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Topic: Folding triangles (Read 803 times) |
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Noke Lieu
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(good to see we're back)
So.
An isosceles triangle.
1. Fold it parallel to the base such that the trapezium and the 'new' triangle have the same area.
Where do you fold it?
2. When you do that, there's going to be two congruent triangles (the overhanging parts of the trap) that are similar to the overhanging part of the triangle.
What's the fewest number of pieces that you can cut the smaller triangles into that then cover the larger?
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Immanuel_Bonfils
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Re: Folding triangles
« Reply #1 on: Jul 13th, 2010, 9:52pm » |
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Assuming the new triangle is the overhanging,
1. For the trapezium S = (2H - h) B/2H, where capital refers to original triangle an h is the trapezium heigth. For the new (or the overhanging) triangle S = B h12/2H, where h1 iis the hanging trianlge height. Equating we get h1 = 3h = 3H/5 , knowing that H = 2h + h1
2. I'm not sure what is the second question, but
since the overhanging triangles base and heigth are three times that of one of the overhanging triangle of the trap, nine of the small tiangles will cover the larger one.
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Noke Lieu
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Re: Folding triangles
« Reply #2 on: Jul 13th, 2010, 10:23pm » |
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sorry abou the confusion on the second question- it felt clumsy typing it.
Let me try again:
Given 3 similar isosceles triangles- 2 of area A, one of area 2A- what is the minimum number of pieces the 2 smaller triangles need to be cut into such that they fit (jigsaw style) on the bigger triangle.
And it appears that you've answered a different (perhaps better?!) question on the first. 
I was only after the normal
"Where do you cut a triangle (parallel to one side) such that the resulant triangle has half the area of the of the original?"
but was trying to disguise it a little.
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towr
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Re: Folding triangles
« Reply #3 on: Jul 14th, 2010, 12:19am » |
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The minimum number of pieces is 1 (i.e. no cuts), when the triangles are right-angled isosceles triangles.
If the base is [sqrt(2)-1] times the leg, you need to cut one in two.
In general. Eh, who knows.. Maybe I'll think of something later.
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Noke Lieu
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Re: Folding triangles
« Reply #4 on: Jul 14th, 2010, 4:30pm » |
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*grin*
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towr
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If you are allowed to turn pieces over (mirroring), then for the general case I can make do with at most 9 pieces. See attachment.
You can use the same principle without mirroring, but it'd take more pieces.
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| « Last Edit: Jul 29th, 2010, 5:09am by towr » |
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