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10

I think the answer is First of all, Then, the last of which is pretty interesting because

6

Assuming "sector beams" are the five kite-shaped things joined in the centre one solution would be

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

It seems that Here are some ways These were found with an exhaustive algorithmic search. My code in R: # Create the grid grid<-data.frame(x=1:4,b=1) %>% left_join(data.frame(y=1:4,b=1)) %>% mutate(b=1:16) squaredDiff<-function(a) (a-lag(a))^2 squaredDist<-function(n,p){grid %>% filter(b %in% c(n,p)) %>% summarise(across(.cols = c(x,...

2

Complement to loopy walt answer: There are 8 'distinct' solutions apparently (if I coded well in Julia), each solution having 10 variants (rotation and mirror -wise). The program runs for any n=numVertices and only provides solutions for n=4 and n=5. Perhaps, for n=3, 3^2=9 is too small, and, for n=6, 6^2=36 is too big. code: using Printf using Combinatorics ...

2

This is an attempt to answer your secondary question: [I]f I choose a random (solvable) starting problem, what's the probability that it will have a unique solution? One answer is found in the paper On Determining Paint by Numbers Puzzles with Nonunique Solutions by Ryan Mullen (Journal of Integer Sequences, vol. 12, 2009, Article 09.6.5). A natural ...

2

On New Year's Eve Like @Glorfindel said, The only reasons not to play on any given day are: Given that Freddy won on the 22nd and the 25th, Therefore

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