# Tag Info

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I didn't have a knife with me, so I only used my unit circle cookie cutter to split each square like this: I then rearranged the parts into this shape: Since the angle covered by this shape is exactly 120 degrees (see the final spoiler block to confirm), three of these make a nice circle, with some white shining through the gaps: Since the fit of the ...

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I think this is as simple as I could get it.

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Consider the following diagram: To improve on the lower bound using the same model, one option is to Here is a plot of the sin function (blue), the original lower bound of $\alpha/2$ (green), and the updated lower bound (red):

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Here is a purely trigonometric one: This uses the angle doubling formula for the sine and the facts that in the first quadrant (argument between $0$ and $\frac \pi 2$) the cosine is an absolutely decreasing function and the tangent is at least as large as its argument.

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A visual solution.

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From OEIS: A000448 For a circle centered at the origin, this gives these lattice points: For a total of

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Increasing the bound:

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UPDATED WITH A COMPLETE REWRITE A minimum velocity does First observe the hard lower bound This is straightforward to establish: I will now show simulations of a strategy that strongly suggest that The strategy has to phases: Phase 1: Phase 2: Few more examples with

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The strategy:

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Answer (sorry for the crude drawing):

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The image is derived from a and appears to have been produced via the following algorithm:

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A little experimentation with Euclids formula, demonstrates that there are only primitive Pythagorean triples with largest element less than or equal to 80. If, from these, we choose the sets and note that 65=5*13, then we can produce the triples Using them, plus their reflections about the lines y=x, y=-x, y=0, and x=0, plus the four extra points sitting ...

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Sure, a 1×4√5 rectangle and a 1×√5 rectangle.

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Here's one way, without rotating any rectangles:

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For the case of dominos, you can see it as a maximum cardinality bipartite matching problem. The domino covering problem can be seen as the problem of pairing black squares with neighboring white squares in a way to maximize the number of pairs. There are fast algorithm to solve that. If you manage to pair all squares then a solution exists. This doesn't ...

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