You have 8 wooden right-angle isosceles triangles (the lengths of the two lines that make up the right angle are equal). These are numbered from 1 to 8. Every single triangle is equal in size. Using these triangles how many DIFFERENT squares can you make. (Not neccesarily all at once)
Note: A square must be entirely wooden in order for it count. A square cannot have the exact same combination of triangles that have already been used.e.g if you used triangles 1 and 2 in one combination you can never have a square made from only triangles 1 and 2. (However triangles 1 and 3 would make a distinct square)
(In reply to
Solution? by fwaff)
It's not clear from the problem, but I think for a type 2 square, order is important. That is, I get the impression that looking at the blocks in clockwise order, 1-2-3-4 is distinct from 1-2-4-3. I'm assuming that rotations don't generate a new square (otherwise there would be an infinite number!), and that mirror images do count, unless they can be produced by a rotation. This would give (8P4)/4 combinations rather than 8C4.
Type 3 squares get even hairier. Each quarter square can have its diagonal connect the center point with a corner or it might connect two edges. Each quarter can have two configurations independently of the others, and the remaining blocks can be assigned 8! different orderings. However, this would include different rotations, so there also needs to be a division by 4.
My count gives 28 (as per fwaff) + (8P4)/4 + 8! x 2^4 / 4 = 28 + 420 + 161,280 = 161,728.
re: Solution?
