $FG$ , $HI$ are concurrent, but not at the center of the circle. Each of these yields a degenerate point triangle. Thus, there are $28-4-8 = 16$ triangles in this case.

.

Because each triangle must have between $0$ and $3$ vertices (inclusive) on the circle, these cases are both mutually exclusive and exhaustive. Thus, the total number of triangles is $56+280+280+16 = \boxed{632}$ .

This is very similar to a problem of a few weeks back, when we had eight points on the circle, and we needed to know the maximum possible number of triangles that could be seen. The answer was ${}^8C_3 + 4\times{}^8C_4 + 5\times{}^8C_5 + {}^8C_6 \;=\; 644$ where the sum has been broken down into the number (between $3$ and $6$ ) of vertices on the circle that used to define the lines which form each triangle.

The only difference in this case is that some of these $644$ triangles do not exist, since the triangles that exist in the general case do not exist in this more symmetric case.

The only triangles that disappear are those which (in the general case) are formed by three interior vertices (so that we use $6$ circumference vertices to define them). In the symmetric case, some of these triangles "collapse to a point" since the three vertices are all the same. This can only happen at vertices inside the circle where three or more chords meet. There are only $9$ such points: the centre (where $4$ diameters meet) and another $8$ points where three chords meet. Thus there are $4$ choices of three chords which all meet at the centre, and one choice of three chords which meet at a point for each of the other $8$ points. Thus a total of $12$ of the ${}^8C_6=28$ possible cohoices of triangles formed by $3$ internal vertices (and hence defined by $6$ circumference vertices) do not exist in the symmetric case. Thus there is a total of $644-12=632$ triangles altogether.