Twelve points are marked on a circle at equal intervals. Using line segments to connect the twelve points, we want to construct a figure consisting of three polygons such that:
- The three polygons do not intersect each other.
- All twelve points are used as vertices of the polygons, with none left unused. Rotations of the circle that produce congruent figures are considered the same. Find the number of distinct figures that satisfy these conditions.oindent
[THOUGHT] To solve this problem, we need to partition 12 points into three non-intersecting polygons. Each polygon must use a subset of the 12 points, and no two polygons should share any vertices. This is equivalent to partitioning the 12 points into three sets where the sum of the sizes of these sets is 12, and the sizes of the sets must be at least 3 (since a polygon must have at least 3 sides). Furthermore, since rotations of the circle producing congruent figures are considered the same, we must account for rotational symmetry.
Let's denote the sizes of the three polygons as (a), (b), and (c), where (a + b + c = 12) and (a, b, c \geq 3). The number of ways to partition 12 points into three polygons is given by the number of ways to choose the points for each polygon, divided by 3! (since the polygons are indistinguishable) and by the number of rotations.
First, we list all possible partitions of 12 into three parts, each at least 3:
- (3, 3, 6)
- (3, 4, 5)
- (4, 4, 4)
For each partition, we calculate the number of distinct ways to form the polygons:
-
Partition (3, 3, 6):
- Choose 3 points out of 12 for the first polygon: (\binom{12}{3})
- Choose 3 points out of the remaining 9 for the second polygon: (\binom{9}{3})
- The remaining 6 points form the third polygon.
- Divide by 3! because the polygons with the same number of sides are indistinguishable.
- Divide by 12 because of rotations: (\frac{1}{12})
[ \text{Number of ways} = \frac{1}{3! \cdot 12} \binom{12}{3} \binom{9}{3} = \frac{1}{72} \times 220 \times 84 = \frac{18480}{72} = 260 ]
-
Partition (3, 4, 5):
- Choose 3 points out of 12 for the first polygon: (\binom{12}{3})
- Choose 4 points out of the remaining 9 for the second polygon: (\binom{9}{4})
- The remaining 5 points form the third polygon.
- Divide by 12 because of rotations: (\frac{1}{12})
[ \text{Number of ways} = \frac{1}{12} \binom{12}{3} \binom{9}{4} = \frac{1}{12} \times 220 \times 126 = \frac{27720}{12} = 2310 ]
-
Partition (4, 4, 4):
- Choose 4 points out of 12 for the first polygon: (\binom{12}{4})
- Choose 4 points out of the remaining 8 for the second polygon: (\binom{8}{4})
- The remaining 4 points form the third polygon.
- Divide by 3! because the polygons are indistinguishable.
- Divide by 12 because of rotations: (\frac{1}{12})
[ \text{Number of ways} = \frac{1}{3! \cdot 12} \binom{12}{4} \binom{8}{4} = \frac{1}{72} \times 495 \times 70 = \frac{34650}{72} = 475 ]
Finally, we sum these numbers to get the total number of distinct figures:
[ 260 + 2310 + 475 = 3045 ]
Therefore, the number of distinct figures that can be formed is (\boxed{3045}). [/THOUGHT]