Problem 295

[harryeakins]

I'm having difficulty with the first part of this problem... - link
It is stated that L(10) = 30, but I can't get that answer. I think there are 31 lenticular holes for r <= 10, and here's why:

There are 7 radii which have arcs spanning a single cell in the lattice (r = 1, root(5), root(13),..., root(85)). These can be combined together the for (7*8)/2 = 28 lenticular holes.

There are 2 radii which have arcs spanning a rectangle of width 3x1 (r = 5, root(65)). These can be combined together for (2*3)/2 = 3 lenticular holes.

28+3 = 31 != 30

I can't figure this one out for the life of me! Any hints?

Thank you

[niino]

We must count distinct lenticular pairs.
Don't you count one lenticular pair twice?

[jaap]

There are 31 different lenticular shapes that could be formed, but only 30 distinct lenticular pairs.

[harryeakins]

Ahhh i got it

Cheers!

[albert_nik]

Continuing with the same reasoning I count:

• for N=10 --> 31 shapes and 30 distinct pairs.
  for N=100 --> 3330 shapes and less than 3330 distinct pairs.

Are there some other pairs that I'm missing?

[ffff0]

What about rectangles of other sizes?

[albert_nik]

Both sides >1 ?

[ffff0]

Indeed.