34

The problem is equivalent to Now,


27

I'll get things started with:


20

The possible side lengths are And here's why:


18

Alternate answer:


12

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,...


10

I'm not sure if this is fine, but


7

The usual heuristic for this sort of thing is that at any given point there's some number of possible solutions -- e.g., for the puzzle discussed here there are 5!=120 possible arrangements of the pieces -- and each question will, once answered, reduce that number (to the number of configurations for which that answer is correct), and you want to reduce as ...


6

The answer is All other triangles have


5

First I counted polygons, then I counted congruence classes of polygons. First, counting polygons, I found 76 triangles, 94 quadrilaterals, 164 pentagons, 158 hexagons, 36 heptagons, 0 octagons, and 0 nonagons, for a total of 528 distinct polygons. Next, counting congruence classes, I found 8 triangles up to congruence, 16 quadrilaterals up to congruence, 23 ...


4

This is a question about puzzles, so it's not inherently off-topic for the site. But there's not much of a meaningful answer we can give to some of these questions. 1. According to what strategies could one ask the two types of questions? 2. Where to begin? Does it make sense to ask Questions A first for squares on the rim, as opposed to central squares ...


4

Inspired by athin's answer and DrD's response to it. This can be achieved like so:


3

Sketch: The general strategy is to overwhelm Bob combinatorially with simultaneous threats. To keep our argument simple we will assume that all our constructions (and all Bob's counter measures!) below yield distinct points. We can expect this to be the case when "free" points, i.e. points which are not constructed from other points are chosen &...


3

I have no proof that this solution is optimal, but it is better than the previous ones. Step 1: Step 2: The total cost is: We can decrease the efficiency very slightly by: which increases the cost by: As a bonus, here is the second most efficient dissection. Step 1: Step 2: Total cost:


2

It can be done quite nicely like this:


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