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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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@hexomino's answer is correct and well-reasoned, as always. Here's another approach, which to me feels much more.. "axe-to-the-head" is what I'd call it in my native language, so I thought it might be interesting enough to warrant posting. Lower bound: (This is what @trolley813's encrypted comment is saying.) Upper bound:

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This paper by Joel Haddley and Stephen Worsley answers a slightly different question - finding monohedral disc dissections where not all pieces touch the centre - but the results generally apply to this problem too. My favourite answer is this one: Note that this one has interior pieces that don't even have a single point on the boundary. There is an ...

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This is a minor upgrade on @sybog64's answer: One way of thinking about it is to start with this configuration and then taking groups of 2 slices and rotating each group by 120°.

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The problem is equivalent to Now,

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I haven't found a perfect solution, my pieces are symmetrical but not identical

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I'll get things started with:

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There has been some mathematical research done on this subject, and it turns out that: as shown in this paper by Sándor Bozóki, Tsung-Lin Lee, and Lajos Rónyai. The paper was also discussed in for example Huffpost.

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The Circles Covering Circles page of Erich Friedman's "Packing Center" shows… After learning the above information, I set out to try and find a solution myself. It would be possible to search for a set of sprinkler positions that maximize the amount of the lawn covered, however I chose a slightly different metric. Given a set of sprinkler ...

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I think this is the only solution. (Edit: It isn't. See teedyay's answer for another solution.) I found this solution mostly by trial and error, but I did keep in mind that

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The construction: I don't know if it's the only solution, but that's the only one I found.

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

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Here is an alternative solution to the one already found:

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The possible side lengths are And here's why:

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I believe the set of points (x,y) such that $x y = ±1$ satisfies the conditions. The set consists of 4 segments of hyperbolas. Any straight line crosses at least 2 of these segments resulting in 2 to 4 intersections. Except for the lines x=0 or y=0 which cross none. Here's a graph:

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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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The claim is

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Here is one solution. I came up with it pretty quickly so I'm guessing there are multiple solutions out there: Here is a second solution which is similar to the first, but not an exact mirror/rotation of it: Here is a third solution not at all like the previous two: What seems to be a common theme in these solutions is: A 4th solution, using the same ...

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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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1 cut across both slices She sliced them fillet-style, 1/3 and 2/3 the thickness of each of the original two slices. This maintains the cross-section, frosting and all. Dividing a line into 3 parts can be done by construction, so no measurement is needed. As @Jeffrey notes, only one cut per slice of cake is needed, dividing each slice into one thin slice 1/3 ...

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Another symmetric solution: In fact, there are two more similar solutions:

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It will This kind of generalises.

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The proof is in two parts, corresponding to the two linkages which are joined to each other at a single point. For each part, I'll try to both explain in words and illustrate on the picture you've provided.

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A visual solution.

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Suppose we have a rectangle made with $h$ horizontal lines and $v$ vertical lines. We can assume without loss of generality that $h\le v$. The number of squares with side length $s$ is $(h-s)(v-s)$. The total number of squares is therefore f(h,v) = \sum_{s=1}^{h-1} (h-s)(v-s)\\ =\sum_{s=1}^{h-1} (hv - (h+v)s + s^2)\\ =\sum_{s=1}^{h-1} hv - (h+v)\sum_{s=1}^{...

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The maximum is For this solution, the squared distances are You can solve the problem via integer linear programming as follows. Let binary decision variable $x_{i,j}$ indicate whether a pawn is placed on square $(i,j)$. For each pair $(i_1,j_1)$ and $(i_2,j_2)$, let binary decision variable $y_{i_1,j_1,i_2,j_2}$ indicate whether \$x_{i_1,j_1} \land x_{i_2,...

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I think the answer is Reasoning As trolley813 mentions in the comments

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