[THOUGHT] We need to count the number of distinct figures formed by partitioning 12 equally spaced points on a circle into three polygons, where polygons are non-intersecting (except possibly at vertices? Actually "do not intersect each other" means their interiors don't intersect, but they can share vertices? Wait: "All twelve points are used as vertices of the polygons" means each point is a vertex of exactly one polygon? Or could a point be a vertex of two polygons? Typically, if polygons are non-intersecting, they can share vertices only if those vertices are common to both polygons, but then the point would be a vertex of two polygons. The condition says "All twelve points are used as vertices of the polygons" but doesn't say each point is used exactly once. However, "with none left unused" suggests each point is used as a vertex, but possibly in multiple polygons? But if a point is used as a vertex of two polygons, then that point belongs to two polygons. But then the polygons would intersect at that point? Possibly allowed if they are non-intersecting except at vertices. However, typical problems about dividing points into polygons often mean partitioning the set of points into disjoint vertex sets for each polygon. Also "the three polygons do not intersect each other" suggests they are disjoint in interior, but can share vertices? Possibly yes if they are arranged such that polygons are separated by chords that don't cross. But if they share a vertex, then at that vertex the polygons meet, but that might be considered intersecting? Usually "do not intersect" means interiors are disjoint, boundaries may touch at vertices. However, the problem likely expects that each polygon uses a distinct subset of points, i.e., partition the 12 points into three sets of sizes corresponding to polygons. Because otherwise counting becomes messy.

Let's interpret: We have 12 points on a circle, labeled 0,...,11. We want to draw chords connecting these points to form three polygons (simple polygons, not self-intersecting). The polygons are non-intersecting (their interiors don't overlap). All points are used as vertices. Rotational symmetry considered same.

We need to count distinct figures up to rotation.

First, we need to understand possible polygon sizes. The sum of vertices across three polygons is 12. Each polygon must have at least 3 vertices (triangle or more). So we need to partition 12 into three integers each ≥3. Possible partitions: (3,3,6), (3,4,5), (4,4,4). Also (3,3,6) and (3,4,5) and (4,4,4). But note polygons are non-intersecting: they must be placed on the circle without crossing. Since points are on a circle, any polygon formed by connecting its vertices in order around the circle will be convex and inside the circle. However, if we take three subsets of points and form convex polygons for each subset, they might intersect if the subsets interleave. For example, if we take points {0,1,2} for triangle and {3,4,5} for triangle, they are separate arcs, so polygons are disjoint (they lie in disjoint arcs of the circle). But if we take points {0,2,4} for one polygon and {1,3,5} for another, the polygons will intersect because chords cross. So the condition "non-intersecting" imposes that the vertex sets of the polygons must appear in contiguous blocks around the circle (i.e., the subsets are intervals of the circular sequence). Because if they are not contiguous, chords connecting non-consecutive vertices will cross chords of other polygons.

Thus, we need to partition the 12 points into three contiguous arcs (blocks) of sizes corresponding to polygon vertices. Each block will be a convex polygon inscribed in the circle (since points are on a circle, connecting them in order yields a convex polygon). However, note that a polygon with vertices on a circle is not necessarily convex if we connect them in some order? But if we connect them in the order they appear on the circle, we get a convex polygon (since all vertices lie on a circle and we take them in cyclic order). So each polygon is the convex polygon formed by its vertices in the circular order.

Thus, the problem reduces to: Count the number of ways to partition the 12 points into three contiguous arcs (cyclic intervals) of sizes a,b,c (each ≥3) such that a+b+c=12, up to rotation.

But wait: Are there other configurations where polygons are not convex? For example, a polygon could be non-convex but still have all vertices on the circle? Actually if all vertices lie on a circle, any simple polygon with those vertices is either convex (if vertices are taken in order around the circle) or self-intersecting if not. So non-convex polygons would have vertices not in order, but then chords would cross? Possibly they could still be non-intersecting? But if vertices are on a circle and we connect them in some order that is not the circular order, the polygon will be self-intersecting (star-shaped). The problem likely expects simple polygons (non-self-intersecting). So each polygon must be convex (or at least simple with all vertices on circle). For a simple polygon inscribed in a circle, the vertices must appear in the polygon in the same cyclic order as on the circle. So each polygon's vertices are a contiguous block on the circle.

Thus, the three polygons correspond to three contiguous arcs partitioning the circle.

Now, we need to count distinct partitions up to rotation.

Let’s denote the sizes of the three arcs as (a,b,c) with a,b,c ≥3, a+b+c=12.

We can list all integer triples:

1. (3,3,6)
2. (3,4,5)
3. (4,4,4)

Now, for each triple, we need to count distinct cyclic orderings up to rotation.

For a given triple (a,b,c), the number of distinct arrangements on a circle up to rotation is the number of distinct cyclic permutations of the multiset {a,b,c}. Since the circle is labeled by points, but rotations are considered identical, we can fix one point at position 0 (say the first point of the first polygon) to break rotational symmetry. However, because the polygons are contiguous arcs, we can think of choosing starting points for each arc. But since rotations are considered same, we can fix the first point as vertex 0. Then we need to choose lengths of arcs in cyclic order.

But careful: The arcs are contiguous and partition the circle. If we fix point 0 as the start of the first polygon, then the first polygon occupies a contiguous block of a points starting at 0. Then the second polygon occupies next b points, and third polygon c points. This gives a unique representation for each triple (a,b,c) with a,b,c ≥3. However, note that different cyclic permutations of (a,b,c) yield different arrangements? Since we fixed point 0 as start of first polygon, the triple is ordered. But rotations that make the first polygon not start at 0 are considered same. So by fixing point 0 as start of first polygon, we have eliminated rotational symmetry. However, we must also consider that the polygons are unlabeled? The problem says "three polygons" — are they considered distinct? Typically, the polygons are unlabeled (just three polygons). So arrangements that differ by permuting the polygons are considered the same if they have the same geometric figure. But since the polygons are non-intersecting and placed on arcs, swapping polygons corresponds to different cyclic orderings. However, if the triple has repeated sizes, swapping identical-sized polygons yields the same figure. So we need to count distinct cyclic arrangements of the multiset {a,b,c} up to rotation.

Standard approach: For a cyclic sequence of length 3 (three arcs), the number of distinct arrangements up to rotation is given by the number of necklaces with beads of sizes a,b,c. Since there are only 3 beads, we can enumerate.

Case 1: (3,3,6). Multiset: two 3's and one 6. Distinct cyclic arrangements: Since the necklace has 3 beads, we need to count distinct cycles up to rotation. Possible distinct arrangements: (3,3,6) and (3,6,3) and (6,3,3) — but these are rotations of each other? Let's check: Starting with (3,3,6), rotate by one position gives (3,6,3), rotate again gives (6,3,3). So all three are rotations of each other. Therefore, only 1 distinct arrangement up to rotation.

But wait: Could there be a reflection symmetry? The problem only considers rotations, not reflections. So only rotations matter. So (3,3,6) yields 1 distinct figure.

Case 2: (3,4,5). All sizes distinct. For three distinct sizes, the number of distinct cyclic arrangements up to rotation is (3-1)! = 2 (since necklaces with distinct beads: (a,b,c) and (a,c,b) are not rotations of each other because rotating (a,b,c) gives (b,c,a) and (c,a,b), but not (a,c,b)). So for distinct a,b,c, there are 2 distinct necklaces. So (3,4,5) yields 2 distinct figures.

Case 3: (4,4,4). All equal. Only 1 distinct arrangement.

Thus total = 1+2+1 = 4.

But wait: Are there any other configurations where polygons are not necessarily convex? Could polygons be star-shaped? Possibly, but then vertices are not contiguous? However, if vertices are not contiguous, chords cross. So likely not allowed.

But also, could polygons share vertices? If polygons share a vertex, then that point is a vertex of two polygons. But then the polygons would intersect at that point, which might be allowed if interiors don't intersect. However, sharing a vertex means the point is used twice, but the condition "All twelve points are used as vertices" doesn't forbid a point being vertex of multiple polygons. But then the polygons would share that point, and they might intersect along a side? Actually if they share a vertex, they could still not intersect interiorly. But then the arcs would not be contiguous? Because if polygons share a vertex, then the vertex sets are not disjoint intervals; they interleave. That would cause chords to cross? Possibly not if the polygons are arranged such that they share only a single point and are otherwise separated. For example, consider two triangles sharing a vertex but otherwise in disjoint arcs. However, if they share a vertex, then that point belongs to both polygons, but the arcs of the circle for each polygon would overlap at that vertex. That would mean the vertices of each polygon are not a contiguous block because there is a gap? Actually, if we have a polygon with vertices A,B,C and another with A,D,E, then the vertices of first polygon are not contiguous on the circle because between B and C there might be other points not in first polygon? But if all points are used, then the remaining points are used by third polygon. Possibly the vertices of each polygon are not contiguous. But would that cause crossing chords? Let's test with a small example: 6 points on a circle, partition into two triangles sharing a vertex. Suppose points 0,1,2 for triangle1 and 0,3,4 for triangle2, and point 5 unused? But all points must be used. So third polygon must use point 5 and maybe others. This is messy.

Given the problem likely expects contiguous arcs interpretation, we'll stick with that.

But let's verify by brute force enumeration of all possible partitions of 12 points into three polygons (non-intersecting). We can write a program to generate all ways to assign each point to one of three polygons, then check if polygons are non-intersecting. However, checking non-intersection is complex. But we can use the known fact: For polygons inscribed in a circle, they are non-intersecting iff their vertex sets are non-interleaving (i.e., they form intervals on the circle). So we can enumerate all ways to choose three disjoint intervals (contiguous arcs) covering all 12 points, each of length at least 3. Then count distinct up to rotation.

Let's do that.

We have 12 points labeled 0..11. Choose three arcs (contiguous blocks) that partition the circle. Since it's a circle, we can fix point 0 as the start of first arc to break rotation symmetry. Then we need to choose lengths a,b,c ≥3, a+b+c=12. Then the arcs are: arc1: points 0..a-1; arc2: points a..a+b-1; arc3: points a+b..11.

But note that arcs are contiguous and cover the circle exactly. This representation is unique for each ordered triple (a,b,c). However, we must consider that rotating the whole figure yields a different triple in this representation? But we fixed point 0 as start of first arc, so rotations are already accounted for. However, we also need to consider that polygons are unlabeled. In our representation, the polygons are labeled by order of arcs. But the actual polygons are not labeled; swapping arcs yields the same figure if the arcs are identical in size? Actually, if the polygons are considered unlabeled, then arrangements that differ by permuting the arcs are considered the same figure. In our representation with fixed point 0, we have labeled arcs in order. So we need to identify arrangements that are equivalent under permutation of arcs.

Thus, we need to count distinct partitions of the circle into three unlabeled contiguous arcs of lengths ≥3.

That is equivalent to counting distinct cyclic sequences of lengths a,b,c up to rotation and permutation of the entries.

So we need to count orbits of the set of ordered triples (a,b,c) with a,b,c≥3, a+b+c=12, under the dihedral group of order 3? Actually, permutations of the triple correspond to relabeling polygons. Since the circle is unlabeled except for rotational symmetry, we need to consider the combined symmetry group: rotations of the circle and permutations of polygon labels. However, the problem statement says "Rotations of the circle that produce congruent figures are considered the same." It does not mention reflections. It also says "three polygons" — likely the polygons are unlabeled (just three polygons). So we need to consider that the polygons are indistinguishable. Therefore, we need to count equivalence classes under rotation of the circle and permutation of the three polygon labels.

But careful: Permuting polygon labels is equivalent to permuting the arcs. However, since the arcs are contiguous and cover the circle, a permutation of arcs corresponds to a different cyclic ordering of lengths. But some permutations might be realizable by a rotation? For example, (3,3,6) rotated by one gives (3,6,3), which is a permutation (swap first and second? Actually (3,3,6) permuted to (3,6,3) is a cyclic shift, which is a rotation of the circle. So for (3,3,6), the two 3's are identical, so any permutation yields the same cyclic sequence up to rotation. So that case already accounted.

For (3,4,5), the two distinct necklaces (3,4,5) and (3,5,4) are not rotations of each other, but they might be related by a permutation of labels? Actually, if we permute labels, (3,4,5) can be mapped to (4,3,5) etc. But these are rotations of each other? Let's check: (3,4,5) rotated by one gives (4,5,3), which is same multiset. But (3,5,4) rotated by one gives (5,4,3) and (4,3,5). So (3,5,4) is a rotation of (5,4,3) but not of (3,4,5). However, if we allow permutation of polygon labels, then (3,4,5) and (3,5,4) are considered the same figure because we can relabel polygons. But careful: Relabeling polygons corresponds to permuting the arcs, but does that preserve the cyclic order? If we permute arcs arbitrarily, we can get any ordering of the three lengths. So effectively, for distinct lengths, all 6 permutations are possible via relabeling. However, rotations of the circle correspond to cyclic shifts of the sequence. So the full symmetry group is generated by cyclic shifts and permutations of the three arcs. But since the circle is unlabeled, we can rotate so that the smallest length polygon is at a fixed position? Not necessarily.

We need to count distinct configurations up to rotation and relabeling of polygons.

Alternatively, we can think: We have three unlabeled polygons. Their vertex sets are three contiguous arcs partitioning the circle. So the configuration is determined by the multiset of arc lengths. However, the cyclic order of lengths matters because arcs are arranged around the circle. But if polygons are unlabeled, then the cyclic order of lengths up to rotation and reversal? But reversal corresponds to reflection of the circle, which is not considered same unless the problem allows reflection symmetry. The problem only says rotations are considered same, not reflections. So we must consider only rotations.

But polygon labels are unlabeled, so we can permute arcs arbitrarily. However, permuting arcs is not a symmetry of the circle; it's a relabeling of polygons. Since the polygons are unlabeled, two configurations that differ only by permuting which polygon is which are considered identical. Therefore, we need to count equivalence classes of ordered triples (a,b,c) under two actions: rotation of the circle (cyclic shift) and permutation of the three entries (any permutation). But note that cyclic shift is a particular permutation (cyclic permutation). So the full symmetry group is the symmetric group S3 acting on the triple, but with the additional constraint that we consider two triples equivalent if they differ by a cyclic shift? Actually, the circle is rotated, which corresponds to cyclically shifting the order of arcs. But if we also allow arbitrary permutation of arcs, then any permutation of arcs can be achieved by first permuting arcs arbitrarily and then rotating? Not exactly: Rotating corresponds to cyclic shift, which is a permutation in S3. The group generated by all permutations is S3 itself. So if we allow arbitrary permutation of arcs, then any permutation of the triple is allowed. However, is that correct? Because permuting arcs corresponds to swapping which polygon is called first, second, third. Since the polygons are unlabeled, we can indeed permute them arbitrarily. So the symmetry group is S3 (all permutations of the three arcs). But note that the circle rotation is already a cyclic shift, which is an element of S3. So allowing full permutation means we consider two triples equivalent if they differ by any permutation in S3. That would mean that for (3,4,5), all 6 permutations are equivalent, so only 1 distinct figure. But wait, is that correct? Let's think: Suppose we have a configuration with arcs of lengths 3,4,5 in that order around the circle. If we permute the polygons, we get a configuration with arcs of lengths 4,3,5 in that order. But is that configuration the same as the original? Only if there exists a rotation of the circle that maps the first configuration to the second. But a rotation corresponds to a cyclic shift of the sequence. Can we achieve swapping first and second arcs by a rotation? For distinct lengths, rotating (3,4,5) gives (4,5,3) or (5,3,4). It does not give (4,3,5). So (3,4,5) and (4,3,5) are not rotations of each other. However, if we allow relabeling of polygons, we consider them the same figure because the polygons are not labeled. So indeed, if we permute arcs arbitrarily, (3,4,5) and (4,3,5) are considered the same. Therefore, all permutations of the triple are equivalent under polygon relabeling. So for distinct lengths, there is only 1 figure.

But wait: The problem statement: "Rotations of the circle that produce congruent figures are considered the same." It does not mention polygon labeling. Typically in such combinatorial geometry problems, the polygons are considered unlabeled. So we should treat them as unlabeled.

Thus, we need to count the number of unlabeled partitions of the circle into three unlabeled contiguous arcs of lengths ≥3.

That is: count the number of multisets {a,b,c} with a,b,c≥3, a+b+c=12, but also considering that the cyclic order matters? Actually, if the polygons are unlabeled, then the cyclic order of arc lengths is not important because we can permute arcs arbitrarily. However, the cyclic order is part of the geometric arrangement: arcs are arranged around the circle in some order. But if we can permute arcs arbitrarily, then any cyclic permutation of the arcs yields the same set of arcs? But careful: Permuting arcs means we reassign which polygon corresponds to which arc. Since the polygons are unlabeled, the assignment is arbitrary. Therefore, the only invariant is the multiset of arc lengths. However, is that sufficient? Consider two different cyclic orders of the same multiset: (3,4,5) and (3,5,4). Are they geometrically distinct? They are different arrangements of arcs around the circle. But if we relabel polygons, we can map one to the other? Let's see: In (3,4,5), polygon A has size 3, polygon B size 4, polygon C size 5, in that order. In (3,5,4), polygon A size 3, polygon B size 5, polygon C size 4. If we swap polygons B and C, we get (3,4,5). So indeed, with polygon relabeling, (3,4,5) and (3,5,4) are the same because we can swap the polygons of sizes 4 and 5. Therefore, any permutation of the multiset yields the same configuration. So the only invariant is the multiset of sizes.

Thus, the number of distinct figures is simply the number of multisets {a,b,c} with a,b,c≥3, a+b+c=12.

Let's list all partitions of 12 into three parts each ≥3:

1. {3,3,6}
2. {3,4,5}
3. {4,4,4}

That's 3 possibilities.

But earlier we got 4 when considering only rotations and distinct cyclic orders. With polygon relabeling, we get 3.

Which is correct? The problem statement: "Rotations of the circle that produce congruent figures are considered the same." It does not mention reflections or polygon relabeling. However, "three polygons" likely are unlabeled because otherwise they'd be considered distinct polygons. Typically, when counting configurations of polygons on a circle, the polygons are considered unlabeled unless specified. But we need to check the context: "construct a figure consisting of three polygons" — the figure is what we count. The polygons are part of the figure, but the figure is defined by the arrangement of vertices and edges. If we swap two polygons, the figure might look different if the polygons have different shapes/sizes. But if the polygons have different sizes, swapping them yields a different arrangement of arcs around the circle? Actually, swapping polygons means we exchange which polygon occupies which arc. Since the polygons are unlabeled, the figure is defined solely by the set of arcs. So if we have arcs of lengths 3,4,5 in some order, swapping the polygons of sizes 4 and 5 yields arcs of lengths 3,5,4, which is a different cyclic order. But if we consider the figure as just the drawing of chords, the drawing is symmetric under relabeling of polygons? No, the drawing is not symmetric because the arcs are in different positions. However, if we consider the polygons as unlabeled, then the figure is determined by the multiset of arc lengths and the cyclic order? But if we allow arbitrary permutation of arcs, then any cyclic permutation of the multiset is considered the same figure. That would mean we only care about the multiset, not the cyclic order. But is that plausible? Let's think of a concrete example: 12 points, arcs of lengths 3,4,5 in that order. If we rotate the circle so that the arc of length 3 starts at a different point, that's a rotation. But if we permute arcs, we get a different arrangement of lengths around the circle. Are these two arrangements congruent under rotation? Only if the cyclic order can be matched by rotation. For distinct lengths, different cyclic orders are not rotations of each other. However, if we consider the polygons unlabeled, then the figure with arcs in order (3,4,5) is the same as the figure with arcs in order (4,3,5) because we can just swap the polygons of sizes 4 and 3? But swapping polygons of sizes 4 and 3 changes the cyclic order from (3,4,5) to (4,3,5). But is that allowed? Since the polygons are unlabeled, yes. So the figure is defined by the cyclic sequence of arc lengths up to permutation of the labels. That means we consider two cyclic sequences equivalent if one can be obtained from the other by a permutation of the entries. But note that any permutation of three entries can be realized by a combination of rotation and swapping? Actually, the group of symmetries we consider is generated by cyclic shifts (rotations) and arbitrary permutations of the polygons (which correspond to permuting the entries). However, arbitrary permutation of three entries is not necessarily a rotation. So the equivalence relation is: two ordered triples (a,b,c) are equivalent if there exists a permutation σ in S3 such that (a,b,c) = σ(a',b',c') after possibly rotating the circle? Wait, rotation of the circle corresponds to cyclic shift of the sequence. So if we allow arbitrary permutation of polygons, then any permutation of the triple is allowed. But rotation is already a permutation (cyclic shift). So the full symmetry group is S3 (all permutations). However, is rotation of the circle already included? Yes, because rotating the circle corresponds to a cyclic shift of the arcs. But if we allow arbitrary permutation of polygons, then we can map any permutation to any cyclic shift? Not exactly: The group S3 includes both cyclic shifts and transpositions. So if we allow both rotations and polygon relabeling, then indeed any two triples that are permutations of each other are equivalent. Because we can first permute the polygons arbitrarily, then rotate if needed. But note: Permuting polygons is not a physical rotation of the circle; it's a relabeling. Since the polygons are unlabeled, we consider that as the same figure. So indeed, the figure is determined by the multiset of arc lengths.

Thus, the number of distinct figures is the number of multisets {a,b,c} with a,b,c≥3, a+b+c=12.

That is 3.

But wait: Are there any other configurations where polygons are not convex? Possibly polygons could be self-intersecting? But the problem says "polygons" likely meaning simple polygons. Also "do not intersect each other" suggests they are non-intersecting simple polygons.

Thus, answer might be 3.

However, I recall a known problem: "Twelve points on a circle divided into three polygons" — I think the answer might be 4. Let's search memory: There is a known problem: "How many ways to divide a circle into 3 polygons using chords connecting 12 points?" Possibly answer is 4. But we need to be sure.

Let's write a brute-force enumeration to verify. We'll generate all ways to partition 12 points into three subsets (each subset size ≥3) such that the convex hulls of the subsets are non-intersecting. For points on a circle, the convex hulls are just the polygons formed by the points in order. Non-intersecting condition: the convex hulls are disjoint. For points on a circle, the convex hulls are non-intersecting iff the subsets are non-interleaving (i.e., they appear as contiguous blocks). So we need to count partitions of 12 points into three contiguous blocks of sizes ≥3. Then we need to count distinct up to rotation and polygon relabeling.

We'll write a Python script to enumerate all possible triples of contiguous blocks (with fixed starting point 0) and then factor out rotations and polygon relabeling.

Let's do that.

We'll generate all ordered triples (a,b,c) with a,b,c≥3, a+b+c=12. For each, we have a specific arrangement: blocks of sizes a,b,c in order. Then we consider the multiset of sizes {a,b,c}. But also consider the cyclic order? Since polygons are unlabeled, any permutation of the blocks yields the same multiset. However, is the cyclic order irrelevant? Consider two different cyclic orders of the same multiset: e.g., (3,4,5) and (3,5,4). Are they distinct figures? If polygons are unlabeled, swapping the polygons of sizes 4 and 5 yields the other cyclic order. So they are the same figure. Therefore, only the multiset matters.

Thus, count distinct multisets.

But wait: What about the arrangement of blocks on the circle? The blocks are contiguous and cover the circle. The order of blocks around the circle is part of the geometry. However, if we can permute the blocks arbitrarily (since polygons unlabeled), then any ordering of the blocks is equivalent. So indeed, only the multiset matters.

Thus, answer should be 3.

But let's verify by brute force enumeration of all possible partitions into three contiguous blocks, then apply equivalence under rotation and permutation of blocks.

We'll write a script to enumerate all ways to choose starting indices for three blocks (non-overlapping, covering all points) with block lengths ≥3. Since points are on a circle, we can fix point 0 as the start of the first block to break rotational symmetry. Then we need to choose lengths a,b,c ≥3, a+b+c=12. Then the blocks are: block1: points 0..a-1, block2: a..a+b-1, block3: a+b..11. This gives a unique representation for each ordered triple (a,b,c). However, this representation already fixes rotation by fixing point 0 as start of block1. Now, we need to consider polygon relabeling: two ordered triples that are permutations of each other correspond to the same figure. So we need to group ordered triples into equivalence classes under permutation of entries.

Thus, we can generate all ordered triples (a,b,c) with a,b,c≥3, a+b+c=12. Then compute their multisets. Count distinct multisets.

Let's do that in Python.

But also note: Could there be configurations where the blocks are not contiguous but still non-intersecting? As argued, likely not. So our enumeration should be complete.

Let's code.

We'll generate all integer solutions to a+b+c=12 with a,b,c≥3. Then for each, compute the sorted triple (a,b,c). Then count distinct sorted triples.

That yields 3.

But wait: What about the case where a polygon is a quadrilateral but vertices are not necessarily contiguous? Could we have a polygon with vertices not contiguous but still non-intersecting with others? For example, take points 0,1,2,3 for polygon A, and points 4,5,6,7 for polygon B, and points 8,9,10,11 for polygon C. That's contiguous blocks of 4 each. That's (4,4,4). That's included.

What about a polygon with vertices 0,2,4? That's triangle but with vertices every other point. Then the other polygons would have to use the remaining points 1,3,5,7,8,9,10,11? But we need three polygons total. That seems messy and likely leads to intersecting chords. So probably not allowed.

Thus, our contiguous blocks assumption is reasonable.

Let's run a quick brute force to check if there are any other configurations that satisfy the conditions. We'll generate all possible assignments of points to three polygons (each point assigned to exactly one polygon) such that each polygon has at least 3 points. Then check if polygons are non-intersecting. But checking non-intersection is nontrivial. However, we can use the fact that for points on a circle, polygons are non-intersecting iff the vertices of each polygon form a contiguous arc (cyclic interval). So we can check for each assignment whether the vertices of each polygon are contiguous on the circle. That is a necessary and sufficient condition for non-intersecting convex polygons? Actually, if polygons are convex (which they are if all vertices lie on a circle and taken in order), then they are non-intersecting iff their vertex sets are non-interleaving. Non-interleaving means that on the circle, the vertices of each polygon appear consecutively. So we can check contiguity.

Thus, we can enumerate all ways to partition 12 points into three subsets (each size ≥3) such that each subset is contiguous on the circle. Then count distinct up to rotation and polygon relabeling.

We'll implement that.

We'll label points 0..11. We'll iterate over all choices of three start indices and three lengths (≥3) such that the blocks are disjoint and cover all points. Since it's a circle, we need to consider blocks that wrap around? But because we require contiguity and covering, the blocks will be contiguous arcs that partition the circle. So we can fix the starting point of the first block to be 0 (to break rotational symmetry). Then the blocks are determined by lengths a,b,c. So the enumeration is exactly the ordered triples (a,b,c) with a,b,c≥3, a+b+c=12. So that's what we already did.

Thus, the number of distinct figures is the number of equivalence classes of these ordered triples under permutation of entries (since polygons unlabeled) and rotation (already fixed by setting first block start at 0). But careful: Setting first block start at 0 already factors out rotation. However, does permuting entries preserve the condition that first block starts at 0? Not necessarily. When we permute entries, we get a different ordered triple. But if we allow polygon relabeling, we consider those equivalent. So we need to group ordered triples into equivalence classes under permutation of entries. That is exactly counting distinct multisets.

Thus, answer is 3.

But wait: Could there be configurations where the polygons are not necessarily convex? For example, a polygon could be a star polygon? But star polygons self-intersect, which likely violates "polygons do not intersect each other" (self-intersection is not allowed). So they must be simple polygons.

Thus, answer likely 3.

However, I recall a similar problem: "Twelve points on a circle are divided into three polygons" might have answer 4. Let's search memory: There is a known problem: "How many ways to divide a regular 12-gon into three polygons by drawing non-intersecting diagonals?" Possibly different.

Given the problem explicitly says "Using line segments to connect the twelve points", meaning we can draw chords between points. The polygons are formed by these chords. The condition "The three polygons do not intersect each other" means their interiors are disjoint. That implies the polygons are separated by chords that don't cross. That essentially means the vertices are partitioned into three contiguous arcs.

Thus, I think 3 is correct.

But let's double-check by writing a small script to enumerate all partitions of 12 points into three contiguous arcs of sizes ≥3, and then apply equivalence under rotation and permutation of arcs. We'll also consider that polygons are unlabeled, so we treat two partitions as same if they have the same multiset of sizes.

We'll generate all ordered triples (a,b,c) with a,b,c≥3, a+b+c=12. Then compute sorted triple. Count distinct sorted triples.

We'll also consider that rotations are already accounted for by fixing the first block start at 0. However, is it possible that a configuration with a different rotation yields a different multiset? No, because rotating doesn't change the multiset of sizes. So our enumeration of ordered triples with first block start at 0 already covers all rotationally distinct configurations? Wait: By fixing the first block start at 0, we are essentially choosing a specific rotation for each configuration. But different rotations of the same configuration would correspond to different ordered triples? For example, consider a configuration with blocks of sizes 3,4,5 in that order. If we rotate the circle so that the block of size 3 starts at a different point, we get a different ordered triple in our representation? Because we always set the first block (the one starting at point 0) to be the block that starts at the point we fix as 0. But if we rotate the circle, the point that was originally 0 might now be in the middle of a block. So our representation with first block start at 0 may not capture all rotations. Actually, we need to consider that the labeling of points is fixed. Points are marked on the circle at equal intervals. They are labeled 0..11 in some order. When we rotate the circle, the labels are permuted cyclically. So if we consider configurations up to rotation, we identify configurations that are related by a cyclic shift of the labels. In our enumeration with first block starting at point 0, we are essentially choosing a representative from each rotation equivalence class by requiring that point 0 is a vertex of the first polygon and is the start of its block? But is it necessary that point 0 is the start of a block? Not necessarily. The block could start at any point. However, we can always rotate the circle so that the start of the first polygon is at point 0. Since the polygons are contiguous arcs, we can rotate so that the first polygon's arc starts at point 0. That yields a unique representative for each rotation equivalence class, provided we label the polygons in some order. But the polygons are unlabeled, so we cannot consistently label the first polygon. However, we can still fix that point 0 is a vertex of some polygon, but not necessarily the start. But we can rotate so that point 0 is the first vertex of some polygon in the cyclic order. That is equivalent to fixing the start of the polygon that contains point 0. Since the polygons are contiguous arcs, the arc containing point 0 is a contiguous block. We can rotate so that point 0 is the first vertex of that arc. That gives a unique representative for each rotation class, provided we decide which polygon's arc is considered first. But since polygons are unlabeled, we can choose any polygon containing point 0 as the first block. However, to avoid ambiguity, we can fix that the polygon containing point 0 is the smallest polygon? Or we can consider all possibilities.

Thus, our earlier enumeration with first block starting at 0 and block sizes a,b,c corresponds to a specific ordering of polygons: the polygon containing point 0 is the first polygon, and we take its arc starting at point 0. This enumerates all rotationally distinct configurations where the polygon containing point 0 is considered the first polygon. But since polygons are unlabeled, we might be overcounting or undercounting? Actually, for each configuration, there are three choices of which polygon is considered first (if all three polygons are distinct). However, since we fix the polygon containing point 0 as the first polygon, that polygon is uniquely determined by the configuration (the polygon that contains point 0). So each configuration has exactly one polygon containing point 0. Therefore, we can fix that polygon as the first polygon, and rotate so that point 0 is the first vertex of that polygon. That yields a unique representative for each rotation equivalence class. So our enumeration of ordered triples (a,b,c) with a,b,c≥3, a+b+c=12, where the first block is the polygon containing point 0, and point 0 is its first vertex, indeed enumerates each rotationally distinct configuration exactly once, provided we consider polygons unlabeled? Wait, but we still have labeled polygons: first polygon (contains point 0), second polygon (next arc), third polygon (remaining). However, the polygons are unlabeled, so we might consider two configurations that are the same up to permutation of polygons as identical. In our enumeration, we have fixed the first polygon to be the one containing point 0. That is fine because that polygon is uniquely determined by the configuration (the one containing point 0). However, the other two polygons are not labeled. But in our ordered triple, we assign the second polygon to the next arc and third polygon to the last arc. That imposes an ordering on the second and third polygons based on their position around the circle. But since the polygons are unlabeled, swapping the second and third polygons should yield the same configuration. However, does swapping them correspond to a different arrangement of arcs? Yes, because the arcs are in different positions. But if we swap the polygons, we get a different cyclic order of lengths. However, since the polygons are unlabeled, we consider those as the same figure only if there exists a rotation that maps one cyclic order to the other. But as we argued earlier, for distinct lengths, different cyclic orders are not rotations of each other. However, if we allow swapping polygons, we are effectively considering the two cyclic orders as equivalent. But is that allowed? The problem statement does not say polygons are labeled. So we should consider them unlabeled. Therefore, we need to consider equivalence under permutation of the polygons (any permutation). But note: The polygon containing point 0 is fixed as the first polygon in our representative. So if we permute polygons, the polygon containing point 0 might not be first anymore. But since we are using representatives where the polygon containing point 0 is first, we cannot directly apply arbitrary permutations. However, we can still consider that two configurations are equivalent if there exists a permutation of the polygons that maps one configuration to the other. But since the polygon containing point 0 is unique in each configuration, a permutation that maps the configuration to another must map the polygon containing point 0 in one to the polygon containing point 0 in the other. Because point 0 is a distinguished point on the circle (since we fixed the labeling). Actually, point 0 is fixed on the circle. So in both configurations, point 0 is a vertex of some polygon. That polygon is the one containing point 0. So any permutation of polygons must map that polygon to the corresponding polygon in the other configuration. Therefore, the permutation can only permute the other two polygons. So the equivalence under polygon relabeling reduces to whether the second and third polygons can be swapped. That is, configurations (a,b,c) and (a,c,b) are considered the same. So we need to identify ordered triples that differ by swapping the last two entries.

Thus, the number of distinct figures is the number of ordered triples (a,b,c) with a,b,c≥3, a+b+c=12, modulo the equivalence (a,b,c) ~ (a,c,b). But note that (a,b,c) and (a,c,b) are distinct unless b=c.

So we need to count orbits under this equivalence.

Let's enumerate all ordered triples (a,b,c) with a,b,c≥3, a+b+c=12. There are finitely many. We'll generate them and group by (a, min(b,c), max(b,c)) essentially.

Let's do that in Python.

We'll also consider that rotations are already accounted for by fixing the polygon containing point 0 as first. So our enumeration is complete.

Now, let's write the code to generate all ordered triples and count distinct under the equivalence (a,b,c) ~ (a,c,b).

But also note: Could there be configurations where the polygon containing point 0 is not the first polygon in the cyclic order? In our representation, we always take the polygon containing point 0 as the first polygon. That's fine.

Thus, we'll generate all ordered triples (a,b,c) with a,b,c≥3, a+b+c=12. Then for each, compute the canonical representation: sort the last two: (a, min(b,c), max(b,c)). Then count distinct canonical triples.

Let's do that.

We'll also consider the possibility of blocks that are not in order? Actually, the blocks are in cyclic order. So our representation is correct.

Let's code.

[PYTHON] from itertools import product

solutions = [] for a in range(3, 13): for b in range(3, 13): c = 12 - a - b if c >= 3: solutions.append((a,b,c))

print("All ordered triples (a,b,c):") for sol in solutions: print(sol)

Now group by equivalence (a,b,c) ~ (a,c,b)

canonical = {} for a,b,c in solutions: # canonical form: sort b and c key = (a, min(b,c), max(b,c)) canonical[key] = canonical.get(key, 0) + 1

print("\nDistinct canonical forms:") for key in sorted(canonical.keys()): print(key)

print(f"\nNumber of distinct figures: {len(canonical)}") [/PYTHON]

Let's run this.

But before running, we should also consider that the polygons are unlabeled, so we might also need to consider swapping the first polygon with one of the others? But as argued, the polygon containing point 0 is distinguished. However, is it possible that point 0 is not a vertex of the smallest polygon? But that doesn't matter. The polygon containing point 0 is unique. So we cannot swap it with another polygon because that would move point 0 to a different polygon, which is not a symmetry of the configuration (since point 0 is a fixed point on the circle). Wait: The points are marked at equal intervals, but the labeling of points is not physically meaningful? The problem says "Twelve points are marked on a circle at equal intervals." It does not say the points are labeled. So we are not given specific labels. The points are indistinguishable except for position. So point 0 is just a reference. However, when we consider rotations, we consider the whole configuration up to rotation. So the absolute position of point 0 is not fixed; we can rotate the circle. But in our enumeration, we fixed point 0 as the start of the first polygon. That is a choice of a reference point. But since the points are indistinguishable, we can always rotate so that a vertex of some polygon is at a reference position. However, we also need to decide which polygon contains that reference point. But since polygons are unlabeled, we can arbitrarily choose which polygon is considered first. But to avoid overcounting, we need to consider that two configurations that differ only by which polygon contains the reference point are considered the same? Actually, the reference point is not a distinguished point on the circle; it's just a marker we introduce to break symmetry. Since the points are equally spaced, any vertex could be chosen as the starting point. So when we fix point 0 as the start of the first polygon, we are effectively choosing a vertex of the configuration as the origin. But since the configuration has 12 vertices, any vertex could be chosen as the start. However, because the polygons are unlabeled, we might overcount by a factor of 3 if the first polygon is not distinguished. Let's think carefully.

We want to count distinct configurations up to rotation. One way to count is to count all configurations modulo rotation. A standard method is to fix one vertex to be at position 0 (i.e., choose a vertex as the reference). Then count all configurations with that vertex as the start of some polygon. However, since the polygons are unlabeled, we must consider that the vertex at position 0 could belong to any of the three polygons. But in our enumeration, we assumed that the polygon containing point 0 is the first polygon, and we take the arc starting at point 0. That means we are only considering configurations where the polygon containing point 0 is taken as the first polygon. But is that without loss of generality? For any configuration, we can rotate so that the polygon containing point 0 becomes the first polygon in the order, and point 0 is the first vertex of that polygon. That is possible because we can rotate the circle so that point 0 is at the start of that polygon's arc. However, we must also decide which polygon is considered first. Since polygons are unlabeled, we can arbitrarily choose the polygon containing point 0 to be the first polygon. That yields a unique representative for each rotation equivalence class. But is it unique? Given a configuration, the polygon containing point 0 is unique. So there is exactly one polygon containing point 0. So we can label that polygon as the first polygon. Then we rotate so that point 0 is the first vertex of that polygon. That gives a unique ordered triple (a,b,c) where a is the size of that polygon, b and c are the sizes of the other two polygons in the order they appear around the circle (starting after the end of the first polygon). This ordered triple is unique for each rotation equivalence class. Therefore, our enumeration of ordered triples (a,b,c) with a,b,c≥3, a+b+c=12, where the polygon containing point 0 has size a, and the other polygons are ordered as they appear, enumerates each rotation equivalence class exactly once. However, this still treats the second and third polygons as labeled (by their order). But since the polygons are unlabeled, swapping the second and third polygons yields a different ordered triple (a,c,b). Are these two ordered triples representing the same configuration? Only if there is a rotation that maps one to the other while keeping the polygon containing point 0 as the first polygon. But rotating the circle would change the polygon containing point 0? No, because point 0 is fixed as the start of the first polygon. If we swap the second and third polygons, the polygon containing point 0 remains the same (first polygon). However, the order of the other two polygons around the circle is reversed. But is that configuration obtainable from the original by a rotation? No, because rotation preserves cyclic order. So (a,b,c) and (a,c,b) are not rotations of each other (unless b=c). However, if polygons are unlabeled, we consider the configuration with arcs of sizes a,b,c in cyclic order the same as with arcs of sizes a,c,b because we can just swap the labels of the second and third polygons. But note: swapping labels does not change the geometry; it's just renaming polygons. So indeed, they represent the same geometric figure. Therefore, we need to identify (a,b,c) and (a,c,b).

Thus, our canonical representation (a, min(b,c), max(b,c)) is correct.

But wait: What about the case where the polygon containing point 0 is not uniquely determined? It is always unique because point 0 is a vertex of exactly one polygon. So that's fine.

Thus, the number of distinct figures is the number of distinct canonical triples.

Now, let's run the Python code to see what we get.

But before that, we should also consider that the polygons might be considered labeled? The problem statement does not specify. Typically, when counting "figures", the polygons are part of the figure and are not labeled. So unlabeled is appropriate.

However, let's check the count: Our code will output number of distinct canonical triples.

We'll run it.

[PYTHON] from itertools import product

solutions = [] for a in range(3, 13): for b in range(3, 13): c = 12 - a - b if c >= 3: solutions.append((a,b,c))

print("All ordered triples (a,b,c):") for sol in solutions: print(sol)

Now group by equivalence (a,b,c) ~ (a,c,b)

canonical = {} for a,b,c in solutions: # canonical form: sort b and c key = (a, min(b,c), max(b,c)) canonical[key] = canonical.get(key, 0) + 1

print("\nDistinct canonical forms:") for key in sorted(canonical.keys()): print(key)

print(f"\nNumber of distinct figures: {len(canonical)}") [/PYTHON]

Let's run this.

But note: The loops generate triples where a,b,c can be any integers >=3 summing to 12. However, we are iterating a from 3 to 12, b from 3 to 12, and compute c. That will generate duplicates? Actually, each triple (a,b,c) is generated exactly once for each a,b pair. But note that (a,b,c) and (a,c,b) are generated separately if b != c. That's fine.

Now, after grouping by canonical form, we count distinct canonical forms.

Let's see the output.

We'll run the code.

[PYTHON] from itertools import product

solutions = [] for a in range(3, 13): for b in range(3, 13): c = 12 - a - b if c >= 3: solutions.append((a,b,c))

print("All ordered triples (a,b,c):") for sol in solutions: print(sol)

Now group by equivalence (a,b,c) ~ (a,c,b)

canonical = {} for a,b,c in solutions: # canonical form: sort b and c key = (a, min(b,c), max(b,c)) canonical[key] = canonical.get(key, 0) + 1

print("\nDistinct canonical forms:") for key in sorted(canonical.keys()): print(key)

print(f"\nNumber of distinct figures: {len(canonical)}") [/PYTHON]

Now, let's execute.

But wait, we need to output only the final answer. However, we are in the middle of reasoning. We'll run the code to see the result.

Let's do it. </think>

[PYTHON] from itertools import product

solutions = [] for a in range(3, 13): for b in range(3, 13): c = 12 - a - b if c >= 3: solutions.append((a,b,c))

print("All ordered triples (a,b,c):") for sol in solutions: print(sol)

Now group by equivalence (a,b,c) ~ (a,c,b)

canonical = {} for a,b,c in solutions: # canonical form: sort b and c key = (a, min(b,c), max(b,c)) canonical[key] = canonical.get(key, 0) + 1

print("\nDistinct canonical forms:") for key in sorted(canonical.keys()): print(key)

print(f"\nNumber of distinct figures: {len(canonical)}") [/PYTHON]

[THOUGHT] The Python output shows that the distinct canonical forms are (3,3,6), (3,4,5), (4,3,5), (4,4,4), (5,3,4), (6,3,3), which are 6 in number. However, note that the problem says "Rotations of the circle that produce congruent figures are considered the same." This means we must count distinct figures up to rotation of the circle, not just up to permutation of polygons.

But careful: The polygons are determined by the number of sides (a,b,c). If we arrange the 12 points around a circle and assign consecutive blocks of points of sizes a, b, c to three polygons, the figure is determined up to rotation by the cyclic order of the three block sizes. So (a,b,c) and (b,c,a) and (c,a,b) are the same figure.

Thus we must count distinct cyclic triples (a,b,c) with a+b+c=12, a,b,c≥3, up to cyclic permutation.

Let's compute properly:

We have all ordered triples (a,b,c) with a,b,c≥3, a+b+c=12. We found 10 such ordered triples.

Now group by cyclic equivalence: (a,b,c) ~ (b,c,a) ~ (c,a,b).

Let's compute that.

[PYTHON]

Find all ordered triples

ordered = [] for a in range(3, 13): for b in range(3, 13): c = 12 - a - b if c >= 3: ordered.append((a,b,c))

print("Ordered triples:", ordered)

Group by cyclic equivalence

cyclic_groups = {} for t in ordered: # canonical cyclic form: lexicographically smallest among its rotations rotations = [t, (t[1], t[2], t[0]), (t[2], t[0], t[1])] canonical = min(rotations) cyclic_groups[canonical] = cyclic_groups.get(canonical, []) + [t]

print("\nCyclic equivalence classes:") for can, members in cyclic_groups.items(): print(f"{can}: {members}")

print(f"\nNumber of distinct cyclic classes: {len(cyclic_groups)}") [/PYTHON]

[THOUGHT] The Python output shows that there are 4 distinct cyclic equivalence classes of triples (a,b,c) with a,b,c ≥ 3 and a+b+c=12, up to cyclic permutation. These correspond to the distinct figures up to rotation of the circle.

Thus the number of distinct figures is 4.

\boxed{4}