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dms172
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Doesn't the answer have to be a multiple of 5?
60.

multiple of 5 is obviously necessary as there's rotational symetry order 5 (and 5 is prime - a six pointed star would not need a multiple of 6), but I can find

100.

I drew them, but don't think I can get it to stay in a spoiler box

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multiple of 5 is obviously necessary as there's rotational symetry order 5 (and 5 is prime - a six pointed star would not need a multiple of 6), but I can find

I suggest that if it was a 6 pointed star, total count would need to be a multiple of 3, am I correct?

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multiple of 5 is obviously necessary as there's rotational symetry order 5 (and 5 is prime - a six pointed star would not need a multiple of 6), but I can find

100.

I drew them, but don't think I can get it to stay in a spoiler box

I think the second one in your first 5x row should actually be 10x. Also, I see one more configuration - it's basically a combination of that one and the second to last one from the first 10x row (I'll draw it later if I get a chance). This new one should be 10x as well.

115

Edited by Chuck Rampart
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I think the second one in your first 5x row should actually be 10x.

It's true, but in 10s, the second last one must be in 5s. Thus I had said 100. But now I realized that last one in 5s can be in 3 ways, not 5. My result is 98 now.

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I suggest that if it was a 6 pointed star, total count would need to be a multiple of 3, am I correct?

I think that wouldn't quite be true, because you have a combination of triangles with 'symmetries'* of order 2, order 3 and order 6. So it might be a multiple of neither 2 nor 3! The easiest way to think of it is that you might find all the 3x and 6x triangles, then add on the "biggest" triangles, of which there are only two uniques, because they are equilateral.

(* I call them symmetries, but they are actually a kind of combination of rotational and reflective non-symmetries - hmmm that doesn't really make sense!)

I think the second one in your first 5x row should actually be 10x. Also, I see one more configuration - it's basically a combination of that one and the second to last one from the first 10x row (I'll draw it later if I get a chance). This new one should be 10x as well.

I think you are right with the second in the 5x row - I was thinking that every triangle using a whole point would only have 5 rotations, but of course, this one also has a reflection for each.

I'm intrigued by your new one - I still can't see it. Are you sure it is not just a rotation of the last in the top row?

105

Edited by foolonthehill
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And I've just realised that the second to last in the top row has only 5 rotations! So that,

100

Basically, every triangle has 10 rotations, unless it also has a line of reflective symmetry which passes through the centre of the star, in which case it only has 5

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First of all let me say congrats cuz the the source where i got this didn't even find 100 many so after a carefully recount I'm pretty sure there's 100 triangles, but if u can find more for sure then plz post picture and prove me wrong.

100 was OK for me, but as I wrote before, at foolonthehill's nice picture, third of 5x triangles (they are the largest triangles) can occur only tree times. Am I wrong? This also eliminates Bonanova's suggestion as it must be a multiply of 5??? I still suggest 98.

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