Problem 091

[Georg]

Is it possible that there are less triangles with 3-coordinates than triangles with 2-coordinates?

[Oliver1978]

I don't think so, and that's bothering me. I've made up a 3x3 square grid. And this is what I get:

Code: Select all

0-0 / 0-3 / 1-3
0-0 / 0-3 / 2-3
0-0 / 0-3 / 3-3
0-0 / 0-2 / 3-2
0-0 / 0-1 / 3-1
0-0 / 1-0 / 1-3
0-0 / 2-0 / 2-3
0-0 / 3-0 / 3-1
0-0 / 3-0 / 3-2
0-0 / 3-0 / 3-3

Well, it seems I'd miscounted  :? Now I've got 10. Still less than 11 which are 2-coordinate right-angled triangles...

[i][update][/i]

I've arrived at 12:
0-0 / 1-2 / 3-0
0-0 / 2-2 / 3-1

[jaap]

[update]

I've arrived at 12:
0-0 / 1-2 / 3-0
0-0 / 2-2 / 3-1

Apart from the two you just added (I assume the first was meant to be 0,0/1,2/3,1) and their mirror images, you are also missing all those with the right angle at the 0,0 corner.

[Oliver1978]

[update]

I've arrived at 12:
0-0 / 1-2 / 3-0
0-0 / 2-2 / 3-1

Apart from the two you just added (I assume the first was meant to be 0,0/1,2/3,1) and their mirror images, you are also missing all those with the right angle at the 0,0 corner.

That would make it 11, not 12, my mistake. So additionally there are
0,3 / 0,0 / 3,0
0,2 / 0,0 / 3,0
0,1 / 0,0 / 3,0

A total of 14 then after all. Awefully confusing...

[Georg]

A total of 14 then after all.

And still not enough.

[Oliver1978]

This is encouraging I've checked again and I don't have any clue which ones I'm missing...

[jaap]

... and their mirror images ...

With that you should get all 19.

[Oliver1978]

Still going through and trying to find all of them...

[itteerde]

Hi,

hope this is still in the acceptable range of questions...

I wrote an algorithm for Problem 91 and it solves the only example given and I believe the idea to be sound (especially as this seems to be a pretty easy one..). Nevertheless it does not work for limit=50. Can anyone (who wrote the correct solution with a flexible limit, too, so just would need to run his correct solution once more) give me the value for some high enough limit other than 50 to help me with looking for my oversight?

maybe limit=67 or something like that which does not easily lead to the real solution?

Basically I just would like to check if my result is to low (so an oversight in the possibilities and back to thinking) or if I create some multiples counting them.

[DJohn]

Your post seems to have been edited to remove your smaller numbers while I was replying, so I'll edit my reply to remove them too. The one you give for 4 is too low.

[itteerde]

Your post seems to have been edited to remove your smaller numbers while I was replying, so I'll edit my reply to remove them too. The one you give for 4 is too low.

yeah, 61, right?

but if so I still get limit=50 wrong, also someone shared limit=67 with me and I have that the same way too...

[itteerde]

Whoa, that is interesting. Now I have it (just a little simplification that seems to have removed an oversight) and it produces the same vor limit=67 but the correct (and surely different) one for 50... How are the odds for that..!