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:


13

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. The first step is, given a list of the small circles' centers, determine the minimum radius required for the small circles to cover the large circle. I did this by by constructing the ...


12

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}^{...


9

The answer is The easiest way to see that is


8

It is not possible with any finite or countably infinite set. Take any point P in the set. For each other point Q in the set, draw a line between P and Q. Then take the angle between line PQ and the horizontal axis. This forms a set of angles A. If S is countable, then the set of angles A must also be countable (since there is at most once angle per point ...


7

For any circle touching the perimeter, the longest possible chord between the two contact points is the diameter of the circle - 2 meters. This is shorter than the side length of an inscribed dodecagon but not that of an inscribed tredecagon, so any valid sprinkler setup must have at least 13 sprinklers on the perimeter.


6

Assuming the rectangle is meant to be filled we can do bricks or studs. This is optimal because


6

Here's an approximate solution, obtained by randomly generating 10,000 points and solving the corresponding set cover problem. There are still some gaps, but maybe it can be tweaked to cover them. It is different than the others posted so far because it has a circle that is concentric with the outer circle.


5

The first rebus, figured out by Feryll, is a hint to: The second rebus shows: Then, based on the Morse code deciphered by Feryll, we are supposed to: This is clued by the axis labels and the first half of the Morse code: The result is the string: Then using Feryll's correct identification of the image hosting site in question, one can convert this to a ...


4

With the new hints this has become much simpler. In fact, OP has (unintentionally?) changed the character of the puzzle. At least some solutions can be step-by-step (like a sudoku for want of a better simile) reconstructed from the hints. For example number 8 (I'll only do one, so people still have a chance to earn the bounty by solving one of the others): ...


4

To rule out @hexomino's trivial solution (empty set) let us require that every straight line intersects S in fintitely many and at least 2 points. Then one simple way to construct S is


4

Partial Answer Let me start with a baseline - 27 sprinklers, and try to build from there: Desmos Link With some manual tweaks, I am able to bring it down to 25 sprinklers. Desmos Link


3

It is relatively easy to show that Now To find the number of different ways to make a particular number $N$ of squares


3

Observations so far:


3

The general formula for the number of squares in a $n \times (n+k-1)$ grid ($n>0,k\ge 0$) is The smallest solution is


2

Answer: Reasoning Let's try for an upper bound on possible solutions.


2

I can do ten lines. It starts nicely but ends funny. Or eleven lines. Or even twelve lines. Xelia owes Alice three magnets now...


2

I believe you cannot go further down then sprinklers. I can't show a solution that it's possible to do it with this number yet. But here is my reasoning why I believe it's not possible to go any lower. Approach: Necessary formulas: Then: Therefore, the fewest number of sprinklers is:


1

I'm going to say because you said: and coupled with Of course, you could take the unconventional approach of Which would mean, if you have an appetite for such wizardry, I can comfortably assert *it can be marginally smaller but all values are approximate given the spirit of the answer


1

Here is my intuition on it, which is not mathematically rigorous. Answer Explanation As stated by Hint 2, running between 2 columns of zombies can't be done infinitly as zombies get too close to us after some distance. Yet it seems a good starting point. Let's consider we get to point 50, 50 and then run upwards (between the 2 columns of zombies). We will ...


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