
Jane Street Puzzles - dmmalam
https://www.janestreet.com/puzzles/current-puzzle/
======
gre
IBM posts monthly puzzles as well.

[http://www.research.ibm.com/haifa/ponderthis/index.shtml](http://www.research.ibm.com/haifa/ponderthis/index.shtml)

[http://www.research.ibm.com/haifa/ponderthis/challenges/July...](http://www.research.ibm.com/haifa/ponderthis/challenges/July2018.html)

------
kej
These are pretty good puzzles, the full archive is at
[https://www.janestreet.com/puzzles/archive/](https://www.janestreet.com/puzzles/archive/)

------
knappa
I haven't thought it through yet, but Pick's theorem (
[https://en.wikipedia.org/wiki/Pick%27s_theorem](https://en.wikipedia.org/wiki/Pick%27s_theorem)
) might be useful for the current puzzle.

~~~
mikhailfranco
Not sure why, but that is _unexpected_.

And the proof is so simple and elegant.

------
jschulenklopper
Interesting. For a mainly numerical superset of similar puzzles, I guess that
[https://projecteuler.net/](https://projecteuler.net/) still is the best known
collection.

------
extra__tofu

       n=1, A=2:
       h    b   #tri  #non-congruent tri
       1    4   b-1   roundup(#tri/2)
       2    2   b-1   roundup(#tri/2)
    
       .
       .
       .
       
       n=4, A=16:
       h    b      #tri  #non-congruent tri
       1    2A/h   b-1   roundup(#tri/2)
       2    2A/h   b-1   roundup(#tri/2)
       4    ...    ...
       8    ...
       16   ...          ...
    

summing up #non-congruent tris for each n I see

    
    
       n  tri
       1  3
       2  7
       3  15
       4  31
    

so I think that answer may be 2^(n+1) - 1.

Could be wrong because I forgot what congruent really means.

~~~
tylerhou
You're undercounting some triangles and missing the acute condition. For n =
4, there is more than one triangle with base 4 and height 4, for example: [(0,
0), (4, 0), (1, 4)]; [(0, 0), (4, 0), (2, 4)]; (and [(0, 0), (4, 0), (3, 4)]
but it's congruent to the first so we don't count it). Furthermore, a triangle
with base 16 and height 1 might be non-acute.

~~~
dilatedmind
for the first 8 im getting 1,3,6,14,28,60,120,248

~~~
yantrams
Just a hunch but I think you might be skipping isosceles triangles. And
possibly adding right angled triangles because of floating-point arithmetic.

~~~
dilatedmind
how about 3,6,14,28,60,120,248,496?

~~~
yantrams
Nope :( The sequence starts with 1 because there is just one triangle with an
area of 2 that satisfies the conditions. I ended up with a sequence similar to
your previous one when I forgot to take the absolute value of area -- x1
_(y2-y3) + x2_ (y3-y1) + x3*(y1-y2))/2 . That could be a possibility in case
you were using the same formula. In your first result - 1, 3, 6, 14.... ,
first two are right, third over-counted and rest are undercounted.

------
21
Nope: [https://www.janestreet.com/puzzles/turn-based-strategy-
game/](https://www.janestreet.com/puzzles/turn-based-strategy-game/)

