1. Joined
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    26 Nov '12 00:173 edits
    Prove that if a hexagon is dissected into identical equilateral triangles, and these triangles are rearranged into smallr hexagons (6 triangles to a small hexagon) then you always get a square number of small hexagons.
  2. Standard memberforkedknight
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    26 Nov '12 17:05
    As far as I know, the only way to disect an equilateral triangle into smaller equilateral triangles is to cut it into a triforce, resulting in 4 smaller triangles.

    4 is a square number.

    The simplest way to dissect a hexagon into equilateral triangles results in 6 triangles.
  3. Standard memberwolfgang59
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    26 Nov '12 19:11
    Originally posted by forkedknight
    As far as I know, the only way to disect an equilateral triangle into smaller equilateral triangles is to cut it into a triforce, resulting in 4 smaller triangles.

    4 is a square number.

    The simplest way to dissect a hexagon into equilateral triangles results in 6 triangles.
    An equilateral triangle also divides into 9 smaller ones, ... or 16, ... or 25.

    How to prove?

    Consider an equilateral triangle made up of rows of smaller triangles. Any row will have as many triangles as the one above plus 2. So the rows (satrting at top) contain 1,3,5,7,9 etc triangles.

    The sum of any number of odd numbers is always a square so any equilateral triangle can be made from a square number of smaller triangles.

    nth odd number is (2n=1)

    S = 1+3+5+ .... +(2n+1)
    S= (2n+1) + (2n-1) + (2n-3) + .....+ 5+3+1

    simply add the two lines above

    2S = 2n + 2n + 2n + ... +2n
    there are n terms so

    2S =2n^2
    S=n^2

    Rubbish proof .... sorry 😳