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Alexandre Muñiz
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Alexandre Muñiz

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The fact that there are eight different sets of four integers between 1 and 8 that sum to 18 invites the creation of a sort of magic square where each set occurs once as a row or column. There are even a few ways of doing this so that the diagonals also sum to 18. This seems like the sort of idea that is obvious enough that it would have to have been looked at before, but I have no idea how to find out where it may have been previously discussed.
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Long shot, and not the sort of answer you're looking for, but maybe this fact is related to some of the bijections between bipartitions of a set of size 8 and 2-dimensional subspaces of (F_2)^4 that are used to construct the Golay code, Mathieu groups, etc.  I know that in other parts of those constructions weight enumerator things come up where i has weight i, i.e. you might care about sums of subsets.
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Another 1- through 4-oct tiling. Also, I  missed this, but George Sicherman's Polyform curiosities site has some good polyoct constructions. For instance, the Baiocchi figures: http://userpages.monmouth.com/~colonel/obaiocchi/index.html
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Alexandre Muñiz

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Here are some crossed stick configurations where the lengths of the two subsegments of a stick have ratio φ. I don't think any of these will make good lasercut puzzles, but I'm sure something interesting can be done with some of them.
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Hmm, I didn't even notice that the one on the lower left has a couple of points on three segments. Oops.
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New blog post with some notes on possible syntax for a Python-based Interactive fiction authoring system. I might play around with this some more, not so much with the intention that I will actually eventually have a finished product, but because it involves digging into interesting but less frequently used capabilities of Python, and there are some cute intellectual puzzles involved.
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Here's a tiling of the 12 hexiamonds with 16 symmetries and a little distortion. This isn't the most symmetrical thing you can tile with the hexiamonds. You can tile an octahedron, which has 48 symmetries. You can also tile a pair of tetrahedra, which have 24²·2=1152 symmetries. But it's easy to run into the problem of pieces being ambiguous where all three of the cells at a vertex belong to the same hexiamond. (this can happen with the octahedron too, but the situation is easier to avoid since it takes four cells at a vertex.) I haven't found a solution for the pair of tetrahedra with no ambiguity, but I assume one exists.
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Have him in circles
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Alexandre Muñiz

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It turns out that there are four partitions of the integers 1-8 into two sets of 4 that each sum to 18. We can put the numbers from each pair of partitions onto strategically placed sites on opposite sides of popsicle stick shaped pieces so that the pieces can overlap at the number sites as on the figure below. (If the pieces have the right combination of flexibility and resilience, they will hold together in this configuration. You can try it with craft popsicle sticks.) We still have to decide on a principle for ordering the numbers on either side of the sticks, but for one reasonably simple principle, there is a unique solution to the problem of placing the sticks such that each line of four numbers possesses the same sum. Remarkably, that sum is not 18, but 17. (There is another principle which produces a puzzle with 8 solutions, which seems a bit fairer to solve. Maybe someday I'll produce a run of these.)
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All of the polyocts of size 1 to 4.
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Gravitationally propelled mathematical sculpture at Gathering for Gardner. (Video by Joseph Devincentis.)  #G4G11
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This is a cute geometry problem that I might be able to use as a basis for a crossed stick puzzle design. If we reflect the four line segments below over the x axis, we have all eight segments with congruent intersection points, which is what we need for a crossed stick puzzle.

But I'd need to solve the problem. (Solve for what? Um, the exact value of any angle, I guess.) I'm pretty sure there is a unique solution. I can make some headway by using the fact that the triangle with both b segments is isosceles, but then I get stuck trying to figure out how to use the equality of the a segments or the c segments.

Any thoughts on how to proceed?
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And now, by making a better drawing in Geogebra (where the b segments are constructed to be equal by using a circle) I'm convinced that you are right. Ah, well.
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A new blog post, with some solutions that were sent in to me, is at http://puzzlezapper.com/blog/2013/10/some-contributed-solutions/
(But the one shown below is a solution that I sent in to somebody else!)
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Have him in circles
52 people
Jack Welch's profile photo
Anthony Pryor's profile photo
Christos Dimitrakakis's profile photo
Carl Muckenhoupt's profile photo
Jesse McGrew's profile photo
Ed Pegg's profile photo
Strick Yak's profile photo
Matthew Murray's profile photo
David Welbourn's profile photo
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