Questão 67

**Explaining color permutations** I need to solve for the straightforward problem of 36 choose 2 divided by 3, which gives 210. The rotations at 120° and 240° need explaining, especially since a pair cannot remain invariant unless each individual piece is fixed — which isn't possible here with multiple fixed elements. By articulating that some configurations involving cycles will remain unchanged under rotation, I can clarify the equivalence class concept. I should ensure all cycles are noted, since the center placement matters.**Describing painting invariance** I'm considering that no configuration remains the same under a 120° rotation since it cycles through three squares, with none fixed in their positions. I need to explain this in terms of the center intersection vertex figure. The total combinations are 630, which divided by 3 gives me 210. My output should be concise, maybe around 200 words in Portuguese but focused on explaining the concept without overcomplicating it. I’ll organize the explanation into three paragraphs with relevant formulas.

**Explaining direct no pair invariant** I'm diving into the concept of direct no pair invariant. I see that it relates to how each cell moves in an orbit of three. So, if I think about the rotation of a pair, it can't leave the same set because it's stuck in a cycle of three. It's curious how these movements interact! I think I’ll keep this in mind for future discussions on invariants and cycles in permutations.

**Exploring triangular rotations** I'm considering the image showing two yellow rotations, possibly relating to a nuanced equivalence of rotation with 120 degrees and a board of 36 triangles. The answer seems to be 210. I need to explain and ensure that perhaps some triangles stay fixed under rotation. In geometry terms, with an order of 6 on a triangular lattice, I’m mapping how triangles position around the central vertex. The cycles and orbits also involve counting distinct colored cells under rotation for Burnside's identity and how permutations might work with fixed pairs.**Analyzing cell identities** I’m contemplating the situation where if both cells are fixed individually, then swapping them creates an action of order 2, which contradicts the scenario where cubing a transposition remains the same. So, I conclude there would be none. Alternatively, the orbit of a subset under rotation could have a size of 3, leading to 630 divided by 3, which gives me 210. It seems I also need to translate my thoughts and explain this in Portuguese!

I'm counting the ways to pick 2 of the 36 cells, which gives 630 total colorings, and I'm checking which of those stay the same under a 120° rotation. Any pair fixed by the rotation has to be made of whole orbits of size 3, unless it includes the center cell, so I'm checking whether the center cell is a single fixed point. I'm checking whether the triangle's centroid, at (3, √3), lands on a grid vertex, and it does, so no single cell sits at the center. That means every cell belongs to an orbit of size 3, so no non-identity rotation fixes a 2-cell subset. Dividing 630 by 3 gives 210 orbits.

I'm checking whether the 2-cell choices can be fixed by a 120° rotation, which would mean both cells have to be the center cell, and I'm asking whether a cell actually contains the centroid of a side-6 triangle. For n=6, the centroid lands on a grid vertex rather than inside a cell, so no cell is fixed by the rotation. That makes all orbits size 3, giving 630/3 = 210, which is answer A.

I'm working out the count for a triangle split into 36 small triangles: choosing 2 of 36 cells gives 630 colorings, and I'm applying Burnside with the 120° and 240° rotations, which fix nothing since no cell is fixed. I'm checking the orbit argument: since the centroid is a lattice vertex and 6 is divisible by 3, the 36 cells split into 12 orbits of size 3, so no pair of cells is fixed by a rotation. That leaves (630+0+0)/3 = 210.

The problem is in Portuguese and comes from the 2027 FUVEST exam. The board is a large equilateral triangle divided into 36 smaller triangles. Two cells are painted yellow. Paintings are equivalent if one can be rotated 120° around the intersection of the bisectors. The task is to count how many distinct paintings exist. I am analyzing the triangular grid structure to determine how 36 small equilateral triangles are arranged. The total number of smallest equilateral triangles equals n². For n=2, the grid contains 3 upward-pointing and 1 downward-pointing triangles, totaling 4. The side of the large triangle is divided into six equal parts, confirming 36 unit triangles. The 120° rotation is around the centroid, where the angle bisectors intersect. The problem reduces to counting orbits of pairs of distinct cells under the rotation group. The group has three elements: identity, 120°, and 240°. I will apply Burnside's lemma. The group has order three. I need to count how many colorings remain unchanged under each rotation. The identity fixes all 630 pairs of cells. For the 120° rotation, the permutation splits the 36 cells into cycles of length three. I am checking whether any cell is fixed by this rotation. The 120° rotation has order three. A pair of distinct cells cannot be invariant under it, because that would require a swap that would force the rotation to have order two. The only fixed pairs are those where both cells coincide, which is impossible since the cells must be distinct. I am reconsidering whether the 120° rotation fixes any cells. If it does, a pair of fixed cells would be preserved by the rotation. I need to visualize the board geometry more carefully. I am re-examining the subdivision. The 36 unit triangles may not all be upward-pointing. I need to verify the exact count of upward and downward triangles in a standard triangular grid. The total number of smallest equilateral triangles is n². The orientations alternate, but the exact count of upward and downward triangles is not needed for this problem. I am checking whether the 120° rotation fixes any small triangles. For n=6, the centroid is a point, not the center of a small triangle, so no cell should be fixed by this rotation. The 36 cells form 12 orbits of three triangles each. I am checking whether any small triangle is centered exactly at the centroid, which would make it fixed by the 120° rotation. The image shows two equivalent paintings. One has yellow cells in the lower left and upper middle. The other has them in the upper right and lower right, rotated by 120 degrees. I am calculating for which values of n the centroid coincides with the center of a unit triangle. I also notice that 2027 is a future year, so this problem may be hypothetical rather than from an actual past exam. I am now estimating the side length from the image description. The figure appears to show horizontal lines and blue points at vertices, suggesting the large triangle might be divided into four rows rather than six. The side length is confirmed as six, since n=4 would yield only sixteen unit triangles. I am now reasoning purely through group theory rather than visualizing the board. The centroid lies at two-thirds of each median. I placed the large triangle with vertices at (0,0), (6,0), and (3, 3√3) for n=6. The centroid is at (3, √3). I am checking whether any unit triangle has its center at this point. I derived the centroid of a unit upward-pointing triangle at (x + 0.5, y + √3/6). The rotation center is at (3, √3). These coordinates do not match for any integer grid point, so no small triangle is fixed by the 120° rotation. The x-coordinate of the rotation center matches a possible half-integer base position for an upward-pointing triangle. However, the required y-coordinate does not equal any valid row height, since it is not a multiple of √3/2. Therefore no unit triangle is centered exactly at the centroid. Down-pointing unit triangles are also ruled out. Their centroids would require a row height of (7/6)√3, which is not a multiple of √3/2. No small triangle is fixed by the 120° rotation. A triangle mapped to itself must have its centroid fixed at the rotation center. No unit triangle satisfies this condition, so the 120° rotation permutes all 36 cells in cycles of length three. A pair of distinct cells cannot be invariant under this permutation, because a two-element subset would require a swap rather than a three-cycle. A subset is invariant under a rotation if and only if it is a union of cycles. Since each cycle has length three, the size must be a multiple of three. A pair of cells has size two, so no two-element subset is fixed by either the 120° or 240° rotation. Applying Burnside's lemma gives 210 distinct paintings, which is option (A). I am now questioning whether every orbit actually has size three. The problem text specifies equivalence via a 120° rotation, not the full group. I am re-examining whether this narrower definition changes the count. The wording "por uma rotação de 120°" is ambiguous. It could mean only direct 120° rotations, or the full equivalence relation generated by that rotation. The images likely show two paintings related by exactly 120°. This distinction affects whether the count should use Burnside's lemma on the full group or a narrower orbit calculation. The equivalence relation is asymmetric. If painting B is obtained by rotating A exactly 120°, then A cannot be obtained by rotating B by 120°; that would require a 240° rotation instead. This means the relation does not satisfy symmetry. The 120° rotation alone does not define a symmetric or transitive equivalence relation. In contest problems, the intended interpretation is usually the full group of rotations, which yields 210 distinct paintings. The relation is not reflexive, since no painting is obtained from itself by a 120° rotation. A three-cycle of paintings P, rot(P), and rot²(P) shows that P is equivalent to rot²(P), because rotating rot²(P) by 120° yields P. The symmetry of this equivalence relation is now being questioned. The equivalence relation is symmetric. The phrase "uma deverá ser obtida a partir da outra" means either painting can be obtained from the other by a 120° rotation. In a three-cycle, P and rot(P) are equivalent because each is the 120° rotation of the other. The symmetric phrasing implies the equivalence relation includes both directions. The relation is not reflexive because no painting equals itself under a 120° rotation. The intended interpretation is likely the full orbit of rotations, grouping paintings that can be transformed into one another. The total number of unit triangles is confirmed as 36. There are 21 upward-pointing and 15 downward-pointing triangles, which sum to 36 and match the standard formula n². Option C (420) would arise from treating the two painted houses as ordered pairs, but the problem specifies unordered sets of two identical yellow cells. The correct count remains 210. The relation is not an equivalence relation because it is not reflexive. A three-cycle of paintings shows transitivity, but since no painting equals itself under a 120° rotation, the relation fails the reflexivity requirement. The phrase "consideradas equivalentes" is interpreted as the full orbit generated by 120° rotations, since a relation that excludes reflexive pairs is not a true equivalence relation. This matches standard Brazilian contest problems, where the count is typically divided by three. The relation forms a graph of disjoint triangles. The maximum number of paintings with no two equivalent is three, since each orbit contains exactly three elements. The phrase "rotação de 120°" does not specify direction, so both clockwise and counterclockwise rotations may be allowed. This would generate a larger orbit than the three-cycle of direct 120° rotations alone. The symmetric phrasing "uma deverá ser obtida a partir da outra" covers both clockwise and counterclockwise 120° rotations. The orbit of three paintings is fully connected, so all three are mutually equivalent. This confirms the answer remains 210. A smaller example with n=3 confirms the orbit-counting method: 9 cells yield 36 pairs, and dividing by 3 gives 12 distinct equivalence classes. This validates the approach. The thought then questions why the problem might present larger numbers, suggesting the intended answer could differ. The Burnside formula may be too advanced for high school students. A simpler argument works: every painting belongs to a group of exactly three mutually equivalent rotations, so the total count is 210. The images likely show a concrete example of two equivalent paintings to illustrate the rotation. The triangular grid is invariant under 120° rotation because the sides cycle and parallel lines map to other directions. The coordinate check confirms lattice points map to lattice points under this rotation. The rotation permutation has order three, so its cycle lengths must divide three. A two-element subset cannot be invariant under this permutation because no two-cycle exists. This confirms that every orbit contains exactly three distinct paintings. The center point (3,√3) is confirmed as a grid vertex where six small triangles meet. The 120° rotation cycles these surrounding triangles, so none of them is fixed.

We need solve. Need understand problem. Portuguese. Board equilateral triangle subdivided into 36 equilateral triangles. Two cells painted yellow. Equivalence if one obtained from other by rotation 120° about intersection of bisectors (center). Count non-equivalent paintings. Need choose among. We need parse grid size. Equilateral triangle subdivided into 36 small equilateral triangles. Standard side length n subdivided has n^2 small equilateral triangles? Wait number of small equilateral triangles of unit side in triangular grid side n equals n^2? Let's verify n=1 ->1, n=2 ->4 (3 up +1 down), n=3 ->9 (6 up +3 down), n=4 ->16, n=6 ->36. Yes n=6. The images show side length 6? Count rows likely 6. We need color exactly two of 36 small triangles yellow. Equivalence under rotations by 120° around center (order 3). Need number of orbits of 2-element subsets under C3 action. Since rotations by 120 and 240; no reflection. Burnside: (1/3)(total subsets + fixed by 120 + fixed by 240? Actually rotations 120 and 240 have same fixed sets? Need compute). Number of orbits = (1/3)(N + F + F?) where F = number of 2-subsets invariant under 120 rotation = subsets with elements fixed individually (since set size 2, cannot swap pairs unless cycles length? rotation of cells may fix some cells, and cycles of length 3; a 2-subset invariant under 120 must consist of cells each fixed by rotation, because any nonfixed orbit size 3 requires 3 elements). Rotation 120 of triangular board: cells? Center at centroid. There is one small triangle at center? Need determine under 120 rotation about centroid: for n=6 triangular grid, are there cells fixed? A small equilateral triangle can be mapped to itself under 120 rotation only if its center is at rotation center? Wait a small equilateral triangle centered at origin can be invariant under 120 rotation if its orientation? The triangular grid has up/down triangles. Under 120 rotation, an upward small triangle remains upward? Equilateral triangle rotated by 120 about its own center maps to itself. A downward triangle rotated by 120 maps to itself too (as set). So if a small triangle is centered at board center, it fixed. For n=6, center of big triangle coincides with center of one small triangle? Let's examine. The centroid of triangular grid side 6 may be a vertex intersection? In images, red bisectors meet at a point. There is a small upward yellow triangle right above center? Wait images show red medians meet at a vertex? Let's inspect: big triangle rows. The center of an equilateral triangle is at distance from base. For n=6, subdivision has horizontal lines at integer heights 0..6. Centroid height = altitude/3? From base? Actually centroid at 1/3 altitude from base. If altitude measured from base, center y = h/3 = (√3/2 *6)/3 = √3. So horizontal line y=√3, which is at row line 2 from base? Unit small triangles have height √3/2. y=√3 = 2*(√3/2). So center lies on a horizontal grid line. In triangulation, center is intersection of medians, maybe vertices. In images red lines meet at a vertex. Looking: red from top to base center, and from lower left vertex to side midpoint, etc. The three red meet at a point in interior, appears vertex of grid, not center of a small triangle. Let's count: big side 6. Center is at row height 2 units from base, at midpoint horizontally. This is a grid vertex (since row y=2, horizontal positions). So no cell centered at center. A 120 rotation may fix no small triangle? Wait could a small triangle surrounding center? Center is a vertex shared by 6 triangles; rotation cycles them. No cell invariant because area small triangle centered not at center. For n multiple of 3? Need be careful. The rotation center might be center of a small downward/up triangle depending n mod 3. For n=6, center is a vertex, so no fixed cells. Then F=0, orbits = N/3 = C(36,2)/3 = 630/3=210. That's option A. But likely too easy? Options include 630 (if ignoring rotation?) and 210. But question might have nontrivial because rotations of 120 can map pair to itself swapping two cells if rotation by? For order 3, a 2-subset cannot be invariant by swapping two under a 120 rotation (permutation has order 3, cannot have transposition). unless pair includes one fixed and one fixed. So if no fixed cells, no fixed pair, orbits 210. Option A. But wait images: First configuration: two yellow triangles. Second is rotation 120? Need count non-equivalent. Maybe rotations not only 0,120,240, but "uma rotação de 120°" could mean one rotation to be equivalent, maybe also repeated? It says "por uma rotação de 120°" meaning obtained by a 120° rotation; but equivalence is transitive? If A equiv B via +120, B equiv C via +120, so A,C via +240. So group. But options 315,420 etc suggest maybe they counted ordered pairs? Let's verify. Total unordered pairs C(36,2)=630. Divide by 3 =210. Options include 210. Why other numbers? Maybe if there are fixed cells, Burnside yields (630 + 2f choose?)/3. If f=?? For n=6, if rotation fixes 3 or etc, result maybe 210 + something. Let's recalc possible f. Number of fixed cells under 120 rotation on triangular board side n=6? Center is vertex, none fixed. But wait small triangles at corner? Under rotation, top corner cell maps to bottom left etc. no fixed. So f=0. Thus 210. But then why option 315? If erroneously include rotations by 0 and 120 only? (630+F)/2=315 if F=0. Option B. They include. Option C 420? For C(36,2)=630, /? 630*2/3=420. Maybe applying rotation by 120 but not distinguish direction? Hmm. Let's not jump. Need fully parse image: "Duas casas de um tabuleiro no formato de um triângulo equilátero, subdividido em 36 triângulos equiláteros, são pintadas de amarelo. Para que duas pinturas sejam consideradas equivalentes, uma deverá ser obtida a partir da outra por uma rotação de 120° em relação ao encontro das bissetrizes, conforme ilustrado nas imagens a seguir:" Then images. "Considerando as imagens apresentadas, quantas pinturas não equivalentes existem?" Options. Could "36 triângulos equiláteros" mean a triangular board subdivided into 36 **congruent** equilateral triangles? Standard side 6 gives 36 small triangles. Yes. Need maybe count of "pinturas" could mean choose two **casas** (triangles). The board has 36 cells. Two cells painted. Equivalent under rotation 120. We did. But wait: Are triangles all considered "casas"? Upwards and downwards; rotation by 120 about center maps orientation? A 120 rotation preserves upward/downward? It rotates by 120; an equilateral triangle pointing up after 120° remains? An equilateral triangle pointing up rotated 120° around its own center remains up; around external center, orientation vector rotates by 120. Upward triangle orientation vector angle 90° from base; rotated 120 becomes 210°, which points down-left? But the triangular tiling's upward triangle vertices: orientation can be classified as "up" (one vertex up) and "down" (one vertex down). Rotation by 120 of an up-pointing small triangle yields? Let's test big triangle has 3-fold; if one small up triangle rotates by 120 around centroid, its orientation (a vertex direction) rotates from up to down-left, but in the triangular lattice there are only up/down relative to page? Actually an equilateral triangle can have 6 orientations as cells? In triangular tiling, all small triangles are either up (apex upward) or down (apex downward), only two. Rotating an up triangle by 120° about an external point could yield a triangle with apex? Example upright equilateral triangle: vertices at angles 90,210,330. Rotate vertices by 120 -> angles 210,330,90; same set of vertices orientation? Wait as a shape centered at its center, rotating by 120 maps to itself because equilateral. But orientation vector apex rotates 120, yet set of vertices same. However relative to page, an upright triangle rotated 120 around its own center remains upright (same set). That seems contradictory: a triangle pointing up at center; vertices at 90 (top), 210 (bottom-left), 330 (bottom-right). Rotate by 120: top vertex moves to 210 (bottom-left), bottom-left moves to 330, bottom-right moves to 90. The set unchanged. So orientation unchanged as set. For a cell not centered at rotation origin, rotating its vertices by 120 gives a translated triangle whose vertices relative to its center are same as original set, so an up cell maps to up cell? But in triangular tiling there are also down cells. Let's use equilateral triangular cell pointing up, vertices (0,0),(1,0),(0.5,√3/2). Rotate by 120 around origin: vertices (0,0),(-0.5,-√3/2),(-? compute 0.5,0.866 rotated -> cos120*0.5 - sin120*.866 = -0.25-0.75=-1, y=0.433-0.433=0?) Wait result (-1,0), plus. Coordinates: (0,0),(-0.5,-0.866),(-1,0). This triangle has apex? Vertices form downward? It is a down triangle maybe? Hmm the orientation set relative to translation: centroid (-0.5,-0.288); vertex differences: (0.5,0.288), (0,-0.577), (-0.5,0.288). Apex downward. So rotating about external point changes up to down? Let's re-evaluate. Rotating an equilateral triangle around its own center by 120 gives same set; but that's an orientation-preserving symmetry mapping vertices cyclically, not translating. Under rotation about arbitrary origin, the image of a cell may have center rotated; the cell shape at its center is rotated by 120 around its own center too? Decomposition: rotation about O = translation of center + rotation about new center by same angle. Thus image cell shape is original rotated about its own center by 120, which is same set. Wait if an up cell rotated about its center by 120 equals itself (set upright). So image should be upright. My coordinate rotation about origin of a cell including origin as a vertex: the cell center is (0.5,0.288). Rotation about origin by 120: center -> (-0.5,-0.288)? Let's compute center (0.5,0.28866) rotate: x'=cos*0.5 - sin*0.28866= -0.25-0.25=-0.5, y'=sin*0.5+cos*0.28866=0.433-0.144=0.2887. Not (-0.5,-0.288) I guessed. Vertex set: (0,0); (1,0)->(-0.5,-0.866); (0.5,0.866)->(-1,0). These relative to new center (-0.5,0.2887): (0.5,-0.2887), (0,-1.1547), (-0.5,-0.2887). That's an upward triangle? Apex? The unique top? Points at y -0.2887,-1.1547,-0.2887; apex bottom. Hmm if center y? Let's identify: This has horizontal base at y=0? Vertices transformed: (0,0),(-0.5,-0.866),(-1,0). That's downward-pointing triangle (apex bottom? Actually top base from (0,0) to (-1,0), third below: apex down). It is down. But original cell was up with base from (0,0) to (1,0), apex up. Why decomposition fails? Because rotating about its center by 120 maps the set to itself, yes, but the "cell" as an unlabeled set after translation? Wait rotated image of all points in cell = {R(O + x): x in cell}. If decompose around cell center c: R(c+y)=R(c)+R(y). R(y) rotates vector y by 120. For an equilateral triangle centered at c, the set of y is invariant under 120 rotation. Therefore R(cell)=R(c)+cell. So shape remains up at center R(c). My coordinate map should yield an up triangle centered at R(c). Did I choose vertices wrong? Original up cell with vertices (0,0),(1,0),(0.5,0.866). Its center is average (0.5,0.2887). The set y = vertices - center: (-0.5,-0.2887),(0.5,-0.2887),(0,0.5774). Rotate y by 120: (-0.5,-0.2887)->(0.5,-0.2887)? Let's compute: cos=-0.5, sin=0.866. For (-0.5,-0.2887): x'=0.25+0.25=0.5, y'=-0.433+0.144=-0.2887. yes maps to (0.5,-0.2887). The set invariant. Good. Then R(c)=(-0.5,0.2887). Add y set: vertices = (-1,0),(0,0),(-0.5,0.866). That's an upward triangle with base from (-1,0) to (0,0), apex (-0.5,0.866). But I computed vertex (0.5,0.866) rotated to (-1,0) yes; (1,0) rotated ->(-0.5,-0.866), not in set? Wait R(1,0)=(-0.5,-0.866), that's not among new triangle vertices. Hmm original vertices were (0,0),(1,0),(0.5,0.866). Rotate all: (0,0),(-0.5,-0.866),(-1,0). That's downward shape. But decomposition says R(c)+set y = vertices? Let's compute R(c) for c=(0.5,0.2887): R(c)=(-0.5,0.2887). Add rotated y set: y vertices: (-0.5,-0.2887)->(0.5,-0.2887); (0.5,-0.2887)->(-0.5,-0.2887)? compute: x'= -0.25+0.25=0; y'=0.433+0.144=0.577? Let's carefully. Rotation matrix 120: (x,y)->(-1/2 x -√3/2 y, √3/2 x -1/2 y). For y1=(-0.5,-0.288675): x'=0.25+0.25=0.5; y'=-0.433+0.144=-0.288675. For y2=(0.5,-0.288675): x'=-0.25+0.25=0; y'=0.433+0.144=0.57735. For y3=(0,0.57735): x'=0-0.5=-0.5; y'=0-0.288675=-0.288675. Set is (0.5,-0.2887),(0,0.5774),(-0.5,-0.2887), not original y set. Ah because rotating y by 120 of an **upward triangle centered at origin** gives? Wait original y set vertices: top (0,0.577), bottom-left(-0.5,-0.289), bottom-right(0.5,-0.289). Rotate each by 120: top -> (-0.5,-0.289), bottom-left -> (0.5,-0.289), bottom-right -> (0,0.577). This is same set! My computed for y2? y2=(0.5,-0.2887) rotate ->? x' = -0.25 +0.25=0; y'=0.433+0.144=0.577 yes top. y3=(0,0.577) rotate ->(-0.5,-0.289). y1=(-0.5,-0.289) rotate ->(0.5,-0.289). Set same. Good. Then R(c)+set y = (-0.5,0.2887)+ {(0,0.577),(-0.5,-0.289),(0.5,-0.289)} = (-0.5,0.866),(-1,0),(0,0). That is up. But direct rotate original vertices gave (0,0),(-0.5,-0.866),(-1,0). Are these different? Because R(original vertex) = R(c+y) = R(c)+R(y). Since R(y) set = y set, result = R(c)+y set. Direct compute R(0.5,0.866): x'=-0.25-0.75=-1, y'=0.433-0.433=0. So (-1,0). R(1,0)=(-0.5,-0.866). R(0,0)=(0,0). Those are (0,0),(-1,0),(-0.5,-0.866), which is down. But according to decomposition, R(c)+y set includes (-0.5,0.2887)+(0,0.577)=(-0.5,0.866), not (-0.5,-0.866). Which is correct? Wait c=(average) = (0.5,0.2887). But is that the center of equilateral triangle? Average is (0.5,0.2887) yes. Rotate c: x'=-0.25-0.25=-0.5, y'=0.433-0.144=0.2887. Add original y set vertices relative to center: Original vertex (0.5,0.866) relative to c: (0,0.5774). That's top. R(c)+ relative top (not rotated) = (-0.5,0.866). But actual rotated top vertex R(0.5,0.866)=(-1,0). Why R(c)+ (relative rotated) differs? Wait R(c+y) = R(c)+R(y). R(y for top) = top rotated = (-0.5,-0.2887). Then R(c)+R(y)=(-1,0). Ah I used y set invariant, but y set is invariant as set, yes includes element (0,0.5774) but R(top)=bottom-left. Adding R(c)+bottom-left = (-0.5,0.2887)+(-0.5,-0.2887)=(-1,0). The set R(c)+R(S) = R(c)+S since S invariant. Thus R(c)+S should equal original rotated vertices. But R(c)+S = {(-0.5,0.866),(-1,0),(0,0)}. Contains (-0.5,0.866), but direct doesn't. The error: S (vertices relative to center) for original up triangle is not invariant under 120? Let's list S: original vertices (0,0),(1,0),(0.5,0.866), center c=(0.5,0.2887). Relatives: (-0.5,-0.2887),(0.5,-0.2887),(0,0.5774). This set is an **upward triangle centered at origin**? Actually an upward triangle centered at its centroid has one vertex at top and base bottom. Yes. Rotation by 120 of an equilateral triangle **set** about its centroid maps vertices cyclically: top (0,0.577) -> (-0.5,-0.2887) (bottom-left). Yes invariant. Let's rotate S: top -> (-0.5,-0.2887) in S; bottom-left (-0.5,-0.2887) rotate ->? x'=0.25+0.25=0.5, y'=-0.433+0.144=-0.2887, in S; bottom-right -> (0,0.577). invariant. So R(c)+S = S shifted to R(c). This set is same shape orientation as S: upward. But direct rotating all points should also map the triangle to a triangle of same orientation (because rotation preserves orientation). So why direct vertices gave down? Recompute direct R(0.5,0.866): x' = -0.5*0.5 -0.866*0.866 = -0.25-0.75=-1.0; y'=0.866*0.5 + (-0.5)*0.866 =0.433-0.433=0. So (-1,0). R(1,0)=(-0.5,0.866)? y'=0.866*1 -0 =0.866, not -0.866! Formula above had y' = √3/2 x -1/2 y, yes for (1,0) y'=0.866. I earlier wrote -0.866 (using angle 240?). So rotated vertices: (0,0),(-0.5,0.866),(-1,0) = up. Good. Rotation by 120 preserves up/down? It maps up to up, down to down? But a triangle pointing up under 120 around its center remains up because set invariant; yes. Under rotation around external center, orientation remains up. So up cells map to up cells, down to down? Wait an equilateral triangle pointing down rotated 120 about its center also set invariant (down). So yes orientation (up/down) preserved under rotations by multiples 120? Actually a 60° rotation would flip up/down; 120 preserves. But big triangular board itself: rotating by 120 maps the set of up cells to up cells? The board has 21 up cells and 15 down cells for n=6? Formula: n(n+1)/2 up =21; n(n-1)/2 down=15. 21 not divisible by 3? 21 divisible by3 yes; 15 divisible by3. Good. No issue. Now about fixed cells: A small triangle (up or down) fixed as set under 120 rotation about center if its center is at center? Because rotation by 120 around its own center preserves cell. If its center not at O, can a small triangle map to itself? A rotation about O has no other fixed point; for set invariant, its centroid maps to centroid, so centroid=O. Thus only cell centered at O could be fixed. For n=6, O is vertex; no cell centered. For n multiple of? So f=0. However, wait big triangle side length 6; centroid at height h/3 from base = (n√3/2)/3 = n√3/6. For n=6 =√3. Grid vertices y coordinates are k√3/2, k=0..6 if base horizontal y=0. y=√3 = 2*(√3/2), indeed vertex row. x coordinate center: base midpoint? At y=2, grid vertices spacing 1? Big triangle at height 2 has width? Need coordinate. Base length 6, apex at x=3,y=6h? Let's set base vertices (0,0),(6,0), apex (3,3√3). Centroid (3,√3). At y=√3 (row k=2 from base), vertices x =? lines slope; x positions: from left edge x = y/(2√3?) Wait horizontal row has vertices at x = ? For y=2*(√3/2)=√3, row from base is k=2, segment endpoints left x = k/2? Let's set big base from (0,0) to (6,0), apex (3,3√3). Rows horizontal at y=k√3/2. Left boundary x = k/2, right = 6 - k/2. For k=2: left=1, right=5. Vertices at x=1,2,3,4,5? Center x=3 yes vertex. So center is vertex, not cell centroid. No fixed. Thus answer 210. But let's consider if "rotação de 120°" could include only clockwise, and equivalence defined directed? "uma deverá ser obtida a partir da outra por uma rotação de 120°" If only one direction, the relation is not symmetric? But "equivalentes" implies if A can become B by +120, then B can become A by -120 (=240), which is also a rotation of 120? Hmm -120 is not 120? In Portuguese "uma rotação de 120°" might mean angle magnitude 120, including both directions? Usually rotation of 120° clockwise or counterclockwise? They illustrate one arrow? Need check. The two images: first shows two yellows (one central above, one lower left). Rotating first by 120° clockwise or counter? Let's map. Center is vertex. First yellow: an upward triangle above center? Actually central yellow just above center? And lower left downward? Rotating 120 clockwise: top central moves lower right? Second image: central downward? Wait in second, one yellow near top center (downward?) and one lower right (upward?); Need see. Let's identify image coordinates with side n=6. Need understand rotation direction. First image: There is a yellow triangle at center top? Let's label rows. Big triangle apex top. Center intersection red at row from top? In first, yellow central is just above intersection? It appears a small upward triangle whose bottom vertex is the center intersection (red lines meet at its bottom vertex). There is another yellow lower-left: a downward triangle on row 3? In second, yellow near top center is a downward triangle? It lies above left? Actually second image: yellow central? It seems just above intersection too, but pointing down? Wait rotation by 120 should rotate entire pattern; the central cell adjacent to the center intersection cycles among the three triangles around the center: there are six small triangles around the center vertex (center is a vertex). Rotating by 120 cycles them in two orbits of 3: upward? Need. If the center is a vertex, small triangles meeting at center: 6 around. Rotating 120 gives two orbits of 3 triangles around center. In first, yellow central might be one of these; second after rotation is another. The other yellow likewise. So no fixed pair. Total arrangements = choose 2 cells =630. Since group of 3 acts freely on pairs? Need if any pair invariant under 120? Since no cell fixed, but could a pair be mapped to itself by 120 by swapping the two cells? As said permutation of two elements under a 120 rotation? Wait a 120 rotation as a permutation of cells has cycles length 3 or 1. Restricted to a 2-element set invariant as a set, both cells must be fixed or form a 2-cycle. Cycles length 2 impossible because 3-cycle permutation. But note on a set of two labeled cells, applying 120 twice can swap? The group element of order 3 cannot induce transposition on a 2-set. So invariant pair only fixed cells. f=0 => no. Action free on ordered? For unordered pairs, stabilizer trivial. Orbit size 3. So 210. Could there be rotations 0,120,240 and pairs where configuration invariant under 240 but not 120? If invariant under 240 (order 3), also under inverse? 240 generates group, so invariant under 120 as 240^2. Same. Thus A. Need provide explanation in Portuguese. But perhaps need address possible alternative: If one uses Burnside. We can explain board has 36 casas; total ways choose two = C36,2=630. Rotation by 120 around center; for side 6, the center is a vertex of grid, so no small triangle is fixed by rotation. Therefore no painting of two cells remains identical after 120 rotation (because a fixed painting would require the two cells individually fixed or swapped in 2-cycle, impossible). Thus every equivalence class has exactly 3 distinct paintings. Count=630/3=210. Need ensure "center is a vertex not center of cell" verified maybe in image red lines meet at a vertex. The image shows red bisectors meet at a vertex. So rotation has no fixed "casa". We can also mention no configuration fixed under 120. Thus. Let's craft final concise with calculations. Include answer line exactly. Final.

**Exploring Equivalence Transformations** I'm currently delving into the symmetry operations of an equilateral triangular grid. My focus is on defining equivalence based on 120-degree rotations. I'm considering how these rotations permute the 36 smaller triangles and how that impacts the coloring of two specific cells. **Calculating Non-Equivalent Colorings** I've been working through the combinatorial aspect, specifically applying Burnside's Lemma or direct combinatorial reasoning. My current focus is on determining the total number of ways to choose two smaller triangles from the 36, which I've calculated as 630. This sets the stage for applying the group action. **Applying Group Theory Formulas** I'm now concretely applying Burnside's Lemma. I've established the identity element fixes all 630 possible pairings. My current focus is on identifying the orbits of the 36 smaller triangles under 120 and 240-degree rotations to determine the fixed points for these group elements. **Analyzing Rotational Orbits** I've analyzed the center of rotation and found it coincides with a vertex of some smaller triangles. This is crucial for understanding how 120 and 240-degree rotations permute the 36 smaller triangles, which will directly inform the number of fixed colorings. **Pinpointing Rotational Center** I've determined the rotational center lies on a grid vertex, specifically where six small triangles meet. This insight is key to accurately mapping how 120 and 240-degree rotations permute the 36 smaller triangles and will directly aid in counting fixed colorings. **Concluding Fixed Points** I've confirmed that for both 120 and 240-degree rotations, zero pairs of triangles remain fixed. This is because the orbits of individual triangles under these rotations are of size 3, and a pair cannot be formed from such orbits. **Concluding Burnside's Lemma** I've successfully applied Burnside's Lemma, combining my previous findings. With 630 total colorings and zero fixed points for 120 and 240-degree rotations, the calculation yields 210 unique non-equivalent colorings. This confirms the result derived from the formula. **Confirming Orbit Sizes** I'm now confirming that each equivalence class, or orbit, must consist of either 1 or 3 triangles. I'm exploring whether an orbit can have a size of 1. For a coloring to be invariant under a 120-degree rotation, the set of two yellow triangles must be fixed by that rotation. Given the rotational center is on a grid vertex, the 36 triangles are partitioned into orbits of size 3 under 120-degree rotations. **Validating Orbit Sizes** I'm solidifying the understanding that orbits under $\mathbb{Z}_3$ must be size 1 or 3. I've proven that an orbit of size 2 for the coloring is impossible, as it would imply a contradiction where $R^2(T_1) = T_1$ and ultimately $T_1 = T_2$. This confirms that all sets of two yellow triangles are either fixed or permuted in sets of three.

**Considering rotational symmetry** I'm currently working through the problem of counting distinct colorings. My focus is on how the rotational symmetry of the larger triangle affects the equivalence of different ways to color two of the smaller cells. I'm exploring the orbits generated by the $120^\circ$ rotation. **Analyzing rotation's impact** I'm currently calculating the total colorings before considering symmetry, which is 630. My focus has shifted to the group of rotations ($0^\circ, 120^\circ, 240^\circ$) and how they affect invariant sets. I've noted that the center of the large triangle is a vertex, meaning no single small triangle is invariant under these rotations. **Applying rotational invariance** I've realized that since I'm only coloring two triangles, configurations cannot be invariant under $120^\circ$ or $240^\circ$ rotations. This means every orbit under the rotation group has size 3, leading to 210 distinct colorings. However, a re-reading of the problem and illustration suggests a more specific interpretation of rotational equivalence. **Clarifying rotational axes** I'm reconsidering how the rotations actually map the two chosen triangles. It seems my previous interpretation of which triangles are being rotated was too general. I need to precisely identify the initial pair and then trace their positions under the $120^\circ$ and $240^\circ$ rotations. **Tracking yellow triangle rotations** I'm now carefully examining how the specific yellow triangles are oriented. It appears my previous thinking about which red ray bisects which rotated yellow triangle was inaccurate. I need to meticulously follow the path of the initial yellow triangle under $120^\circ$ rotation to see how it aligns with the red rays. **Mapping rotated yellow triangles** I'm now carefully tracing the specific positions of the yellow triangles after rotation. It seems I was misinterpreting the intersection points with the red rays. I'm meticulously mapping how the initial yellow triangle's vertices align with the red rays under a $120^\circ$ rotation. **Tracing ray intersections** I'm currently verifying the precise locations of the yellow triangles relative to the fixed red rays. My current analysis suggests the yellow triangle's position is defined by its angle relative to the ray, and I'm using coordinates to confirm whether a rotated triangle will align with any of the existing red rays. **Calculating barycentric coordinates** I'm now using barycentric coordinates $(a, b, c)$ to uniquely identify each small triangle based on its distance from each side of the larger triangle. This approach allows me to precisely track the transformations of specific triangles under rotation. **Tracking triangle transformations** I'm now leveraging barycentric coordinates to precisely track how each small triangle transforms under rotation. This geometric approach is proving crucial for accurately mapping their positions and ensuring I correctly account for their final orientation and placement. **Mapping barycentric permutations** I'm verifying how barycentric coordinates permute under rotation. A $120^\circ$ rotation cyclically shifts $(a, b, c)$ to $(b, c, a)$, preserving triangle orientation (UP/DOWN). This confirms UP triangles map to UP, and DOWN to DOWN, under rotation. **Verifying yellow triangle coordinates** I'm now meticulously verifying the barycentric coordinates of the yellow triangles in both images. My aim is to confirm their precise locations and ensure my coordinate assignments accurately reflect their positions relative to the red rays, which is crucial for tracking transformations. **Confirming coordinate mappings** I've confirmed that a $120^\circ$ rotation permutes barycentric coordinates as $(a, b, c) \mapsto (c, a, b)$. This transformation correctly maps the coordinates of the yellow triangles between the left and right images, validating the rotational equivalence. **Confirming invariant sets** I'm double-checking that no single triangle is invariant under $120^\circ$ rotation, both for UP and DOWN types, based on barycentric coordinate sums. This confirms that all orbits under the rotation group have size 3.