Give every object a complete signature—size, orientation, position, vertex set, first split, or side total—so no shape is missed and none is counted twice.
A drawn figure can be large or small, upright or tilted, and placed in many locations. A tiling can split at a joint or cross it. A square can be assembled from different segment sets.
Choose a signature that ignores only the differences the problem says do not matter.
Only squares whose sides follow drawn grid lines count here. For side length s, the top-left corner has (3-s+1)(6-s+1) positions.
Fix the square size first, then count its possible top-left positions.
This diagram includes both the five-pointed star and its outer pentagon boundary. Only triangles whose three entire sides follow drawn straight segments count. Intersections may be vertices. Before viewing every triangle, choose one and trace its three straight sides. Count the small regions it contains, including the central pentagon when present. Record its three vertex labels so the same triangle is not counted again from another corner. The highlighted family below now names its component case.
The worked example groups triangles by how many visible triangular regions—and whether the central pentagon—they contain. The enumeration check independently identifies every triangle by its unordered set of three vertices.
Trace all three sides along drawn segments and use the vertex set to avoid recounting.
| Orientation | Scale | Count |
|---|
Every cell contains both diagonals. Square sides must follow drawn straight segments, including diagonals joined across cells. In the ledger, upright scale means side length in unit cells; tilted scale means the horizontal diagonal length in unit cells. A square is stored by its four vertices. Sorting those vertices gives a canonical key, so the same square cannot be generated four times from four different starting corners.
Separate upright and tilted shapes before subdividing by size.
Count only rhombi formed by two smallest triangles sharing one full edge. For an outer side of 6 unit edges, one orientation has 1+2+3+4+5=15 positions. Rotation gives three orientations, totaling 45.
Choose any three of the twelve grid points as vertices; straight triangle sides may cross cell interiors. Each small cell is a unit square. Count unordered vertex sets with area exactly 3 square units. The two horizontal full-width bases give four triangles each; the two vertical full-height bases give three each. Four corner triangles occur in both lists, so 8+6−4=10. The complete three-point search checks that these lists contain all qualifying triangles.
For a lattice triangle use base × perpendicular height ÷ 2; do not measure pixels.
Choose three of the six labeled vertices and join them; chords are allowed. Count triangles having at least one full side in common with the hexagon boundary, not merely touching a vertex.
Here all three triangle sides must follow the drawn straight segments. The interior intersections are also allowed vertices.
Choose vertices or a forced first piece, then reject configurations that do not close the required shape.
Draw a horizontal dividing line above the bottom 2 × 2 block. Either no domino crosses it, or both columns have a crossing domino. One crossing alone would leave an odd number of cells in the bottom block.
Let T(n) count a 2 × n rectangle. At its left edge, use one vertical domino, leaving T(n−1), or two horizontal dominoes, leaving T(n−2). Thus T(n)=T(n−1)+T(n−2). Start with T(0)=1 (nothing left to fill) and T(1)=1.
| n | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| T(n) | 1 | 1 | 2 | 3 | 5 | 8 | 13 |
No crossing: 13 × 2 = 26. Two crossings: the bottom row is forced horizontal, and the two top-right cells are also forced horizontal. Only a 2 × 4 rectangle remains: 5 ways. Total: 26 + 5 = 31.
For the first-cell split, horizontal first gives 2 × T(4) + T(2) = 12; vertical first gives 2 × T(5) + T(3) = 19. In each expression, the first term has no crossing and the second has two crossings. These disjoint cases again give 31.
13 tilings of a 2×6 rectangle × 2 tilings of a 2×2 square.
Five remaining coverings each use two vertical dominoes across the marked joint.
The cases are disjoint and exhaustive: every tiling either crosses the joint or it does not. Count placements on this fixed board, without merging rotations or reflections. Each domino covers two edge-adjacent cells; equal numbers label its two cells.
The joint has zero or two crossing dominoes. With two, three domino placements are forced.
Before viewing the assemblies, choose a side length and a set of segments. Separate the segments into four groups with that same total. Record the selected set only once: rearranging the same segments within a side does not create a new method under this activity's convention.
There is one rigid segment of each integer length from 1 through 9. Select segments without cutting or bending them, and use each selected segment exactly once, end-to-end along one of four equal sides. Different orders along the same side do not create new methods; a method is determined by its selected set.
The runtime search checks all 511 nonempty segment subsets and tries to partition each qualifying subset into four equal-sum groups.
Fix the side total, then split your selected segment set into four groups with that total.
Correct all eight answers: 32; 35; 46; 45; 10; 18; 20; 31. Count grid squares by side length, X-grid squares by orientation and scale, and rhombi by the direction of the shared small-triangle edge. Count placements separately, even when shapes are congruent.
Answers: 10; 5; 26; 5; 9. The 2×2 grid square has 2×5 positions. The pentagram table has five families. Tilted X-grid counts total 26; joint-crossing tilings total 5. Nine different segment sets can be partitioned into four equal sides.
This certifies that the learner counted figures using complete signatures, exhaustive geometry searches, tiling cases, and equal-side assemblies.