Triangle base division math problem. I might go crazy.?
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Triangle base division math problem. I might go crazy.?

[From: ] [author: ] [Date: 14-02-13] [Hit: ]
. ughhhhhh. Please help if you have a clue, I might just be going insane at this point -Yes, This could drive one crazy. My approach is to count separately the triangle of height 1,......
I've done everything in my power to find out the answer to this question. Honestly, there are so many patterns and such... I can't .find the damn equation that would work for all of the division or rather yet the nth division... ughhhhhh. Please help if you have a clue, I might just be going insane at this point

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Yes, This could drive one crazy. My approach is to
count separately the triangle of height 1,2,3,...,n,
where n is the number of base divisions, which also
corresponds to the number of vertical 'layers' of triangles.

With triangles of height 1, each layer has 2 more height 1
triangles than the previous layer. So for the triangle with
3 base divisions there are 1 + 3 + 5 = 9 height 1 triangles.
For a triangle with n base divisions, there will be
1 + 3 + 5 + .... + (2n - 1) = n^2 height 1 triangles.

For triangles of height 2, we'll look at the 3 base division
triangle for some guidance. The bottom 2 layers contain
2 height 2 triangles, and the top 2 layers contains just 1.
For an n base division triangle, there will be
1 + 2 + .... + (n - 1) = n*(n-1)/2 height 2 triangles.

For triangles of height 3, the 4 base division triangle
will have 2 height 3 triangles in the bottom 2 layers
and 1 in the top 2 layers.

The 4 base division triangle contains just 1 height 4 triangle,
namely itself.

So the 4 base division triangle will have
4^2 + [4*(4-1)/2] + 3 + 1 = 16 + 6 + 3 + 1 = 26 triangles.

The 5 base division triangle will have
5^2 + [5*(5-1)/2] + 6 + 3 + 1 = 25 + 10 + 6 + 3 + 1 = 45 triangles.

Ughhhh.... I almost have a pattern. I'll be back shortly.....
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