Making triangle
Find the maximum number of regular triangles we can make using 6 sticks of the same size.
Assume: The sticks can't overlap and can only touch each other at their ends.
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2 solutions
Yes you are absolutely correct... ;)
It doesn't mention that the triangles must be the same size, nor that they must only touch at the ends, so if we used the 6 sticks to make a Star of David there would be 8 triangles created, (6 smaller ones around the perimeter and 2 larger ones forming the main body).
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Ah, good point, @Brian Charlesworth !
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If the 'same size/only touch at ends' conditions are applicable then your answer hits the mark. If these conditions don't apply, then I'm wondering if there might be a more prolific arrangement than the Star of David.
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@Brian Charlesworth – Yeah, I bet there very well could be, especially since regular triangles were never specified...
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@Geoff Pilling – I've come up with a couple more 8-triangle arrangements, but none with more than 8. I think this will be my doodle-time project for the rest of the day. :) (I've written a report so that Shithil can edit the wording of his question to match the desired solution, if he so chooses.)
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@Geoff Pilling – Indeed! This reminds me of this problem . The pattern starting at 3 sticks seems to go 1 , 3 , 8 , 1 5 , 2 7 , 4 2 , . . . , which has a match here and here .
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@Brian Charlesworth – Hahaha... You're reading my mind... That's where I got it from! :)
But wait a sec. Do you mean that for 5 matches it should be 8? Or 1+3+8=12?
For five matches I have 10 triangles, like in a pentagram.
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@Geoff Pilling – Well, if one of the negative-sloped sticks were removed from your diagram then the resulting arrangement would yield just 8 triangles, unless I'm missing something. But I agree that with a 5-pointed star we would end up with 10 triangles, which is interesting because the 6-pointed star, i.e., David's, only generates 8 triangles. The 7-pointed star generates at least 42 triangles, but I'm only finding 40 in an 8-pointed star and just 30 in a 9-pointed star. There are some variations in how to draw a 10-pointed star, some of which invoke some serious occult implications, soI think I'll stop here before I go too far down this rabbit hole... :)
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@Brian Charlesworth – Also, for four match sticks we can get four triangles by removing one of the negative sloped lines and the base in the figure above.
So, does the series go 1,4,10,... ?
Do we add successive triangular numbers each time?
So, can we generalize it to 1, 4, 10, 20, 35 ... ?
Not sure... Just speculation...
How do you know that 4 gives the maximum number? Why can't it be 5, 6, 7, or ... ?
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As per Geoff's and my discussion below, if no conditions are applied the answer can be as high as 1 5 . I have reported the question so that the conditions necessary for the posted answer to be correct are explicitly stated.
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I have resolved it... We have to make regular triangles..
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@Shithil Islam – This doesn't completely resolve it since, @Brian Charlesworth 's David's Star example contains 8 regular triangles.
@Shithil Islam – O.k., that helps, but in my Star of David example there are 6 regular triangles of one size and 2 larger regular triangles. I think that you'll still need to specify that the sticks cannot overlap, and can only touch each other at their ends.
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@Brian Charlesworth – okkkk I have also resolved it :-)
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@Shithil Islam – O.k., great, that looks good now. :)
@Shithil Islam – Yup, I think it looks good now! :)
@Brian Charlesworth – Yeah, I think that would do it!
Can you prove that the maximum number is above 4?
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No I can't. But the person who posted this solution should show why the answer cannot be larger than 4.
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@Pi Han Goh – I have posted this problem ;) I didn't get any solution that the maximum number is above 4.(I have spend about a month to solution this problem)
The most number of triangles formed by ( 2 n ) edges is ( 3 n ) , as demonstrated in Rivin's paper here . This occurs in a complete graph K n on n vertices. Since K 4 can be drawn in 3D as a tetrahedron, with 4 equilateral triangles, we are done showing that this indeed is the maximum value.
@Geoff Pilling @Pi Han Goh Here is the proof on why a tetrahedron gives the maximum number of equilateral triangles.
You can form them into a tetrahedron, like this to form 4 triangles: