The New Sudoku Players' Forum

[udosuk]

Could anyone help me on this? I suppose it could be worked out using pen & paper...

A 30-60-90 triangle can be cut into 2 such triangles in only one way:


To cut it into 3 such triangles, there are 3 different ways:


So how many ways in total to cut it into 4 such triangles? Their sizes can be either same or different to each other...

[Cec]

"So how many ways in total to cut it into 4 such triangles? Their sizes can be either same or different to each other...

Four?
Cec

[Hud]

Just one question: does each cut have to go completely through the large triangle? In other words, can a cut butt against another cut inside the large triangle? I' not sure it matters, but it may?

[ronk]

Probably more, but my answer is five.

[ab]

You can convert each 3 into a 4 by halving a green, yellow or red triangle, which gives 9 options, but you can also convert each 2 into a 4 by making one of the two triangles into a 3 which gives rise to 2 more (fingers crossed) so I think 11.

[Ruud]

This is what I could find (so far):

My sail designs for 2006



My improved answer is 10.

Ruud.

[ronk]

You can convert each 3 into a 4 by halving a green, yellow or red triangle, which gives 9 options

I totally missed those.

... but you can also convert each 2 into a 4 by making one of the two triangles into a 3 which gives rise to 2 more

But you can do that to each triangle two different ways. And you can make two triangles out of each of the two triangles. Which is what I counted to arrive at my (crappy) answer of 5.

But (I think) each of those 5 is identical to 1 of your 9.

[edit: And Ruud finds a 10th by thinking "outside the triangle" ... and dividing the 90 degree angle by 3. Cool!]

[udosuk]

Actually the answer is more than 11 (by quite a few in fact)...

But ab & Ruud have done a good job so far (especially Ruud for the nice graphics...)

And Hud, I think the answer to your question is yes (although I might not have fully understood your words)...

You definitely have to "think outside the triangle" to find the remaining ones...

[Ruud]

Actually the answer is more than 11 (by quite a few in fact)...

Help! I'm only at 14 now



Tell me when to stop.

Ruud.

[ronk]

Help! I'm only at 14 now

Use other diagonal of 1st figure.

P.S. My screen's barely wide enough to see all at once.

[Ruud]

Use other diagonal of 1st figure.

Thanks, we're at 15 now

My screen's barely wide enough to see all at once.

Ruud.

16, after ab's addition.

[ab]

Ruud in your first downward pointing triangle, you can leave the red triangle alone and flip the other three around their line of symmetry to get the blue triangle at the thin end

[Ruud]

Hi ab.

Good addition. I don't know if Ron's screen is high enough for another picture, so I updated the previous post

Ruud.

[ronk]

Good addition. I don't know if Ron's screen is high enough for another picture, so I updated the previous post

LOL (lots of laughs)! Yes, it's plenty high ... perhaps I exaggerated.

I know this will sound like I'm complaining, but with half the triangles upside down, it's really tough to compare and "see" what we've already got. I'll volunteer to flip half of them over, if you'd rather not do it. (My only tool is MS Paint, tho.)

Ron

[Ruud]

Well, I've had a few laughs too.

Sorry Ron. No need for you to go flippin' I still have the original.



Ruud.