bonanova
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Put a cork in it [original thread]
bonanova replied to bonanova's question in New Logic/Math Puzzles
Snugly means completely fills each of the hole cross sections at some moment as it passes through. Of all the 3-dimensional shapes that do this, one has a maximum and another has a minimum volume. The question is what are these volumes? Been watching TV all evening, so you have the advantage on me. -
Put a cork in it [original thread]
bonanova replied to bonanova's question in New Logic/Math Puzzles
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A polyhedron is a solid with plane faces. It is called convex if each of its faces can be placed flat on a table top. The simplest convex polyhedron is the tetrahedron, with four triangular faces. It can have an endless variety of shapes, but the network of its edges is always the same. Its faces cannot be anything but triangles. If we increase the number of faces to five, we can sketch two different configurations: a pyramid with a quadrilateral base, and a triangular prism. Note that the pyramid has 5 vertices and 8 edges; the prism has 6 vertices and 9 edges. Those are the easy cases, so let's take it to six. How many topologically distinct, convex, six-faced solids [hexahedra] can you sketch? One, of course, is the cube. Note that changing the square faces of the cube into rectangles or trapezoids does not change its topology. Also note that truncating the apex of a square pyramid produces the same topology as a cube. In each case, there are six quadrilaterals sharing an edge with four of the others. To be distinct, the number of edges or vertices or sides of the faces must differ.
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A polygon of how many different sides can be made by passing a plane through a cube?
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A while ago we asked for a shape that fits snugly through three different holes: Actually, the answer is not unique. So now we ask: What is the volume of the smallest shape that fits snugly through the holes? The largest?
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Minimum is easier. OK now I'll work on the maximum.
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Grammar police report: Pole needs to be capitalized. And, wouldn't s/he be more familiar with the shoreline of the Baltic Sea? All seriousness aside, why not give this a shot.
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Hi nrduren, and welcome to the Den. I took the liberty of hiding your spoiler to give responders a true shot at finding the answers on their own. One way of posing questions here is to confirm what people come up with, or not, rather than handing them an answer sheet, even one that's folded up. I'll replace it tho, if if you like. - bn
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Hi futballa16, and welcome to the den. Some perspective on this place: Several of our denizens live so far outside the box they have to go towards town to hunt. On reflection, maybe a lot of us do. I nearly suggested going to the top of the building and tying the string to the barometer. But I realized that might add to the confusion, and one of my tasks here is to reduce it. It's all in good fun; I'm intrigued to hear your answer - should no one get it. - bn
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Yeah ... didn't think of the short case. The perpendiculars fall outside two of the bases.
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x = [-b +/- sqrt(b2 - 4ac)] / 2a etc... is an explicit formula, no? It's double valued, which gives you two y values and then two x values for each of them. But it seems impossible for more than one of the four combinations to lie within the triangle [for any a, b, c.] p.s. You will beat me to it. p.p.s. to OP: Doesn't 0 < b < a guarantee an acute triangle? p.p.p.s. To site admin. It was the span tag. Thanks for the fix. site admin: You are welcome
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Under the same conditions as the previous puzzle,
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Done.
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Spaces, I assume, are compressed out?
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I gave the equations to calculate of A1(x,y) and A2(x,y). Equating them to A/3 lets you solve for x,y. Algebraically, you get a closed form expression. Numerically, you get specific values. In this problem, algebraic is a bit clumsy; but the equations are there. Are there other approaches? Perhaps the equal area condition is stationary point that presents a simpler equation to solve. Don't know, but I'm interested to see what you come up with. I've missed the Aha! solution often enough.
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Done! Nice puzzles.
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Hats on a death row!! One of my favorites puzzles!
bonanova replied to roolstar's question in New Logic/Math Puzzles
Re-read the OP. The first prisoner is not able to determine the color of his own hat. -
A previous relates to this one, but with a difference. Mine is linear, has one pair of each type, and fewer pairs. I've been trying to use it as a first step, without success. Still, it might provide a clue.
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I forgot to point out, so I will now, that the interval method permits one to assign any desired weighted probability to each choice, in addition to providing a fair choice.