122

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 ...


59

How about this? Why does it work? Alternative cut:


48

@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:


48

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 ...


34

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°.


34

The problem is equivalent to Now,


29

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


27

I'll get things started with:


22

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.


22

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 ...


22

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


22

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


22

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


21

Here is an alternative solution to the one already found:


20

The possible side lengths are And here's why:


20

Alternate answer:


18

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:


18

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):


17

The claim is


16

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 ...


16

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.


16

The answer to your question: Proof:


15

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 ...


15

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


14

It will This kind of generalises.


14

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.


14

A visual solution.


13

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}^{...


13

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,...


12

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


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