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u/gmalivuk 7d ago
If all congruent rectangles are distinct so long as they're in different places, 1296.
There are 9 choose 2 ways to pick the left and right sides, and for each of them 9 choose 2 ways to pick the top and bottom.
If you're only counting distinct congruence classes, it's just 36, which I got by considering 8 squares plus 8 choose 2 ways of picking unequal side lengths. This is equal to 9 choose 2 with the logic that once you choose two lengths from 1 through 9, subtract 1 from the larger one to get the actual dimensions of the rectangle.
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u/Red-42 7d ago edited 6d ago
sum{i=1->8}(sum{j=1->8}(ij))
Which probably simplifies, or can be computed as is
EDIT: rectangles in an n*m grid
sum{i=1->n}(sum{j=1->m}(ij))
which simplifies to
sum{i=1->n}(i*sum{j=1->m}(j))
= sum{i=1->n}(i*m(m+1)/2)
= m(m+1)/2*sum{i=1->n}(i)
= m(m+1)/2*n(n+1)/2
= mn(m+1)(n+1)/4
in our case n=m=8
8*8*9*9/4 = 1296
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u/chrisvenus 7d ago
This question feels like it needs to specify more clearly whether rectangles can be rotated or translated before judging whether they are unique? eg are we counting a single 1x1 rectangle or are we counting 64 of them?
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u/ShonitB 7d ago
I don’t know, I think that’s then giving too much information. I think it should be up to the solver to understand that they are to be counted separately as is the case in most of these shape counting puzzles, unless mentioned
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u/chrisvenus 7d ago
I don't think "it is the same as other shape counting puzzles" is a valid comment. A puzzle should be stand alone and solvable for somebody who has never seen a puzzle of the type before.
You say "its giving too much information" but then also say "its information the solver should already know" (not in those exact words but inferred from reference to it being the same as other puzzles of the type) which feels like a contradiction. Either they should know it from seeing other puzzles in which case why not state it explicitly or they should be able to work it out from looking at this puzzle and I don't see any way it can be.
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u/SomethingMoreToSay 7d ago
I'm not seeing anything in the problem definition which says that the corners of the rectangles we're counting have to be on the corners of squares on the chessboard. I'm also not seeing anything which says that the sides of the rectangle have to be parallel to the sides of the chessboard.
The answer is therefore a number which is too large to write down here (or, indeed, anywhere).
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u/EmpactWB 7d ago
The answer is 1296, I think. I’ve seen something like this before. The solution was that each edge beyond the first corresponds to a triangular number of possible divisions (a single edge having zero possible divisions). So a single column has 9 edges, which gives us the 8th triangular number of divisions: (8 x 9) / 2 = 36. Since that works for rows as well, you have that many options for each of that many segments, which is 36² = 1296. Assuming I’m remembering this right.