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        riddles >> easy >> Folding triangles
        
(Message started by: Noke Lieu on Jul 13^th, 2010, 7:09pm)

Title: Folding triangles
Post by Noke Lieu on Jul 13^th, 2010, 7:09pm

(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?

Title: Re: Folding triangles
Post by Immanuel_Bonfils on Jul 13^th, 2010, 9:52pm

Assuming the new triangle is the overhanging,
[hide]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.[/hide]

Title: Re: Folding triangles
Post by Noke Lieu on Jul 13^th, 2010, 10:23pm

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. :D

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.

Title: Re: Folding triangles
Post by towr on Jul 14^th, 2010, 12:19am

[hide]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. [/hide]

Title: Re: Folding triangles
Post by Noke Lieu on Jul 14^th, 2010, 4:30pm

*grin*

Title: Re: Folding triangles
Post by towr on Jul 29^th, 2010, 5:01am

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.

Title: Re: Folding triangles
Post by towr on Jul 29^th, 2010, 9:10am

An alternative needing 7 pieces, and no mirroring.