Tricolored Triangle Tiling Problem: Torsten Sillke Bielefeld, 1992-11-23 Creator: Dirk Hanneforth, Schulstr. 7, D-33647 Bielefeld, Germany Distribution: none Pieces: equilateral triangles * The triangles are dissected by there bisectors a b into six smaller right triangles. These six f c triangles will be colored with the three colors * e d * 1, 2, and 3. Now we select the subset of triangles where each right triangle has a neighbour with equal color. There are 25 different different pieces. Problem: Tile a large triangle with the 25 pices such that all internal edges have matching colors and that each edge of the large triangle is unicolored. Solution: Parity shows that each color must be occur at the boarder. Therefore we can fix the coloration of the boarder. Now there are 28*6 = 168 solutions for such a fixed boarder. As there are no color-symmetries there are 28 total different solutions. 8 solutions are given below. * 1 2 1 2 The only possible boarder * . . * with unicolored edges. 1 . . 2 1 . . 2 * . . * . . * 1 . . . . 2 1 . . . . 2 * . . * . . * . . * 1 . . . . . . 2 1 . . . . . . 2 * . . * . . * . . * . . * 1 . . . . . . . . 2 1 . . . . . . . . 2 * 3 3 * 3 3 * 3 3 * 3 3 * 3 3 * --- * 1 2 1 2 * 1 2 * 1 1 2 2 1 1 1 2 * 1 1 * 1 2 * 1 2 1 1 2 2 1 2 2 3 3 2 * 1 1 * 2 3 * 3 3 * 1 1 3 2 3 3 2 2 1 3 3 2 2 3 2 2 * 1 3 * 1 1 * 2 3 * 2 2 * 1 1 3 3 1 2 3 1 2 2 1 3 3 3 1 2 3 1 2 2 * 3 3 * 3 3 * 3 3 * 3 3 * 3 3 * * 1 2 1 2 * 1 2 * 1 1 2 2 1 1 1 2 * 1 1 * 1 2 * 1 2 1 1 2 2 1 2 2 3 3 2 * 1 1 * 2 3 * 3 3 * 1 1 3 2 3 3 1 2 1 3 3 2 2 3 1 2 * 1 3 * 1 1 * 2 3 * 2 2 * 1 1 3 3 1 2 3 2 2 2 1 3 3 3 1 2 3 2 2 2 * 3 3 * 3 3 * 3 3 * 3 3 * 3 3 * * 1 2 1 2 * 1 2 * 1 1 2 2 1 1 1 2 * 1 1 * 1 2 * 1 2 1 1 2 2 1 2 1 3 3 2 * 2 2 * 1 3 * 3 2 * 1 2 2 1 3 3 2 2 1 2 2 1 3 3 3 2 * 3 3 * 2 1 * 1 1 * 3 3 * 1 1 2 2 1 1 3 3 2 2 1 1 2 3 3 3 3 3 2 2 * 3 3 * 3 3 * 3 3 * 3 3 * 3 3 * * 1 2 1 2 * 1 2 * 1 1 2 2 1 2 2 2 * 1 2 * 2 2 * 1 1 2 2 1 2 1 3 3 2 1 2 * 1 3 * 3 2 * 1 1 * 1 1 3 3 2 1 1 2 1 3 3 3 2 1 1 2 * 3 3 * 2 2 * 2 1 * 3 3 * 1 1 3 3 2 2 1 1 2 2 1 1 3 3 3 3 3 1 2 2 * 3 3 * 3 3 * 3 3 * 3 3 * 3 3 * * 1 2 1 2 * 1 2 * 1 1 2 2 1 3 3 2 * 1 3 * 3 2 * 1 1 3 3 2 2 1 2 2 3 2 2 * 1 2 * 2 3 * 1 1 * 1 1 2 2 3 1 1 2 1 2 2 2 3 1 1 2 * 2 2 * 2 2 * 3 1 * 3 3 * 1 1 1 3 3 3 1 1 2 2 1 1 1 3 3 3 3 1 2 2 * 3 3 * 3 3 * 3 3 * 3 3 * 3 3 * * 1 2 1 2 * 1 2 * 1 1 2 2 1 3 3 2 * 1 3 * 3 2 * 1 1 3 3 2 2 1 1 3 1 1 2 * 2 2 * 3 1 * 1 2 * 1 2 2 3 1 1 2 2 1 2 2 3 1 1 1 2 * 3 3 * 2 3 * 1 1 * 1 1 * 1 1 2 2 3 3 3 2 2 2 1 1 2 3 3 3 3 2 2 2 * 3 3 * 3 3 * 3 3 * 3 3 * 3 3 * * 1 2 1 2 * 1 2 * 1 1 2 2 1 2 2 2 * 1 2 * 2 2 * 1 1 2 2 2 2 1 3 3 3 3 2 * 1 3 * 3 3 * 3 3 * 1 1 3 3 1 1 2 2 1 2 2 3 1 1 2 2 * 2 2 * 2 2 * 1 1 * 1 1 * 1 1 1 2 3 3 1 1 3 2 1 1 1 2 3 3 3 3 3 2 * 3 3 * 3 3 * 3 3 * 3 3 * 3 3 * * 1 2 1 2 * 1 2 * 1 1 2 2 1 3 3 2 * 1 3 * 3 2 * 1 1 3 3 2 2 1 2 2 3 3 2 * 1 2 * 2 2 * 3 3 * 1 1 2 2 1 1 3 2 1 2 2 2 1 1 3 2 * 2 2 * 2 2 * 1 1 * 1 1 * 1 1 1 2 3 3 1 1 3 2 1 1 1 2 3 3 3 3 3 2 * 3 3 * 3 3 * 3 3 * 3 3 * 3 3 * Other Problems for this pieces: * * 2 2 3 3 2 3 2 3 * . . * * . . * 2 . . 3 2 . . 3 2 . . 2 3 . . 3 * . . * . . * . . * . . * 2 . . . . . . . . 3 2 . . . . . . . . 3 * 2 1 * 1 2 * . . * 3 1 * 1 3 * 1 . . 1 2 . . 3 * . . * . . * 2 . . . . 3 1 . . . . 1 * 1 1 * 1 1 * 1 1 * --- * 1 2 1 2 * . . * 1 . . 2 1 . . 2 * . . * . . * 1 . . . . 2 1 . . . . 2 * 3 3 * 3 3 * 3 3 * * 1 1 1 1 * . . * 1 . . 1 2 . . 3 * . . * . . * 2 . . . . 3 3 . . . . 2 * . . * . . * . . * 3 . . . . . . 2 3 . . . . . . 2 * 3 3 * 3 1 * 1 2 * 2 2 * --- * 2 1 2 1 * . . * 2 . . 1 2 . . 1 * . . * . . * 2 . . . . 1 2 . . . . 1 * 3 3 * 3 3 * 3 3 * * 1 2 1 2 * . . * 1 . . 2 1 . . 2 * . . * . . * 1 . . . . 2 1 . . . . 2 * . . * . . * . . * 1 . . . . . . 2 1 . . . . . . 2 * 3 3 * 3 3 * 3 3 * 3 3 * --- 4 S * 1 2 1 2 * . . * 1 . . 2 1 . . 2 * . . * . . * 1 . . . . 2 1 . . . . 2 * 3 3 * 3 3 * 3 3 * * 1 2 1 2 * . . * 1 . . 2 1 . . 2 * . . * . . * 1 . . . . 2 1 . . . . 2 * . . * . . * . . * 1 . . . . . . 2 1 . . . . . . 2 * 3 3 * 3 3 * 3 3 * 3 3 * --- * 3 3 * 3 3 * 3 . . . . 3 3 . . . . 3 * . . * . . * . . * 3 . . . . 3 3 . . . . 3 * . . * . . * 1 . . . 2 1 . . . 2 * . . * . . * 3 . . . . 3 3 . . . . 3 * . . * . . * . . * 3 . . . . 3 3 . . . . 3 * 3 3 * 3 3 * --- * 3 3 * 3 3 * 118 S 1 . . . . 2 16 Sym 1 . . . . 2 * . . * . . * . . * 1 . . . . 2 1 . . . . 2 * . . * . . * 1 . . . 2 1 . . . 2 * . . * . . * 1 . . . . 2 1 . . . . 2 * . . * . . * . . * 1 . . . . 2 1 . . . . 2 * 3 3 * 3 3 * --- * 3 3 * 3 . . 3 3 . . 3 * 2 2 * . . * . . * 1 1 * 2 . . . . . . 1 2 . . . . . . 1 * . . * . . * . . * 1 . . . . . . 2 1 . . . . . . 2 * . . * . . * . . * . . * 1 . . . . . . 2 1 . . . . . . 2 * 1 1 * . . * 2 2 * 3 3 3 3 * --- * 1 1 * 3 . . 2 3 . . 2 * 3 3 * . . * . . * 2 2 * 3 . . . . . . 2 3 . . . . . . 2 * . . * . . * . . * 3 . . . . . . 2 3 . . . . . . 2 * . . * . . * . . * . . * 2 . . . . . . 3 2 . . . . . . 3 * 1 1 * . . * 1 1 * 1 1 1 1 * --- * 2 2 * 2 2 * 2 2 * 2 . . . . . 1 2 . . . . . 1 * . . * . . * . . * 2 . . . . . 1 2 . . . . . 1 * . . * . . * . . * 2 . . . . . 1 2 . . . . . 1 * . . * . . * . . * 93 S 2 . . . . . 1 2 . . . . . 1 2 Sym * 1 1 * 1 1 * 1 1 * --- * 1 1 * 1 1 * 1 1 * 2 . . . . . 3 2 . . . . . 3 * . . * . . * . . * 2 . . . . . 3 2 . . . . . 3 * . . * . . * . . * 2 . . . . . 3 146 S 2 . . . . . 3 * . . * . . * . . * 2 . . . . . 3 2 . . . . . 3 * 1 1 * 1 1 * 1 1 * --- * 2 2 * 2 2 * 2 2 * 1 . . . . . 1 1 . . . . . 1 Sym possible * . . * . . * . . * 1 . . . . . 1 1 . . . . . 1 * . . * . . * . . * 1 . . . . . 1 1 . . . . . 1 * . . * . . * . . * 1 . . . . . 1 1 . . . . . 1 * 3 3 * 3 3 * 3 3 * --- * 2 2 * 2 2 * 2 2 * 1 . . . . . 1 1 . . . . . 1 * . . * . . * . . * 1 . . . . . 1 1 . . . . . 1 * . . * . . * . . * 1 . . . . . 1 34 S 1 . . . . . 1 * . . * . . * . . * 1 . . . . . 1 1 . . . . . 1 * 2 2 * 2 2 * 2 2 * --- 71 S * 1 1 * 1 1 * 1 1 * 1 . . . . . . 1 1 . . . . . . 1 * . . * . . * . . * . . * 1 . . . . . . . . 1 1 . . . . . . . . 1 * . . * . . * . . * . . * . . * 2 . . . . . . . . 3 2 . . . . . . . . 3 * 1 1 * 1 1 * 1 1 * 1 1 * --- 136S * 1 1 * 1 1 * 1 1 * 1 . . . . . . 1 1 . . . . . . 1 * . . * . . * . . * . . * 1 . . . . . . . . 1 1 . . . . . . . . 1 * . . * . . * . . * . . * . . * 2 . . . . . . . . 3 2 . . . . . . . . 3 * 2 2 * 2 2 * 3 3 * 3 3 * --- 23 S * 1 1 * 1 1 * 1 1 * 2 . . . . . . 3 2 . . . . . . 3 * . . * . . * . . * . . * 2 . . . . . . . . 3 2 . . . . . . . . 3 * . . * . . * . . * . . * . . * 2 . . . . . . . . 3 2 . . . . . . . . 3 * 2 2 * 2 2 * 3 3 * 3 3 * --- * 1 1 * 1 . . 1 61 S 1 . . 1 * . . * . . * 1 . . . . 1 1 . . . . 1 * . . * . . * . . * 1 . . . . . . 1 1 . . . . . . 1 * . . * . . * . . * . . * 1 . . . . . . 1 1 . . . . . . 1 * 1 1 * 1 1 * 1 1 * --- * 2 3 2 S 2 3 * . . * 1 . . 2 1 . . 2 * . . * . . * 1 . . . . 2 1 . . . . 2 * . . * . . * . . * 1 . . . . . . 2 1 . . . . . . 2 * . . * . . * . . * . . * 2 . . . . . . . . 3 2 . . . . . . . . 3 * 1 1 * 3 3 * 3 3 * 3 3 * 1 1 * --- * 1 1 1 1 4 S * . . * 2 . . 3 2 . . 3 * . . * . . * 2 . . . . 3 2 . . . . 3 * . . * . . * . . * 2 . . . . . . 3 2 . . . . . . 3 * . . * . . * . . * . . * 3 . . . . . . . . 2 3 . . . . . . . . 2 * 3 3 * 1 1 * 1 1 * 1 1 * 2 2 * --- * 1 1 1 1 21 S * . . * 1 . . 2 1 . . 2 * . . * . . * 1 . . . . 2 1 . . . . 2 * . . * . . * . . * 1 . . . . . . 2 1 . . . . . . 2 * . . * . . * . . * . . * 3 . . . . . . . . 2 3 . . . . . . . . 2 * 3 3 * 3 3 * 3 3 * 3 3 * 2 2 * --- * 1 1 * 1 1 * 2 . . . . 3 12 S 2 . . . . 3 * . . * . . * . . * 2 . . . . . . 3 2 . . . . . . 3 * . . * . . * . . * . . * 2 . . . . . . 3 2 . . . . . . 3 * . . * . . * . . * 2 . . . . 3 2 . . . . 3 * 1 1 * 1 1 * --- * 1 1 * 1 1 * 1 S 1 . . . . 3 1 . . . . 3 * . . * . . * . . * 1 . . . . . . 3 1 . . . . . . 3 * . . * . . * . . * . . * 2 . . . . . . 3 2 . . . . . . 3 * . . * . . * . . * 2 . . . . 3 2 . . . . 3 * 2 2 * 2 2 * --- * / \ Chrismas Tree *-*-* / \ / \ A symmetric solution *-*-*-*-* which changes 1<->2 / \ / \ is possible *-*-*-*-* / \ / \ / \ *-*-*-*-*-*-* / \ / \ / \ *-*-*-*-*-*-* / \ / \ / \ / \ *-*-*-*-*-*-*-*-* / \ *---* -------------------------------------------------------------------------- Pieces: equilateral triangles (2 color version) Torsten Sillke * The triangles are dissected by there bisectors a b into six smaller right triangles. These six f c triangles will be colored with the two colors * e d * 1 and 2. Now we select the subset of triangles where each right triangle has a neighbour with equal color. There are 8 different different pieces. The 8 triangles of this set are * * * * 1 1 1 1 2 2 1 2 1 1 1 1 1 1 1 2 * 1 1 * * 2 2 * * 1 1 * * 1 2 * * * * * 2 2 2 2 1 1 2 1 2 2 2 2 2 2 2 1 * 2 2 * * 1 1 * * 2 2 * * 2 1 * Problems: Octahedron: This is impossible as the number of uni-colored edges is odd for each color. So there will be at least one mismatch. The lowest number of mismatches (at 'x') is shown below. * 1 1 1 1 * 1 1 * x x * 1 1 * 1 1 1 1 1 2 1 2 2 2 1 1 2 1 * 2 2 * 2 2 * 2 2 * 2 2 2 2 * Triangle of order 4 with the double set with uni-colored boarder: * 1 1 1 1 * 2 2 * 1 2 2 1 1 2 2 1 * 1 2 * 2 1 * 1 1 2 2 1 1 1 2 2 2 2 1 * 1 2 * 2 2 * 2 1 * 1 1 2 1 2 2 1 1 1 1 2 1 2 2 1 1 * 1 1 * 1 1 * 1 1 * 1 1 *