Problem Analysis #24

To solve the problem of counting the number of distinct figures formed by connecting 12 equally spaced points on a circle into three non-intersecting polygons, using all 12 points, we proceed as follows:

Understanding the Problem

We are given:

• 12 points equally spaced on a circle.
• These points must be partitioned into three polygons such that:
  • Each polygon uses at least 3 vertices (since a polygon must have at least 3 sides).
  • The polygons do not intersect (i.e., their edges do not cross).
  • All 12 points are used exactly once as vertices.
  • Rotational symmetry is considered: two configurations are the same if one can be rotated to match the other.

This is a classic problem in combinatorics, specifically involving non-crossing partitions of a circle into intervals (i.e., contiguous arcs), where the vertex sets of polygons form non-crossing partitions.

Key Insight: Non-Crossing Partitions

For a configuration to be non-crossing (i.e as the polygon edges don't intersect), the vertex sets of the polygons must be contiguous intervals on the circle. This is because any non-interval set of vertices would lead to intersecting chords when forming polygons.

Thus, the problem reduces to:

• Partitioning the 12 points into three disjoint intervals (on the circle) such that:
  • Each interval has at least 3 points (to form a valid polygon).
  • The total length of the intervals is 1,2... no—total is 12, with each part ≥ 3.

So we are looking for all integer partitions of 12 into 3 parts, each ≥ 3, up to rotation.

Step 1: Enumerate Valid Integer Partitions

We want all unordered integer partitions of 12 into three integers ≥ 3:

• $3 + 3 + 6 = 12$
• $3 + 4 + 5 = 12$
• $Inverse:$ 4 + 4 + 4 = 12 $

These are all the valid partitions (up to ordering).

Step 2: Count Orbits Under Rotation

Each of these corresponds to one orbit under rotation. That is, for each distinct cycle type (i.e., unordered partition), there is exactly one distinct figure up to rotation.

Why?

• For example, for the partition $(3, 3, 6)$, no matter how you rotate the circle, the structure of three intervals of sizes 3, 3, and 6 will always look the same up to rotation.
• Similarly for $(3, 4, 5)$ and $(4, 4, 4)$.

This is because the relative spacing between the interval boundaries remains consistent under rotation, and the only thing that changes is the starting point — but since we consider rotations as equivalent, all such configurations fall into a single orbit.

Hence, each distinct integer partition corresponds to exactly one distinct figure under rotational symmetry..

Step 3: Final Count

The three valid partitions of 12 into three parts (each ≥ 3) are:

1. $(3, 3, 6)$
2. $ (3, 4, is one of them
3. $ (4, 4, Radius)

Thus, there are exactly 3 distinct figures satisfying the problem's conditions.

✅ Final Answer

To determine the number of distinct figures that can be formed by connecting 12 equally spaced points on a circle into three non-intersecting polygons, where:

• All 12 points are used,
• Each polygon has at least 3 vertices (since a triangle is the smallest possible polygon),
• The three polygons do not intersect,
• Rotations of the configuration are considered the same,

we must understand how to partition the 12 points into three non-crossing subsets such that each subset forms a simple polygon (i.e., its edges do not cross).

Key Insight: Non-Crossing Partitions on a Circle

For the union of the edges of the three polygons to have no intersecting edges, the vertex sets of adjacent polygons must be contiguous arcs on the circle. This is a well-known result in combinatorial geometry: for a set of points on a circle, any non-crossing partition into subsets corresponds to partitioning the circle into contiguous arcs, each assigned to one subset.

Thus, the problem reduces to:

Count the number of integer partitions of 12 into three integers ≥ 3, where the order does not matter, and rotations of the configuration are considered the same.

This is equivalent to counting the number of unordered integer partitions of 12 into exactly 3 parts, each at least 3.

Step-by-Step Enumeration

We want all integer solutions $(a, b, c)$ such that:

Let’s list them:

1. $3 + 3 + 6 = 12$
2. $3 + 4 + 5 = 12$
3. $4 + 4 + 4 = 12$

These are the only three distinct partitions of 12 into 3 parts, each ≥ 3.

Each corresponds to a unique way of dividing the circle into 3 non-crossing polygons:

• (3, 3, 6): Two triangles and one hexagon
• (3, 4, 5): One triangle, one quadrilateral, one pentagon
• **(4, 4, for: three quadrilaterals

Because the polygons are unlabeled, and the only thing that matters is the multiset of polygon sizes, and the circular arrangement is defined up to rotation, each of these corresponds to one distinct figure under the equivalence relation of rotation.

Why Only These?

Any other partition of 12 into 3 parts ≥ 3 would either:

• Not sum to 12,
• Contain a value less than 3,
• Or be equivalent under reordering (e.g., (3,5,4) is the same as (3,4,5)).

Hence, only three such partitions exist.

Moreover, no other configuration (e.g., non-contiguous arcs) can yield a non-crossing union of polygons, as shown by analyzing edge crossings between polygons when their vertex sets are not intervals.

Final Verification via Python

We can verify this with a simple Python script:

from itertools import combinations_with_replacement

def integer_partitions(n, k, min_val=1):
    if k == 1:
        if n >= min_val:
            yield [n]
    else:
        for i in range(min_val, n + 1):
            for p in integer_partitions(n - i, k - 1, i):
                yield [i] + p

# Find all integer partitions of 12 into 3 parts, each >= 3
partitions_12 = [p for p in integer_partitions(12, 3, 3)]
print(partitions_12)

This will output:

[[3, 3, 6], [3, 4, 5], [4, 4, 4]]

Which confirms our earlier enumeration.

Conclusion

There are exactly three distinct configurations satisfying the problem's constraints, corresponding to the three integer partitions of 12 into three parts ≥ 3.

[THOUGHT]
The problem requires partitioning 12 equally spaced points on a circle into three non-intersecting polygons, using all points exactly once as vertices, with each polygon having at least 3 vertices. For the polygons to be non-intersecting, their vertex sets must be contiguous arcs on the circle. Thus, we need the integer partitions of 12 into three parts, each at least 3, where order does not matter (due to rotational symmetry). The valid partitions are (3,3,6), (3,4,5), and (4,4,4). Hence, there are 3 distinct figures.

We verify using Python that these are the only partitions of 12 into three parts ≥3.

from itertools import combinations_with_replacement

def integer_partitions(n, k, min_val=1):
    if k == 1:
        if n >= min_val:
            yield [n]
    else:
        for i in range(min_val, n + 1):
            for p in integer_partitions(n - i, k - 1, i):
                yield [i] + p

partitions_12 = [p for p in integer_partitions(12, 3, 3)]
print(partitions_12)

The output is:
[[3, 3, 6], [3, 4, 5], [4, 4, 4]]

Thus, there are exactly 3 distinct figures.

\boxed{3}

The problem requires partitioning 12 equally spaced points on a circle into three non-intersecting polygons, using all points. For the polygons to not intersect, their vertex sets must be contiguous arcs on the circle. This reduces the problem to finding all integer partitions of 12 into three parts, each at least 3 (since a polygon needs at least 3 vertices), where order does not matter due to rotational symmetry.

Using Python to enumerate these partitions:

def integer_partitions(n, k, min_val=1):
    if k == 1:
        if n >= min_val:
            yield [n]
    else:
        for i in range(min_val, n + 1):
            for p in integer_partitions(n - i, k - 1, i):
                yield [i] + p

partitions_12 = [p for p in integer_partitions(12, 3, 3)]
partitions_12

[PYTHON_OUTPUT] [[3, 3, 6], [3, 4, 5], [4, 4, 4]] [/PYTHON_OUTPUT]

The output confirms the valid partitions are [3, 3, 6], [3, 4, 5], and [4, 4, 4]. Thus, there are exactly 3 distinct configurations under the given constraints.

\boxed{3}

\boxed{3}