There are 24 equally spaced points lying on the circumference of a circle. What is the maximum number of equilateral triangles that can be drawn by taking sets of three points as the vertices?

The correct option is 8.

To understand why this is the case, let us break down the geometry of the points on the circle step-by-step.

We are given that there are 24 equally spaced points on the circumference of a circle. Let these points be labeled in order around the circle as $P_1,P_2,...,P_{24}$.

Since the points are equally spaced, the angle subtended by the arc between any two consecutive points at the center of the circle is:
$\frac{360°}{24}=15°$

An equilateral triangle is a triangle in which all three sides are equal, and all three internal angles are 60°. For any three points on the circle to form an equilateral triangle, they must divide the circumference of the circle into three equal arcs. Each arc must contain the same number of intervals between the points.

Since there are 24 intervals in total, the number of intervals between any two vertices of an equilateral triangle must be:
$\frac{24}{3}=8$ intervals.

Therefore, if we choose the first vertex of a triangle as $P_i$, the second vertex must be 8 points away, which is $P_{i+8}$, and the third vertex must be another 8 points away, which is $P_{i+16}$ (with index arithmetic modulo 24).

Let us list the possible unique equilateral triangles by choosing different starting points for the first vertex:
1. $\{P_1,P_9,P_{17}\}$
2. $\{P_2,P_{10},P_{18}\}$
3. $\{P_3,P_{11},P_{19}\}$
4. $\{P_4,P_{12},P_{20}\}$
5. $\{P_5,P_{13},P_{21}\}$
6. $\{P_6,P_{14},P_{22}\}$
7. $\{P_7,P_{15},P_{23}\}$
8. $\{P_8,P_{16},P_{24}\}$

If we try to start a triangle at $P_9$, we would get the set $\{P_9,P_{17},P_1\}$, which is identical to the first triangle we listed. Thus, we see that the pattern repeats after 8 triangles.

Hence, the maximum number of unique equilateral triangles that can be drawn is 8.