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Alexandre Muñiz
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This worm passes through all 12 pentomino shapes. It seems impossible to have a worm like this loop back to the starting pentomino.
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Here's a strict[1] 3,4-coloring of a hexiamond tiling of a torus, where every combination of the colors is present. What's remarkable here is that there is exactly one tiling of this torus that can be colored in this way, out of all of its 328,198 hexiamond tilings. (Even more remarkably, this isn't the first time a reasonably natural polyform tiling problem has produced a unique tiling with a coloring of this type. There is also, out of the 2,339 pentomino tilings of a 6×10 rectangle, a single tiling that can be colored this way.) While this problem could be explored on other tori, this torus has an aesthetic advantage in being the most symmetrical one with the right area.

[1] A strict coloring is one in which regions that meet only at a vertex are considered adjacent, and cannot receive the same color.
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I was noodling with the notion of a geomagic square based on a "magic" square where the rows and columns are all of the sets of four numbers between 1 and 8 that sum to 18. The fact that there are two copies of each number between 1 and 8 in such a square led me to the idea of giving each shape of a given size a different color, and making two colored figures. I first gave up on making the shapes of the same size different colors, and then I gave up on the two coloring. (I also gave up on diagonals at some point.) I think there's something better that can be done here, but I lack the patience.
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The Lo Shu, or 3×3 magic square, was discovered in China in antiquity, and is the only way, (up to symmetry) to place the numbers 1-9 in a 3×3 grid such that the numbers in each row, column, and main diagonal sum to the same number. Well, maybe not. I found a couple of others. Are there more Faux Shu?
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Here's a nice little puzzle. There are seven ways to combine a square and two tans (half-square unit right triangles) such that the resulting shape has two sqrt(2) length sides. Joining those two sides with a curve lets us make paths as we tile with the pieces. Unfortunately, that set of pieces won't let us make a closed loop: there is a parity problem with the directions of the diagonals. (An odd number of pieces have one diagonal in each direction, but to get a loop the total number in each direction must be even.) Luckily, we can fix the problem by adding the 2-tan pieces without a square, and this conveniently gives us exactly 16 for our total area, so we can make a 4×4 square.

This gets more interesting with multiple sets of pieces. You can have puzzles where the goals are based on the topology of the loops instead of or in addition to the geometry of a tiling target. For example, with two sets, you could try to form two concentric loops in a 6×6 square with the corners removed.
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My exchange gift for the next Gathering for Gardner is a "magic" die. All numbers between 1 and 24 occur once. Faces sum to 50, bands around the cube sum to 100, and diagonal bands sum to 75.

After I had the dice in hand I thought of a way that I might have been able to make them more awesome. The upper left number on each face is a number between 1 and 6, so that the die can be used as a standard d6. We could think each of the positions on each face as 'modes' for rolling the dice. It would be really cool if the other three modes formed a set of non-transitive 'dice'. (Non-transitive dice are sets of three[1] dice where A usually beats B, B usually beats C, and C usually beats A.) It happens that the other modes in these dice do not form a non-transitive triple. I intend to search for a numbering of the dice that does make a non-transitive triple. This might require relaxing the constraints I put in for symmetry (of the positions used for the d6 mode) or for the diagonal bands. A "good" non-transitive triple would have margins of victory in the three contests that are balanced and relatively pronounced.

[1] Or more, but for our purposes three.
#g4g12
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There are 12 ways to make an integer length segment of length 7 or less with a mark at a positive integer distance from the endpoints. A reasonable puzzle is to try to form a figure where the pieces are connected as in the lower left. I noodled around a bit, and my best attempt was the one on the lower right, where the remaining piece was the right length, but the mark was in the wrong place. I have no idea if this puzzle is actually solvable.
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Here's a new type of puzzle I'm working on. Given five transparent 15-omino pieces, make a complete set of pentominoes by overlapping them. The solution for this particular set of pieces is shown below.  There are a zillion ways to arrange the pentominoes into five and seven piece clusters that could produce a puzzle of this form.  The real problem is, can we find a particularly good set of 15-ominoes to use?

What makes a particularly good set? As much as possible, the 15-ominoes should be partitionable into three different pentominoes in multiple ways. (These partitions must have a middle pentomino that leaves two other pentominoes when it is cut out.) Also, as many as possible of the pentominoes should be present in dissections of more than one 15-omino. It'd also be nice to have at least one false partial solution, like a pair of 15-ominoes that can make 5 different pentominoes by overlapping that isn't present in the correct solution.

Then there are some practical matters related to making this puzzle as a physical lasercut set. It would be bad if the centers of gravity of the pieces in the pair were not within the pentomino of overlap, because the top piece could fall over. I'd also want the 15-omino pieces to have an efficient packing into a square, in order to minimize wasted material and cut length.
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This is a fragment of a double tessellation. A double tessellation is a set of polygons that covers the plane such that every (non-boundary) point in the plane is in two polygons. Of course, you can create a double tessellation simply by overlaying two single tessellations, but this isn't interesting, so we look for proper double tessellations, which can't be decomposed into single tessellations. (It's also nice if none of the segments overlap.) This tessellation contains two different polygons: 150°–90°–150°–90°–150°–90° hexagons and regular hexagons. There are also two types of vertices: ones where six irregular hexagons meet, and ones where one regular and two irregular hexagons meet. The former are "double vertices" since one goes twice around the vertex before getting to the polygon you started with, and the latter are single vertices.

Puzzles:

Can you find a double tessellation with only double vertices and no single vertices?

Can you find such a double tessellation that is also "vertex transitive"? (i.e. any vertex can be moved to any other vertex by a symmetry of the tessellation?)

Can you find a double tessellation with only one type of polygon? (i.e. all polygons are congruent?)
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I made a laptop. Materials: Samsung Galaxy Note 8 tablet, Keycool 84 keyboard, USB On-The-Go adapter cable, two custom lasercut white delrin pieces, Southco adjustable friction hinges, #8-32 3/8" machine screws and nuts, velcro squares with sticky backs (so that the tablet can be used separately) and high strength molding tape (connecting the lower delrin piece to the keyboard.)
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